Rational Functions are just a ratio of two polynomials (expression with constants and/or variables), and are typically thought of as having at least one variable in the denominator (which can never be 0).
Note that we talk about how to graph rationals using their asymptotes in the Graphing Rational Functions, including Asymptotes section. Also, since limits exist with Rational Functions and their asymptotes, limits are discussed here in the Limits and Continuity section. Since factoring is so important in algebra, you may want to revisit it first. Remember that we first learned factoring here in the Solving Quadratics by Factoring and Completing the Square section, and more Advanced Factoring can be found here.
Introducing Rational Expressions
Again, think of a rational expression as a ratio of two polynomials. Here are some examples of expressions that are and are not rational expressions:
| Expression | Rational? | Explanation |
| $ \displaystyle \frac{{{{x}^{2}}-1}}{{x+3}}$ | Yes | Ratio of two polynomials. |
| $ \displaystyle \frac{1}{x}$ | Yes | Ratio of two polynomials. “1” is a constant monomial. |
| $ \displaystyle \frac{{{{x}^{3}}-8}}{1}$ or $ {{x}^{3}}-8$ | Yes | Note that this polynomial is technically a rational expression, but we typically think of them as having a variable in the denominator. |
| $ \displaystyle \frac{{\sqrt{{2x}}+1}}{{{{x}^{3}}}}$ | No | The numerator isn’t a polynomial, because of the radical. |
| $ \displaystyle \frac{{\frac{1}{x}}}{{x+4}}$ | Not technically, but can be simplified to a rational. | This expression isn’t a rational since the numerator ($ \displaystyle \frac{1}{x}$) isn’t a polynomial, but with some algebra, we can turn it into one: $ \displaystyle \frac{{\frac{1}{x}}}{{x+4}}=\frac{1}{x}\left( {\frac{1}{{x+4}}} \right)=\frac{1}{{x\left( {x+4} \right)}}$. (Flip the denominator and multiply). |
Multiplying, Dividing, and Simplifying Rationals
Frequently, rational expressions can be simplified by factoring the numerator, denominator, or both, and crossing out factors. They can be multiplied and divided like regular fractions.
Here are some examples. Note that these look really difficult, but we’re just using a lot of steps of things we already know. That’s the fun of math! Also, note in the last example, we are dividing rationals, so we flip the second and multiply.
Remember that when you cross out factors, you can cross out from the top and bottom of the same fraction, or top and bottom from different factors that you are multiplying. You can never cross out two things on top, or two things on bottom.
Also remember that at any point in the problem, when variables are in the denominator, we’ll have domain restrictions, since denominators can’t be $ 0$.
Multiplying, Dividing, and Simplifying Rationals
$ \displaystyle \require{cancel} \frac{{\left( {3{{x}^{2}}-5x-2} \right)}}{{{{x}^{2}}-4}}=\frac{{\left( {3x+1} \right)\cancel{{\left( {x-2} \right)}}}}{{\left( {x+2} \right)\cancel{{\left( {x-2} \right)}}}}=\frac{{3x+1}}{{x+2}};\,\,\,x\ne -2,\,\,2$
$ \displaystyle \require{cancel} \begin{align}\frac{{4{{x}^{2}}+12xy+2x+6y}}{{2{{x}^{2}}+6xy+6x+18y}}\,&=\,\frac{{2\left( {2{{x}^{2}}+6xy+x+3y} \right)}}{{2\left( {{{x}^{2}}+3xy+3x+9y} \right)}}\,=\,\frac{{2\left[ {2x\left( {x+3y} \right)+1\left( {x+3y} \right)} \right]}}{{2\left[ {x\left( {x+3y} \right)+3\left( {x+3y} \right)} \right]}}\,\,\,\\&=\,\frac{{\cancel{2}\left( {2x+1} \right)\cancel{{\left( {x+3y} \right)}}}}{{\cancel{2}\left( {x+3} \right)\cancel{{\left( {x+3y} \right)}}}}\,=\,\frac{{2x+1}}{{x+3}};\,\,\,\,x\ne -3,\,\,-3y\end{align}$
$ \displaystyle \require{cancel} \begin{align}\frac{{\frac{{7{{y}^{2}}}}{{{{y}^{2}}-16}}}}{{\frac{{14{{y}^{2}}+49y}}{{3{{y}^{2}}-10y-8}}}}\,&=\,\frac{{\frac{{7{{y}^{2}}}}{{\left( {y-4} \right)\left( {y+4} \right)}}}}{{\frac{{7y\left( {2y+7} \right)}}{{\left( {y-4} \right)\left( {3y+2} \right)}}}}\,=\,\frac{{\cancel{7}{{{\cancel{{{{y}^{2}}}}}}^{y}}}}{{\cancel{{\left( {y-4} \right)}}\left( {y+4} \right)}}\cdot \frac{{\cancel{{\left( {y-4} \right)}}\left( {3y+2} \right)}}{{\cancel{7}\cancel{y}\left( {2y+7} \right)}}\,\,\,\\&=\,\frac{{y\left( {3y+2} \right)}}{{\left( {y+4} \right)\left( {2y+7} \right)}};\,\,\,\,\,\,y\ne -4,\,-\frac{7}{2},\,\,-\frac{2}{3},\,\,0,\,\,4\end{align}$
Finding the Common Denominator
When we add or subtract two or more rationals, we need to find the least common denominator (LCD), just like when we add or subtract regular fractions. If the denominators are the same, we can just add the numerators across, leaving the denominators as they are. We then must be sure we can’t do any further factoring:
$ \require{cancel} \displaystyle \frac{2}{{3x}}+\frac{4}{{3x}}=\frac{{(2+4)}}{{3x}}=\frac{{{{{\cancel{6}}}^{2}}}}{{{{{\cancel{3}}}^{1}}x}}=\frac{2}{x};\,\,\,\,x\ne 0$.
Just like with regular fractions, we want to use the factors in the denominators in every fraction, but not repeat them across denominators. When nothing is common, just multiply the factors.
Let’s find the least common denominators for the following denominators (ignore the numerators for now).
| Rational Expressions | LCD |
| $ \displaystyle \frac{{3x}}{{\left( {2x+1} \right)\left( {x+4} \right)}};\frac{1}{{8{{x}^{2}}}}$ | $ 8{{x}^{2}}\left( {2x+1} \right)\left( {x+4} \right)$ None of the factors are the same, so multiply factors. |
| $ \displaystyle \frac{2}{{\left( {x-4} \right)\left( {x-3} \right)}};\frac{1}{{8{{{\left( {x-3} \right)}}^{2}}}}$ | $ 8\left( {x-4} \right){{\left( {x-3} \right)}^{2}}$ Don’t have to use the $ \left( {x-3} \right)$ in the first fraction, since it’s in the second – but need the whole $ {{\left( {x-3} \right)}^{2}}$! |
| $ \displaystyle \frac{{4x}}{{3{{x}^{2}}\left( {x+2} \right)}};\frac{1}{{9\left( {{{x}^{2}}-4} \right)}}$ Factor denominators: $ \displaystyle \frac{{4x}}{{3{{x}^{2}}\left( {x+2} \right)}};\frac{1}{{9\left( {x+2} \right)\left( {x-2} \right)}}$ | $ \begin{array}{l}9{{x}^{2}}\left( {x+2} \right)\left( {x-2} \right)\\\,\,\,=9{{x}^{2}}\left( {{{x}^{2}}-4} \right)\end{array}$ Don’t have to use the 3 and the $ \left( {x+2} \right)$ in the first fraction, since it’s in the second. |
| $ \displaystyle \frac{1}{{6{{x}^{4}}-3{{x}^{3}}-63{{x}^{2}}}};\frac{x}{{36{{x}^{2}}-126x}}$ Factor denominators: $ \displaystyle \frac{1}{{3{{x}^{2}}\left( {2x-7} \right)\left( {x+3} \right)}};\frac{x}{{18x\left( {2x-7} \right)}}$ | $ 18{{x}^{2}}\left( {2x-7} \right)\left( {x+3} \right)$ Don’t have to use the 3 and the $ \left( {2x-7} \right)$ in the first fraction, since it’s in the second. Don’t have to use the $ x$ in the second since it’s in the first. |
Adding and Subtracting Rationals
Now let’s add and subtract the following rational expressions. Note that one way to look at finding the LCD is to multiply the top by what’s missing in the bottom. For example, in the first example, the LCD is $ \left( {x+3} \right)\left( {x+4} \right)$, and we need to multiply the first fraction’s numerator by $ \left( {x+4} \right)$, since that’s missing in the denominator.
| Adding and Subtracting Rationals | Notes |
| $ \displaystyle \begin{align}\frac{{x+2}}{{x-3}}+\frac{{x-1}}{{x+4}}&=\frac{{x+2}}{{x-3}}\left( {\frac{{x+4}}{{x+4}}} \right)+\frac{{x-1}}{{x+4}}\left( {\frac{{x-3}}{{x-3}}} \right)\\&=\frac{{\left( {x+2} \right)\left( {x+4} \right)}}{{\left( {x-3} \right)\left( {x+4} \right)}}+\frac{{\left( {x-1} \right)\left( {x-3} \right)}}{{\left( {x-3} \right)\left( {x+4} \right)}}\\&=\frac{{{{x}^{2}}+6x+8+{{x}^{2}}-4x+3}}{{\left( {x-3} \right)\left( {x+4} \right)}}=\,\frac{{2{{x}^{2}}+2x+11}}{{\left( {x-3} \right)\left( {x+4} \right)}}\\x&\ne -4,\,\,3\,\end{align}$ | Since there are no common factors, it’s almost like “cross multiplying” to get the numerators (do you see it?).
When you get the final answer, make sure you can’t factor the top and further reduce. |
| $ \begin{align}\frac{{2{{y}^{2}}-16}}{{{{y}^{2}}-4}}-\frac{{y+4}}{{y+2}}&=\frac{{2\left( {{{y}^{2}}-8} \right)}}{{\left( {y-2} \right)\left( {y+2} \right)}}-\frac{{y+4}}{{y+2}}\left( {\frac{{y-2}}{{y-2}}} \right)\\&=\frac{{2{{y}^{2}}-16-\left( {{{y}^{2}}+2y-8} \right)}}{{\left( {y-2} \right)\left( {y+2} \right)}}=\frac{{{{y}^{2}}-2y-8}}{{\left( {y-2} \right)\left( {y+2} \right)}}\\&=\frac{{\left( {y-4} \right)\cancel{{\left( {y+2} \right)}}}}{{\left( {y-2} \right)\cancel{{\left( {y+2} \right)}}}}=\frac{{y-4}}{{y-2}}\,\,\\y&\ne -2,\,\,2\end{align}$ | Multiply the top by what you don’t have in the bottom.
We had to multiply the second term by $ \displaystyle \left( {\frac{{y-2}}{{y-2}}} \right)$ since we didn’t have $ \left( {y-2} \right)$ on the bottom.
Watch out for negatives; don’t forget to push negatives through parentheses! And don’t forget to simplify! |
| $ \displaystyle \begin{align}\frac{{\frac{2}{{x-3}}}}{{\frac{3}{{x+1}}+\frac{1}{{x-3}}}}&=\frac{{\frac{2}{{x-3}}}}{{\frac{{3\left( {x-3} \right)+1\left( {x+1} \right)}}{{\left( {x+1} \right)\left( {x-3} \right)}}}}=\frac{{\frac{2}{{x-3}}}}{{\frac{{4x-8}}{{\left( {x+1} \right)\left( {x-3} \right)}}}}\\&=\frac{{{}^{1}\cancel{2}}}{{\cancel{{x-3}}}}\cdot \frac{{\left( {x+1} \right)\cancel{{\left( {x-3} \right)}}}}{{{{{\cancel{4}}}_{2}}\left( {x-2} \right)}}=\frac{{x+1}}{{2\left( {x-2} \right)}}\,\\x&\ne -1,\,\,2,\,3\end{align}$ Easier way: $ \displaystyle \begin{align}\frac{{\frac{2}{{x-3}}}}{{\frac{3}{{x+1}}+\frac{1}{{x-3}}}}\,&=\left( {\frac{{\frac{2}{{\cancel{{x-3}}}}}}{{\frac{3}{{\bcancel{{x+1}}}}+\frac{1}{{\bcancel{{x-3}}}}}}} \right)\cdot \frac{{\left( {x+1} \right)\left( {\cancel{{x-3}}} \right)}}{{\left( {\bcancel{{x+1}}} \right)\left( {\bcancel{{x-3}}} \right)}}=\frac{{2\left( {x+1} \right)}}{{3\left( {x-3} \right)+1\left( {x+1} \right)}}\,\\&=\,\frac{{2x+2}}{{3x-9+x+1}}=\frac{{2x+2}}{{4x-8}}=\frac{{2\left( {x+1} \right)}}{{4\left( {x-2} \right)}}=\frac{{x+1}}{{2\left( {x-2} \right)}}\\x&\ne -1,\,\,2,\,\,3\end{align}$ | Now let’s combine what we know about adding/subtracting and multiplying/dividing rationals.
Find the common denominator on the bottom first, combine terms, and then flip and multiply to the top. Then simplify.
Note that we could have also just multiplied the original numerator and denominator by the common denominator, as a shortcut. |
| $ \begin{align}\frac{{\frac{1}{{{{x}^{2}}}}-\frac{1}{{{{y}^{2}}}}}}{{x-y}}&=\frac{{\left( {\frac{1}{{\cancel{{{{x}^{2}}}}}}-\frac{1}{{\cancel{{{{y}^{2}}}}}}} \right)\cdot \cancel{{{{x}^{2}}}}\cancel{{{{y}^{2}}}}}}{{\left( {x-y} \right)\cdot {{x}^{2}}{{y}^{2}}}}=\frac{{{{y}^{2}}-{{x}^{2}}}}{{\left( {x-y} \right)\cdot {{x}^{2}}{{y}^{2}}}}\\&=\frac{{\cancel{{\left( {y-x} \right)}}\left( {y+x} \right)}}{{-\cancel{{\left( {y-x} \right)}}\cdot {{x}^{2}}{{y}^{2}}}}=\frac{{-\left( {y+x} \right)}}{{{{x}^{2}}{{y}^{2}}}}=-\frac{{y+x}}{{{{x}^{2}}{{y}^{2}}}}\\x&\ne y;\,\,\,\,x,y\ne 0\end{align}$ | Multiply the original numerator and denominator by the common denominator as a shortcut.
Factor a difference of squares, and remember that $ x-y=-\left( {y-x} \right)$, and the negative sign for a fraction can be moved from the bottom to the top, or in front. Simplify. |
Restricted Domains of Rational Functions
As we’ve noticed, since rational functions have variables in denominators, we must make sure that the denominators won’t end up as “0” at any point of solving the problem.
Thus, the domain of $ \displaystyle \frac{{x+1}}{{2x(x-2)(x+3)}}$ is $ \{x:x\ne -3,0,2\}$. This means if we ever get a solution to an equation that contains rational expressions and has variables in the denominator (which they probably will!), we must make sure that none of our answers would make any denominator in that equation “0”. These “answers” that we can’t use are called extraneous solutions. We’ll see this in the first example below.
Solving Rational Equations
When we solve rational equations, we can multiply both sides of the equations by the least common denominator, or LCD (which is $ \displaystyle \frac{{\text{least common denominator}}}{1}$ in fraction form), and not even worry about working with fractions! The denominators will cancel out and we just solve the equation using the numerators. Remember that with quadratics, we need to get everything to one side with 0 on the other side and either factor or use the Quadratic Formula.
Again, think of multiplying the top by what’s missing in the bottom from the LCD.
Notice that sometimes you’ll have to solve literal equations, which just means that you have to solve an equation for a variable, but you’ll have other variables in the answer. The last example shows this.
| Solving Rational Equations | Notes |
| $ \displaystyle \frac{3}{{x+3}}-\frac{1}{x}\,=\,\frac{{-9}}{{x\left( {x+3} \right)}}$ $ \displaystyle \begin{align}\frac{{x\left( {x+3} \right)}}{1}\cdot \left( {\frac{3}{{x+3}}-\frac{1}{x}} \right)&=\left( {\frac{{-9}}{{\cancel{{x\left( {x+3} \right)}}}}} \right)\cdot \frac{{\cancel{{x\left( {x+3} \right)}}}}{1}\\\frac{{x\cancel{{\left( {x+3} \right)}}\cdot 3}}{{\cancel{{x+3}}}}-\frac{{\cancel{x}\left( {x+3} \right)\cdot 1}}{{\cancel{x}}}&=\left( {\frac{{-9}}{{\cancel{{x\left( {x+3} \right)}}}}} \right)\cdot \frac{{\cancel{{x\left( {x+3} \right)}}}}{1}\\3x-(x+3)&=-9\\2x-3&=-9\\2x&=-6\\\,x&=-3\end{align}$ Does not work! No solution, or Ø | Multiply both sides by the least common (LCD) denominator, or $ \displaystyle \frac{{x\left( {x+3} \right)}}{1}$.
Note that you multiply the numerators with what you don’t have in the denominator. For example, $ \displaystyle \frac{3}{{x+3}}$ doesn’t have $ x$, so you multiply the top 3 by $ x$ to get $ 3x$.
Check your answers to make sure no denominators are 0. Our answer doesn’t work (–3 is an extraneous solution), so there is no solution. |
| $ \displaystyle \frac{{5{{x}^{2}}+8x-4}}{{3{{x}^{2}}+11x+6}}-\frac{1}{{x+3}}\,=\,\frac{x}{{3x+2}}$ $ \displaystyle \begin{align}\frac{{\left( {3x+2} \right)\left( {x+3} \right)}}{1}\cdot \left( {\frac{{5{{x}^{2}}+8x-4}}{{\left( {3x+2} \right)\left( {x+3} \right)}}-\frac{1}{{x+3}}} \right)&=\\\frac{x}{{3x+2}}\cdot \frac{{\left( {3x+2} \right)\left( {x+3} \right)}}{1}\\5{{x}^{2}}+8x-4-\left( {3x+2} \right)\left( 1 \right)&=x\left( {x+3} \right)\\\,5{{x}^{2}}+8x-4-3x-2&={{x}^{2}}+3x\\\,\,4{{x}^{2}}+2x-6&=0\\\,\,2{{x}^{2}}+x-3=\frac{0}{2}&=0\\\,\left( {2x+3} \right)\left( {x-1} \right)&=0\end{align}$ $ x=-\frac{3}{2}\,;\,\,\,\,\,\,\,x=1\,\,\,\,\,\,\,\text{(Both work)}$ | Notice we had to factor the denominator first, so we could get our LCD. We would typically factor the numerator too; we get $ \left( {5x-2} \right)\left( {x+2} \right)$, but it doesn’t really help since we can’t cancel anything out.
Sometimes we end up with a quadratic and we have to factor (“unfoil”) or maybe even use the quadratic formula, if factoring doesn’t work. In this example, we could factor. Then set each factor to 0 to solve for $ x$. Both work!
It’s still a good to check each answer. This isn’t easy; you may want to use the STO> function in your calculator to store the solutions, and then type in the sides of the equations using $ X,{\mathrm T},\theta ,n$. |
| $ \displaystyle \frac{1}{{1-x}}\,=1-\frac{x}{{x-1}}$ $ \begin{align}\frac{{x-1}}{1}\cdot \left( {\frac{{-1}}{{x-1}}} \right)&=\left( {1-\frac{x}{{x-1}}} \right)\cdot \frac{{x-1}}{1}\\-1&=\left( {x-1} \right)-x\\-1&=-1;\,\,\,\,\,\,\,\mathbb{R}\end{align}$ But, $ x\ne 1$, since this would make the denominator 0. Answer is all reals, except $ x\ne 1$, or $ \left( {-\infty ,1} \right)\cup \left( {1,\infty } \right)$. | Multiply the first term by –1 to make the denominators the same: $ \displaystyle \frac{1}{{1-x}}=\frac{{-1}}{{-\left( {1-x} \right)}}=\frac{{-1}}{{x-1}}$. The LCD is $ \left( {x-1} \right)$.
Multiply the numerators by “what’s missing” in the denominator; we end up with $ -1=-1$, which is always true.
The answer would be all real numbers, except we can’t have a 1 in the denominator, so we have to “skip over” the value of 1. |
| Solve for $ f$: $ \displaystyle \,\,\frac{1}{f}+\frac{1}{p}\,=\,\frac{1}{q}\,\,\,\,$ $ \displaystyle \begin{align}\frac{{fpq}}{1}\cdot \left( {\frac{1}{f}+\frac{1}{p}} \right)&=\left( {\frac{1}{q}} \right)\cdot \frac{{fpq}}{1}\\\,\,pq+fq&=fp\\\,fp-fq&=pq\\\,f(p-q)&=pq\\f&=\frac{{pq}}{{p-q}}\\f,p,q&\ne 0;\,\,p\ne q\end{align}$ | When solving for literal equations (equations where your answer will have other variables in it), just pretend the variables are numbers, and perform the same steps.
Here we are solving for $ f$, so, after multiplying both sides by the LCD, we need to get everything with an $ f$ in it to one side.
Then you can factor the $ f$ out, and solve for it. |
Rational Inequalities, including Absolute Values
Solving rational inequalities are a little more complicated since we are typically multiplying or dividing by variables, and we don’t know whether these are positive or negative. Remember that we have to change the direction of the inequality when we multiply or divide by negative numbers. When we solve these rational inequalities, our answers will typically be a range of numbers.
Rational Inequalities from a Graph
It’s not too bad to see inequalities of rational functions from a graph. Look at this graph to see where $ y<0$ and $ y\ge 0$. Notice that we have ranges of $ x$ values in the two cases:
| Rational Graph | To Get Inequalities |
| Find $ \displaystyle \frac{x}{{{{x}^{2}}+4x-5}}<0$, $ \displaystyle \frac{x}{{{{x}^{2}}+4x-5}}\ge 0$
| Remember to start from the left.
When $ y<0$, $ x$ is between $ -\infty $ and the first vertical asymptote (VA), which is –5. The value –5 is not included, since it’s an asymptote). $ y<0$ is also between 0 (not including 0, since we have a $ <$ and not a $ \le $), and 1 (not including 1).
When $ y\ge 0$, $ x$ is between –5 (not including –5), and 0 (including 0, since we have $ \ge $). It also starts again near the 2nd VA, which is 1 (not including 1), and then goes on to $ \infty $.
Thus, for $ \displaystyle \frac{x}{{{{x}^{2}}+4x-5}}<0$, $ x$ is $ \left( {-\infty ,-5} \right)\cup \left( {0,1} \right)$. And for $ \displaystyle \frac{x}{{{{x}^{2}}+4x-5}}\ge 0$, $ x$ is $ \left( {-5,0} \right]\cup \left( {1,\infty } \right)$. |
Solving Rational Inequalities Algebraically Using a Sign Chart
The easiest way to solve rational inequalities algebraically is using the sign chart method, which we saw here in the Quadratic Inequalities section. A sign chart or sign pattern is simply a number line that is separated into partitions (intervals or regions), with boundary points (called “critical values“) that you get by setting the factors of the rational function, both in the numerator and denominator, to 0 and solving for $ x$.
With sign charts, you can pick any point in between the critical values, and see if the whole function is positive or negative. Then you just pick that interval (or intervals) by looking at the inequality. Generally, if the inequality includes the $ =$ sign, you have a closed bracket, and if it doesn’t, you have an open bracket. But any factor that’s in the denominator must have an open bracket for the values that make it 0, since you can’t have 0 in the denominator.
The first thing you have to do is get everything on the left side (if it isn’t already there) and 0 on the right side, since we can see what intervals make the inequality true. We can only have one term on the left side, so sometimes we have to find a common denominator and combine terms.
Also, it’s a good idea to put open or closed circles on the critical values to remind ourselves if we have inclusive points (inequalities with equal signs, such as $ \le $ and $ \ge $) or exclusive points (inequalities without equal signs, or factors in the denominators).
Let’s do some examples; you can always use your graphing calculator to check your answers, too. Put in both sides of the inequalities and check the zeros, and make sure your ranges are correct!
| Rational Inequality | Notes |
| $ \displaystyle \frac{{x+4}}{{x-1}}\ge 0$
The problem calls for $ \ge 0$, so we look for the plus sign(s), and our answers are inclusive (hard brackets), unless, as in the case of $ \left( {x-1} \right)$, the factor is on the bottom. This has to be a soft bracket, since we can’t have a 0 on the bottom.
The answer is $ \left( {-\infty ,-4} \right]\cup \left( {1,\infty } \right]$. | The first thing we need to do is to get everything on the left side, and 0 on the right side, which we already have. We can’t factor further, so we are good to go! We now draw a sign chart. We get the “boundary points” or “critical values” by setting all the factors (both numerator and denominator) to 0; these are –4, and 1. Since the roots are –4 and 1, we put those on the sign chart as boundaries. Then we check each interval with random points to see the rational expression is positive or negative. We put the signs over the interval. Always try easy numbers, especially 0, if it’s not a boundary point! Note that you don’t have to get the result from putting in test points; we just have to get their sign; this will get easier!
Let’s try –5 for the leftmost interval: $ \displaystyle \frac{{\left( {-5} \right)+4}}{{\left( {-5} \right)-1}}=\frac{{-1}}{{-6}}=\text{ positive (+)}$. Let’s try 0 for the middle interval: $ \displaystyle \frac{{\left( 0 \right)+4}}{{\left( 0 \right)-1}}=\frac{4}{{-1}}=\text{ negative (}-\text{)}$. Let’s try 2 for the rightmost interval: $ \displaystyle \frac{{\left( 2 \right)+4}}{{\left( 2 \right)-1}}=\frac{6}{1}=\text{ positive (}+\text{)}$.
We want $ \ge $ from the problem, so we look for the + (positive) sign intervals, so the intervals are $ \left( {-\infty ,-4} \right]\cup \left( {1,\infty } \right]$. We need the hard brackets (inclusion of endpoint) for the –4 since we have $ \ge $ and not $ >$, but we need a soft bracket for the 1, since that factor is on the bottom. |
| $ \displaystyle \frac{{2\left( {x-4} \right)}}{x}<-4$ $ \begin{align}\frac{{2\left( {x-4} \right)}}{x}+4&<0\\\frac{{2\left( {x-4} \right)}}{x}+\frac{{4x}}{x}&<0\\\frac{{2x-8+4x}}{x}&<0\\\frac{{6x-8}}{x}&<0\\\frac{{2\left( {3x-4} \right)}}{x}&<0;\frac{{3x-4}}{x}<0\end{align}$
The problem calls for $ <0$, so we look for the minus sign(s), and our answers are exclusive (soft brackets) since we have $ <$ and not $ \le $. The answer is $ \displaystyle \left( {0,\frac{4}{3}} \right)$. | The first thing we need to do is to get everything on the left side, and 0 on the right side, so we need to add 4 to both sides. We have to find the common denominator ($ x$) and put the two terms together. Then we can factor out the 2 in the numerator, to make things easier ($ \displaystyle \frac{0}{2}=0$, so right side stays the same). We now draw a sign chart. We get the critical values by setting all the factors (both numerator and denominator) to 0; these are $ \displaystyle \frac{4}{3}$ and 0. Note that we can ignore the factor of 2, since it doesn’t have an $ x$ in it. Since the roots are $ \displaystyle \frac{4}{3}$ and 0, we put those on the sign chart as boundaries. Then we check each interval with random points to see if the rational expression (factored or unfactored) is positive or negative. We put the signs over the interval. Always try easy numbers, especially 0, if it’s not a boundary point!
Let’s try –1 for the leftmost interval: $ \displaystyle \frac{{3\left( {-1} \right)-4}}{{-1}}=\frac{{-7}}{{-1}}=\text{ positive (+)}$. Let’s try 1 for the middle interval: $ \displaystyle \frac{{3\left( 1 \right)-4}}{1}=\frac{{-1}}{1}=\text{ negative (}-\text{)}$. Let’s try 2 for the rightmost interval: $ \displaystyle \frac{{3\left( 2 \right)-4}}{2}=\frac{2}{2}=\text{ positive (+)}$.
We want $ <$ from the problem, so we look for the $ -$ (negative) sign intervals. The interval is $ \displaystyle \left( {0,\frac{4}{3}} \right)$. |
And here’s another one where we have to do a lot of “organizing” first, including moving terms to one side, and then factoring:
| Rational Inequality | Notes |
| $ \displaystyle \frac{3}{{x-3}}\ge x-5$ $ \displaystyle \begin{align}\frac{3}{{x-3}}+-x+5&\ge 0\\\frac{3}{{x-3}}+\frac{{\left( {-x+5} \right)\left( {x-3} \right)}}{{x-3}}&\ge 0\\\frac{{3-{{x}^{2}}+8x-15}}{{x-3}}&\ge 0\\\left( {-1} \right)\left( {\frac{{-{{x}^{2}}+8x-12}}{{x-3}}} \right)&\ge 0\left( {-1} \right)\\\frac{{{{x}^{2}}-8x+12}}{{x-3}}&\le 0\\\frac{{\left( {x-6} \right)\left( {x-2} \right)}}{{x-3}}&\le 0\end{align}$
The problem calls for $ \le 0$ (after we multiplied by –1), so we look for the minus sign(s), and our answers are inclusive (hard brackets), except for the boundary point in the denominator (3). The answer is $ \left( {-\infty ,2} \right]\cup \left( {3,6} \right]$. | The first thing we need to do is to get everything on the left side, and 0 on the right side. Then then find the common denominator $ (x-3)$ to put the two terms together. It’s easier to factor with a positive $ {{x}^{2}}$, so we can multiply both sides by –1, but make sure we change the direction of the inequality sign!
After factoring, draw a sign chart. We get the critical values by setting all the factors (both numerator and denominator) to 0; these are 2, 3, and 6.
Now check each interval with random points to see if the rational expression (factored or unfactored – I prefer factored) is positive or negative. We put the signs over the interval. Always try easy numbers, especially 0, if it’s not a boundary point!
Let’s try 0 for the leftmost interval: $ \displaystyle \frac{{\left( {0-6} \right)\left( {0-2} \right)}}{{0-3}}=\frac{{12}}{{-3}}=\text{ negative }(-)$ Now 2.5 for the next interval: $ \displaystyle \frac{{\left( {2.5-6} \right)\left( {2.5-2} \right)}}{{2.5-3}}=\frac{{-1.75}}{{-.5}}=\text{ positive }(+)$ Now 4 for the next interval: $ \displaystyle \frac{{\left( {4-6} \right)\left( {4-2} \right)}}{{4-3}}=\frac{{-4}}{1}=\text{ negative }(-)$ And 7 for the rightmost interval: $ \displaystyle \frac{{\left( {7-6} \right)\left( {7-2} \right)}}{{7-3}}=\frac{5}{4}=\text{ positive }(+)$
We want $ \le $ from the factored inequality, so we look for the $ -$ (negative) sign intervals, so the interval is $ \left( {-\infty ,2} \right]\cup \left( {3,6} \right]$. |
Sometimes we get a funny interval with a single $ x$ value as part of the interval:
| Rational Inequality | Notes |
| $ \displaystyle \frac{{{{x}^{2}}}}{{{{x}^{2}}+4x-5}}\ge 0$
$ \displaystyle \frac{{{{x}^{2}}}}{{\left( {x+5} \right)\left( {x-1} \right)}}\ge 0$
We have a closed circle at 0, since the 0 isn’t in the denominator, and we have a $ \ge .$ We need to include the 0 with the other intervals. The answer is $ \left( {-\infty ,5} \right)\cup \left[ 0 \right]\cup \left( {1,\infty } \right)$. | This one’s a little different since we end up with two minuses in a row – around the 0. Factor the denominator first. We have 3 critical values, including 0, since we have an $ {{x}^{2}}$ in the numerator.
Let’s try –6 for the leftmost interval: $ \displaystyle \frac{{{{{\left( {-6} \right)}}^{2}}}}{{\left( {-6+5} \right)\left( {-6-1} \right)}}=\frac{{36}}{7}=\text{positive }(+)$ Now –1 for the next interval: $ \displaystyle \frac{{{{{\left( {-1} \right)}}^{2}}}}{{\left( {-1+5} \right)\left( {-1-1} \right)}}=\frac{1}{{-8}}=\text{ negative }(-)$ Now .5 for the next interval: $ \displaystyle \frac{{{{{\left( {.5} \right)}}^{2}}}}{{\left( {.5+5} \right)\left( {.5-1} \right)}}=\frac{{.25}}{{-2.75}}=\text{ negative }(-)$ And 2 for the rightmost interval: $ \displaystyle \frac{{{{2}^{2}}}}{{\left( {2+5} \right)\left( {2-1} \right)}}=\frac{4}{7}=\text{ positive }(+)$
We want $ \ge $ from the factored inequality, so we look for the $ +$ (positive) sign intervals. We also need to include 0, so the interval is $ \left( {-\infty ,5} \right)\cup \left[ 0 \right]\cup \left( {1,\infty } \right)$. |
And here’s one where we have a removable discontinuity in the rational inequality, so we have to make sure we skip over that point:
| Rational Inequality | Notes |
| $ \displaystyle \frac{{{{x}^{2}}-5x+6}}{{{{x}^{2}}-9}}\ge 0$ $ \require{cancel} \displaystyle \frac{{\cancel{{\left( {x-3} \right)}}\left( {x-2} \right)}}{{\cancel{{\left( {x-3} \right)}}\left( {x+3} \right)}}\ge 0$ $ \displaystyle \frac{{x-2}}{{x+3}}\ge 0$
The problem calls for $ \ge 0$, so we look for the plus sign(s). We can’t include 3 though; we have to “jump over” it. The answer is $ \displaystyle \left( {-\infty ,-3} \right)\cup \left[ {2,3} \right)\cup \left( {3,\infty } \right)$. | After factoring, we see that we have a removable discontinuity, or hole at $ x=3$. We need to put that on the sign chart number line, with an open hole, since $ x$ can’t be that value. We also have to put an open hole at $ x=-3$, since the factor is in the denominator. We can put a closed hole at $ x=2$, since we have a $ \ge $ sign.
Then we check each interval with random points to see if the factored form of the quadratic is positive or negative. We can use the rational expression after crossing out the hole for these tests, and put the signs over the interval. Make sure to use 0 as a test point if you can!
We want $ \ge $ from the problem, so we look for the $ +$ signs; the interval is $ \displaystyle \left( {-\infty ,-3} \right)\cup \left[ {2,3} \right)\cup \left( {3,\infty } \right)$. |
Here are a couple that involves solving radical inequalities with absolute values. (You might want to review Solving Absolute Value Equations and Inequalities before continuing on to this topic.)
| Rational Inequality with Absolute Value | Notes |
| Let’s do a simple one first, where we can handle the absolute value just like a factor, but when we do the checking, we’ll take into account that it is an absolute value.
$ \displaystyle \,\frac{{\left| {x+4} \right|}}{{x-1}}<0$
The problem calls for $ <0$, so we look for the minus signs, but we can’t include –4. | Even with the absolute value, we can set each factor to 0, so we get –4 and 1 as critical values. We use open circles since we have a $ <$ sign (we’d have to for the 1 anyway, since it’s on the bottom).
Then we check each interval with random points to see if the factored form of the quadratic is positive or negative, making sure we include the absolute value. It’s a little different, since we have 2 minuses in a row without a “bounce” in the graph.
We want $ <$ from the problem, so we look for the $ -$ signs, but can’t include the –4 since it has a circle on it. The interval is $ \displaystyle \left( {-\infty ,-4} \right)\cup \left( {-4,1} \right)$. |
| $ \displaystyle \,\left| {\frac{2}{{x+2}}} \right|\ge 4$ $ \displaystyle \begin{align}\frac{2}{{x+2}}\ge 4\text{ }\,\,\,&\text{or }\,\,\,\frac{2}{{x+2}}\le -4\\\frac{2}{{x+2}}-4\ge 0\text{ }\,\,\,&\text{or }\,\,\,\frac{2}{{x+2}}+4\le 0\\\frac{2}{{x+2}}-\frac{{4\left( {x+2} \right)}}{{x+2}}\ge 0\text{ }\,\,\,&\text{or }\,\,\,\frac{2}{{x+2}}+\frac{{4\left( {x+2} \right)}}{{x+2}}\le 0\\\frac{{-6-4x}}{{x+2}}\ge 0\text{ }\,\,\,&\text{or }\,\,\,\frac{{4x+10}}{{x+2}}\le 0\end{align}$
| This one’s a little more complicated since we don’t have a 0 on the right.
Let’s first separate the absolute value into two separate inequalities. Then we need to get everything to the left side to have 0 on the right first. Simplify with a common denominator.
Our critical values are $ \displaystyle -\frac{5}{2}$, –2, and $ \displaystyle -\frac{3}{2}$. The sign chart is to the left; we can use the original inequality and put T where it works and F where it doesn’t. We see the solution is: $ \displaystyle \left[ {-\frac{5}{2},-2} \right)\cup \left( {-2,-\frac{3}{2}} \right]$.
Note that we can’t include –2 as a solution (soft bracket) since it would make the denominator 0. |
Here are more complicated ones, where the absolute value may need to be multiplied by other variables (think of if you had to cross multiply). Notice how it’s best to separate the inequality into two separate inequalities: one case when $ x$ is positive, and the other when $ x$ is negative. Notice also how we had to use the Quadratic Formula to get the critical points when $ x$ is negative:
| Absolute Value Inequality Problem with Rationals | Notes | ||
| $ \displaystyle \frac{{\left| x \right|}}{{x-1}}>\frac{{x+1}}{{2x+1}}$
The problem calls for $ >0$, so we look for the plus sign intervals. But we have to throw away any intervals where the sign doesn’t agree with our conditions of $ x$ (positive or negative), such as the interval $ \left( {.434,1} \right)$ ($ x$ is supposed to be negative). Tricky! | Separate into two cases, since we don’t know whether $ x$ is positive or negative. After separating into two cases, get all the variables to one side and set to 0, to find the boundary values or critical points using a sign chart).
Look at the negative part of the sign chart when $ x$ is negative, and the positive part of the sign chart when $ x$ is positive.
When $ x$ is positive, the numerator yields no “real” critical points. When $ x$ is negative, use the Quadratic Formula to get the roots (critical points) of the numerator: $ \displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}=\frac{{1\pm \sqrt{{{{{\left( {-1} \right)}}^{2}}-4\left( {-3} \right)\left( 1 \right)}}}}{{-6}}$ $ \displaystyle =\frac{{1\pm \sqrt{{13}}}}{{-6}}\approx -.768,\,\,\,.434$ The answer is $ \displaystyle \left( {.768,-\frac{1}{2}} \right)\cup \left( {1,\infty } \right)$.
Put the inequality in your graphing calculator to check your answer! |
Here’s one more that’s a bit tricky, since we have two expressions with absolute value in it. In this case, we have to separate in four cases, just to be sure we cover all the possibilities.
| Absolute Value Problem with Rationals | Notes | ||||
| $ \displaystyle \frac{{\left| {x-1} \right|}}{{\left| {3x+1} \right|}}>1$
| Separate into four cases, since we don’t know whether $ x-1$ and $ 3x+1$ are positive or negative, since they are absolute values. (We probably could have just used two cases, since the absolute values are on the top and bottom of same fraction, but this way is safer.)
Use sign charts to determine what intervals the rational will be positive or negative. The problem calls for $ >0$, so look for the plus sign intervals, and make sure they work in the original inequality.
The answer is $ \displaystyle \left( {-1,-\frac{1}{3}} \right)\cup \left( {-\frac{1}{3},0} \right)$. We have to “skip over” (asymptote) $ \displaystyle -\frac{1}{3}$ (so we don’t divide by 0), and use soft brackets, since the inequality is $ >$ and not $ \ge $.
Put the inequality in your graphing calculator to check your answer! |
There’s one more that we did here in the Compositions of Functions, Even and Odd, and Increasing and Decreasing section, when we worked on domains of composites.
Applications of Rationals
There are certain types of word problems that typically use rational expressions. These tend to deal with rates, since rates are typically fractions (such as distance over time). We also see problems dealing with plain fractions or percentages in fraction form.
Here is a rational “fraction” problem:
| Fraction Problem | Solution |
The denominator of a fraction is 2 less than twice the numerator.
If 7 is added to both the numerator and denominator, the resulting fraction is $ \displaystyle \frac{4}{5}$.
What is the original fraction? | This problem is a little tricky, since we don’t want to set the variable to what the problem is asking – the original fraction. We want to set $ n=$ the numerator, since we’ll have to get both the numerator and the denominator, and the denominator is in terms of the numerator. So, from the first sentence of the problem, the original fraction is $ \displaystyle \frac{n}{{2n-2}}$. To solve, we just have to add 7 to the numerator and denominator, and set to $ \displaystyle \frac{4}{5}$. Solve for $ n$: $ \displaystyle \begin{align}\frac{{n+7}}{{\left( {2n-2} \right)+7}}\,&=\,\frac{4}{5}\\\,\frac{{n+7}}{{2n+5}}\,&=\,\frac{4}{5}\\\,\left( 5 \right)\left( {n+7} \right)\,&=\,\left( 4 \right)\left( {2n+5} \right)\\\,\,5n+35&=8n+20\\\,3n&=15\\\,n&=5\end{align}$ The original fraction is $ \displaystyle \frac{n}{{2n-2}}\,=\,\frac{5}{8}.$ Let’s check our answer: The denominator of $ \displaystyle \frac{5}{8}$ is 2 less than twice the numerator. If 7 is added to both the numerator and denominator, the resulting fraction is $ \displaystyle \frac{{5+7}}{{8+7}}=\frac{{12}}{{15}}=\frac{4}{5}$. (It was totally coincidental that the answer was the same fraction as in the picture above. Weird :)) |
Here is a rational “percentage” problem:
| Percentage Problem | Solution |
| Bethany has scored 10 free throws out of 18 tries. She would really like to bring her free throw average up to at least 68%.
How many consecutive free throws should she score in order to bring up her average to 68%?
| Let $ x =$ the number of free throws that Bethany should score (in a row) in order to bring up her average. Again, we can use fractions, and this time they will represent the fraction of free throws that she scores. Start out with her current fraction (rate) of consecutive free throws, and then add the number Bethany needs to score to both the numerator (number she scores) and denominator (total number of throws). Solve for $ x$: $ \displaystyle \begin{align}\frac{{10+x}}{{18+x}}&=\frac{{68}}{{100}}\\\left( {100} \right)\left( {10+x} \right)&=\left( {68} \right)\left( {18+x} \right)\\1000+100x&=1224+68x\\32x&=224\\x&=7\end{align}$ Bethany needs 7 more consecutive free throws to bring her free throw percentage up to 68%. Let’s check our answer: If she throws 7 more free throws, she’ll have 17 scores in a row out of 25 tries: $ \displaystyle \frac{{17}}{{25}}=68\%$. |
Distance/Rate/Time Problems
With rational rate problems, we must always remember: $ \text{Distance}=\text{Rate }\times \,\,\text{Time}$. It seems most of the problems deal with comparing times or adding times.
Problem:
Shalini can run 3 miles per hour faster than her sister Meena can walk. If Shalini ran 12 miles in the same time it took Meena to walk 8 miles, what is the speed of each sister in this case?
Solution:
This is a “$ \text{Distance}=\text{Rate }\times \,\,\text{Time}$” problem, and let’s use a table to organize this information like we did in the Algebra Word Problems and Systems of Linear Equations and Word Problems sections. Let $ x=$ Meena’s speed, since Shalini runs faster and it’s easier to add than subtract:
| Rate (Speed) | Time (Fill in Last) | Distance | ||
| Shalini | $ x+3$ | $ \displaystyle \frac{{12}}{{x+3}}$ | $ 12$ | Rate x Time = Distance |
| Meena | $ x$ | $ \displaystyle \frac{8}{x}$ | $ 8$ | Rate x Time = Distance |
| Do Nothing Here | Set Equal | Fill in rates and distances first. Then divide distance by rate to get time. | ||
Put it all together:
| Distance Problem using Rationals | Solution |
| Shalini can run 3 miles per hour faster than her sister Meena can walk. If Shalini ran 12 miles in the same time it took Meena to walk 8 miles, what is the speed of each sister in this case?
$ \displaystyle \begin{align}\text{Time}&=\text{Time}\\\frac{{12}}{{x+3}}&=\frac{8}{x}\\12x&=8\left( {x+3} \right)\\12x&=8x+24\\4x&=24;\,\,x=6\end{align}$ | Since we know that Shalini ran 12 miles in the same time that Meena walked 8 miles, we can set the times equal to each other. Now we can just cross multiply since we only have one term on each side.
Meena’s speed is 6 miles per hour, and Shalini’s speed is $ 6+3=$ 9 miles per hour – pretty fast! |
| Upstream/Downstream Problem using Rationals | Solution |
| The time it takes a canoe to go 3 miles upstream and back 3 miles downstream is 4 hours. The current in the lake is 1 mile per hour. Find the average speed (rate) of the canoe in still water.
$ \begin{align}\,\frac{3}{{x-1}}+\frac{3}{{x+1}}&=4\\\left( {x-1} \right)\left( {x+1} \right)\left( {\frac{3}{{x-1}}+\frac{3}{{x+1}}} \right)&=4\left( {{{x}^{2}}-1} \right)\\\,3\left( {x+1} \right)+3\left( {x-1} \right)&=4{{x}^{2}}-4\\\,\,3x\cancel{{+3}}+3x\cancel{{-3}}&=4{{x}^{2}}-4\\\,\,4{{x}^{2}}-6x-4&=0\\2{{x}^{2}}-3x-2&=0\\\left( {2x+1} \right)\left( {x-2} \right)&=0\\x&=-\frac{1}{2}\,\,\,\,\,\,\text{or}\,\,\,\,\,\,x=2\end{align}$ | In this problem, we use the rates of a boat going upstream (rate in still water rate minus the rate of the water current) and downstream (rate in still water plus the rate of the water current) to add times. Think when you’re going downstream, it’s like you’re going down a hill, so it’s faster.
Let $ x=$ the speed of the boat in still water. We know that the rate of the canoe going upstream (“up” is more difficult so we go slower) is “$ x-1$”, and the rate going downstream is “$ x+1$” (“down is easier, so we go faster). Use that the fact that $ \displaystyle \text{Time}=\frac{{\text{Distance}}}{{\text{Rate}}}$ to add the time upstream to the time downstream; this equals 4 hours.
Then we solve for $ x$; we multiply both sides by the LCD, which is $ \left( {x-1} \right)\left( {x+1} \right)$, or $ {{x}^{2}}-1$. We end up with a quadratic that we can factor, so we get two possible solutions.
Since we can’t have a negative rate in this problem, the average rate of the canoe in still water is 2 miles per hour.
Note: If we were given the rate of the canoe in still water and had to find the rate of the current, we’d do the problem in a similar way. We’d have to add and subtract variables (the rate of the current) from numbers (the rates of the canoe in still water) in the denominators. |
Work Problems:
Work problems typically have to do with different people or things working together and alone, at different rates. Instead of distance, we work with jobs (typically, 1 complete job).
I find that usually the easiest way to work these problems is to remember:
$ \displaystyle \frac{{\operatorname{time} \,\text{together}}}{{\text{time alone}}}\,\,+\,…\,\,+\,\,\frac{{\text{time together}}}{{\text{time alone}}}\,\,=\,\,1$ (or whatever part of the job or jobs is done; if they do half the job, this equals $ \displaystyle \frac{1}{2}$; if they do a job twice, this equals 2).
(Use a “$ +$” between the terms if working towards the same goal, such as painting a room, and “$ -$” if working towards opposite goals, such as filling and emptying a pool.)
“Proof”: For work problems, $ \text{Rate }\times \,\text{Time = }1\text{ Job}$. Add up individuals’ portions of a job with this formula, using the time working with others (time together):
$ \displaystyle \begin{align}\left( {\text{individual rate }\times \text{ time }} \right)\text{ + }…\text{ +}\left( {\text{individual rate }\times \text{ time }} \right)&=1\\\left( {\frac{1}{{\text{individual time to do 1 job}}}\text{ }\times \text{ time}} \right)\text{ + }…\text{ +}\left( {\frac{1}{{\text{individual time to do 1 job}}}\text{ }\times \text{ time}} \right)&=1\\\left( {\frac{{\text{time working with others}}}{{\text{individual time to do 1 job}}}} \right)\text{ + }…\text{ +}\left( {\frac{{\text{time working with others}}}{{\text{individual time to do 1 job}}}} \right)&=1\end{align}$
Also, as explained after the first example below, often you see this formula as $ \displaystyle \frac{\text{1}}{{\text{time alone}}}\,+…+\,\frac{\text{1}}{{\text{time alone}}}\,=\,\frac{\text{1}}{{\text{time together}}}$.
Let’s do some problems:
| Rational Work Problem | Solution |
Erica can paint her room in 5 hours. If she has her friend Rachel help her, they can paint the room together in 3 hours. How long would it take for Rachel to paint the room alone, if Erica wanted to go play tennis for the afternoon?
$ \displaystyle \begin{align}\frac{3}{5}+\frac{3}{R}&=1\\\left( {5R} \right)\left( {\frac{3}{5}+\frac{3}{R}} \right)&=1\left( {5R} \right)\\3R+15&=5R\\2R&=15\\R&=7.5\end{align}$ | Let $ R=$ amount of time (in hours) Rachel can paint the room by herself. We know that the time the girls paint the room together is 3 hours, and the time Erica paints the room by herself is 5 hours.
Use the formula $ \displaystyle \frac{{\text{time together}}}{{\text{time alone}}}\text{+}\frac{{\text{time together}}}{{\text{time alone}}}=1$, since they need to get the whole job done.
Erica’s time to paint is 5 hours, Rachel’s is $ R$, since we’re trying to find it, and their time together is 3 hours. Multiply both sides the LCD, which is $ 5R$.
It would take Rachel 7.5 hours to paint Erica’s room. Let’s hope Rachel comes through to help, so Erica can play tennis! |
Note: The formula above can also be derived by using the concept that you can figure out how much of the job the girls do per hour (or whatever the time unit is), both together and alone. Then you can add the individual “rates” to get the “rate” of their painting together.
We are actually adding the Work they complete (together and alone) using formula $ \text{Rate }\times \text{ }\text{Time }=\text{ Work}$, where the Time is 1 hour (or whatever the unit is).
In this example, Erica’s rate per hour is $ \displaystyle \frac{1}{5}$ (she can do $ \displaystyle \frac{1}{5}$ of the job in 1 hour); Rachel’s rate per hour is $ \displaystyle \frac{1}{R}$; we can add their rates to get the rate of their painting together: $ \displaystyle \frac{1}{5}+\frac{1}{R}=\frac{1}{3}$. If we multiply all the terms by 3, we come up with the equation above! (Sometimes you see this as $ \displaystyle \frac{\text{1}}{{\text{time alone}}}\,\,+…+\,\frac{\text{1}}{{\text{time alone}}}\,=\,\frac{\text{1}}{{\text{time together}}}$.)
| Rational Work Problem | Solution |
| It takes Jill 3 hours to make sandwiches for a charity organization (as many as they need). Allie starts one hour later and (if she were working alone), it would take her 4 hours to make the sandwiches needed. How long will it take both of them working together to complete the job?
$ \displaystyle \begin{align}\frac{T}{3}+\frac{T}{4}&=\frac{2}{3}\\\left( {12} \right)\left( {\frac{T}{3}+\frac{T}{4}} \right)&=\frac{2}{3}\left( {12} \right)\\4T+3T&=8\\7T&=8\\T&=\frac{8}{7}=1\frac{1}{7}\end{align}$ | This one’s a little trickier since “Allie starts one hour later”. Since it takes Jill 3 hours to make sandwiches by herself, after one hour (when Allie starts), she’s already one third of the way through the job (one third of 3 hours is 1 hour). The amount of job left then is two thirds.
Use $ \displaystyle \frac{{\text{time together}}}{{\text{time alone}}}\,\,\text{+}\,\,\frac{{\text{time together}}}{{\text{time alone}}}=\frac{2}{3}$, since they only need to get $ \displaystyle \frac{2}{3}$ of the job done.
Jill’s time to make sandwiches is 3 hours, Allie’s is 4 hours, and we’re trying to find their time together, $ T$. Multiply both sides the LCD, which is 12.
It would take the girls $ \displaystyle 1\frac{1}{7}$ or about 1.14 hours to make the sandwiches. It really helps to have them work together, and it’s more fun that way, too! |
| Rational Work Problem | Solution |
| Two hoses are used to fill Maddie’s neighborhood swimming pool. One hose alone can fill the pool in 10 hours; the second hose can fill it in 12 hours. The pool’s drain can empty the pool in 8 hours. If the two hoses are working, and the drain is open (by mistake), how long will it take to fill the swimming pool?
$ \displaystyle \begin{align}\frac{x}{{10}}+\frac{x}{{12}}-\frac{x}{8}&=1\\\left( {120} \right)\left( {\frac{x}{{10}}+\frac{x}{{12}}-\frac{x}{8}} \right)&=1\left( {120} \right)\\\,12x+10x-15x&=120\\\,7x\,&=\,120\\x&=\frac{{120}}{7}=17\frac{1}{7}\,\,\text{hrs}\end{align}$ | This is the same type of “work” problem where we have “alone” times, but we want to find the “together” time. The difference with this problem is that we have 3 different things working to fill the pool, one of which is working against filling the pool (the drain), which we need to make negative.
Let $ x=$ the time it takes for the 2 hoses (positive work) and the drain (negative work) all together to fill the pool:
Use $ \displaystyle \frac{{\text{time together}}}{{\text{time alone}}}\,\,+\,\,\frac{{\text{time together}}}{{\text{time alone}}}\,\,-\,\,\frac{{\text{time together}}}{{\text{time alone}}}=\,\,1$, since we have 2 hoses coming into the pool, and 1 drain where the water is going out. Multiply both sides the LCD, which is 120.
It would take the 2 hoses coming in, working with the drain with water going out $ \displaystyle \frac{{120}}{7}$ or $ \displaystyle 17\frac{1}{7}$ hours to fill the pool. It’s taking a lot longer having that drain open! |
Here is a systems problem and also a work problem at the same time. We did this problem without using rationals here in the Systems of Linear Equations and Word Problems section (and be careful, since the variables we assigned were different).
| Rational Work Problem | Solution |
| 8 women and 12 girls can paint a large mural in 10 hours. 6 women and 8 girls can paint it in 14 hours. Find the time by 1 woman alone, and 1 girl alone to paint the mural.
$ \displaystyle \begin{align}\frac{8}{w}+\frac{{12}}{g}&=\frac{1}{{10}}\\\frac{6}{w}+\frac{8}{g}&=\frac{1}{{14}}\\\left( {10gw} \right)\left( {\frac{8}{w}+\frac{{12}}{g}} \right)&=\frac{1}{{10}}\left( {10gw} \right)\\\left( {14gw} \right)\left( {\frac{6}{w}+\frac{8}{g}} \right)&=\frac{1}{{14}}\left( {14gw} \right)\\80g+120w=gw;\,\,\,\,\,&84g+112w=gw\\80g+120w&=84g+112w\\8w=4g;\,\,\,&\,\,\,\,g=2w\end{align}$ Substitute in first equation: $ \displaystyle \begin{align}\frac{8}{w}+\frac{{12}}{{2w}}&=\frac{1}{{10}}\\\,\left( {10w} \right)\frac{8}{w}+\frac{{12}}{{2w}}&=\frac{1}{{10}}\,\left( {10w} \right)\\\,80+60&=w;\\\\w&=140\\g=2w&=280\end{align}$ | We can use the approach that we can figure out what per hour rates are (together and alone), and add the individual “rates” to get the “rate” of their painting together. Since a work rate is $ \displaystyle \frac{1}{{\text{time}}}$, this is actually the $ \displaystyle \frac{\text{1}}{{\text{time alone}}}\,\,+…+\,\frac{\text{1}}{{\text{time alone}}}=\frac{\text{1}}{{\text{time together}}}$ method, but with multiple people.
Let $ w=$ the time (in hours) it takes for one woman to paint the mural alone, and $ g=$ the time it takes for one of the girls to paint the mural alone. Each woman can paint $ \displaystyle \frac{1}{w}$ murals in one hour, and each girl can paint $ \displaystyle \frac{1}{g}$ murals in one hour. This means that 8 women can paint $ \displaystyle 8\left( {\frac{1}{w}} \right)$ murals in an hour, and 12 girls can paint $ \displaystyle 12\left( {\frac{1}{g}} \right)$ murals in an hour. Together the 8 women and 12 girls can paint a mural in 10 hours ($ \displaystyle \frac{1}{{10}}$ of a mural in an hour). (We’ve added amounts of work per hour, knowing their total work in an hour is $ \displaystyle \frac{1}{{10}}$.)
Similarly, 6 women can paint $ \displaystyle 6\left( {\frac{1}{w}} \right)$ murals in an hour, and 8 girls can paint $ \displaystyle 8\left( {\frac{1}{g}} \right)$ murals in an hour. Together the 6 women and 8 girls can paint a mural in 14 hours ($ \displaystyle \frac{1}{{14}}$ of a mural in an hour).
Add what they can paint individually together in 1 hour to get how much they can paint together in 1 hour, and, with the 2 parts of the problem, we have a system. Multiply both sides of the equations by their respective LCDs. We try not to work with the $ gw$, so we eliminate it by setting the two equations together (we got lucky here!). Now we can solve to get $ g=2w$. We then use substitution with either one of the equations; we used the first one.
It would take one of the women 140 hours, and one of the girls 280 hours to paint the mural by herself. Whew – that was a tough one!! |
Cost Problems
A third type of problem you might get while studying rationals has to do with average cost, or possibly costs per person (or unit cost) problems. Both of these types of costs can still be thought as rates, as they are an amount over time. Here are some examples:
| Average Cost Problem | Solution |
| Eden bought a new cell phone for $450 and also bought an annual warranty plan for $50 a year.
In what year of ownership will Eden’s average annual cost of the phone be $200?
| This is an average cost problem, so use the formula $ \displaystyle \text{Average Cost}=\frac{{\text{Total Cost}}}{{\text{Total Time}}}$, and the Total Cost includes both fixed costs (one-time costs) and variable costs (costs that depend on time).
Let $ x=$ the year in which Eden’s average annual cost of owning the phone is $200. Then $ 50x$ is the amount Eden has paid for the warranty (variable cost), in addition to the cost of the phone (fixed cost). Thus, we have: $ \begin{align}200&=\frac{{\text{Total Cost}}}{{\text{Total Time}}}\\200&=\frac{{\text{450}+50x}}{x}\\200x&=450+50x\\150x&=450\\x&=3\end{align}$ In the 3rd year of ownership, the average cost of owning the phone will be $200. |
| Cost Problem | Solution |
| A group of girls decided to go on a trip to the beach and the organizer said that the bus would cost $360 to rent.
The organizer also told them that if they got 3 more girls to go on the trip, each girl could pay $6 less (which they ended up doing).
How many girls ended up going on the trip? | Let $ n$ be the original number of girls going on the trip. The cost per girl for the original trip (like a rate) is $ \displaystyle \frac{{360}}{n}$ (use real numbers to see this – if 10 girls were going, it would be $36 per girl). If they added 3 more girls, each girl would have to pay $ \displaystyle \frac{{360}}{{n+3}}$ since the total cost is still $360, but the number of girls is $ \displaystyle n+3$. Now, with $ \displaystyle n+3$ girls, the new cost is 6 dollars less, or the original cost minus 6, so we can set up an equation like this: $ \displaystyle \frac{{360}}{{n+3}}=\frac{{360}}{n}-6$. Now solve; since we have a quadratic, we have to move everything to one side and set it to 0. When we simplify and factor, we need to “throw away” the negative number; this won’t work for the number of girls. $ \begin{array}{c}\frac{{360}}{{n+3}}\,=\,\frac{{360}}{n}-6\\\left( {n\left( {\cancel{{n+3}}} \right)} \right)\left( {\frac{{360}}{{\cancel{{n+3}}}}} \right)=\left( {\frac{{360}}{{\cancel{n}}}-6} \right)\left( {n\left( {n+3} \right)} \right)\\\,360n=360\left( {n+3} \right)-6\left( {n\left( {n+3} \right)} \right)\\\cancel{{360n}}=\cancel{{360n}}+1080-6{{n}^{2}}-18n\\\,\,6{{n}^{2}}+18n-1080=0;\,\,\,6\left( {{{n}^{2}}+3n-180} \right)=0\,\,\,\,\,\\\,\,\,\,{{n}^{2}}+3n-180=0;\,\,\,\,\,\,\left( {n+15} \right)\left( {n-12} \right)=0\\\,\,\,\,\,\,\,\cancel{{n=-15}};\,\,\,\,\,\,\,n=12\,\,\,\,\,\,\,\,\\n+3=15\,\,\,\text{girls}\end{array}$ When we get the answer, we have to be careful and add 3 to that number. We solved for $ n$, which is the original number of girls, but the problem calls for the new number of girls, which is $ n+3$. The original number of girls planning on going on the trip is 12, but the new number is 15. |
Rational Inequality Word Problem
Here is a rational inequality word problem that we saw in the Algebra Word Problems section.
| Inequality Word Problem | Math/Notes |
| A school group wants to rent part of a bowling alley to have a party. The bowling alley costs $500 to rent, plus an additional charge of $5 per student to bowl. The group doesn’t want any student to pay more than $15 total to attend this party.
What is an inequality that could represent this situation?
How many students would need to attend so each student would pay at most $15? | Since there is a one-time cost in addition to a per-person cost, the cost per person will depend on the number of students attending the party: the more students, the lower the cost.
For $ x$ students attending, each would have to pay $ \displaystyle \frac{{500}}{x}$ for the bowling alley rent; try it with real numbers! In addition, each student needs to pay their $5 to bowl.
Therefore, each student will need to pay $ \displaystyle \frac{{500}}{x}+5$, and since we don’t want any student to pay more than $15, the inequality that represents this situation is $ \displaystyle \frac{{500}}{x}+5\le 15$. To see how many students would have to attend to keep the cost at $15 per person, solve for $ x$. In this example, we can multiply both sides by $ x$ without worrying about reversing the inequality sign, since $ x$ can only be positive.
$ \displaystyle \frac{{500}}{x}+5\le 15;\,\,\,\,\frac{{500}}{x}\le 10;\,\,\,\,500\le 10x;\,\,\,\,x\ge 50$. At least 50 students would have to attend. |
Note that there is a Rational Asymptote Application Problem here in the Graphing Rational Functions, including Asymptotes section.
Understand these problems, and practice, practice, practice!
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