Note that we saw how to solve** linear inequalities** here in the **Coordinate System and Graphing Lines** section. Note also that we solve **Algebra Word Problems** without Systems here, and we solve systems using **matrices** in the **Matrices and Solving Systems with Matrices** section here.

## Introduction to Systems

“Systems of equations” just means that we are dealing with more than one equation and variable. So far, we’ve basically just played around with the equation for a line, which is $ y=mx+b$. Let’s say we have the following situation:

You’re going to the mall with your friends and you have **$200** to spend from your recent birthday money. You discover a store that has all jeans for **$25** and all dresses for **$50**. You really, really want to take home **6** items of clothing because you “need” that many new things. How many pairs of jeans and how many dresses you can buy so you use the whole **$200** (tax not included)?

Now, you can always do “guess and check” to see what would work, but you might as well use algebra! It’s much better to learn the algebra way, because even though this problem is fairly simple to solve, the algebra way will let you solve any algebra problem – even the really complicated ones.

The first trick in problems like this is to figure out what we want to know. This will help us decide what variables (unknowns) to use. What we want to know is how many pairs of jeans we want to buy (let’s say “$ j$”) and how many dresses we want to buy (let’s say “$ d$”). Always write down what your variables will be:

Let $ j=$ the number of jeans you will buy

Let $ d=$ the number of dresses you’ll buy

Like we did before, let’s translate word-for-word from math to English:

Now we have the **2** equations as shown below. Notice that the $ j$ variable is just like the $ x$ variable and the $ d$ variable is just like the $ y$. It’s easier to put in $ j$** **and $ d$ so we can remember what they stand for when we get the answers.

This is what we call a system, since we have to solve for more than one variable – we have to solve for **2** here. The cool thing is to solve for **2** **variables**, you typically need **2** **equations**, to solve for **3** **variables**, you need **3** **equations**, and so on. That’s easy to remember, right?

We need to get an answer that works in **both equations**; this is what we’re doing when we’re solving; this is called **solving simultaneous systems**, or **solving system simultaneously**. There are several ways to solve systems; we’ll talk about **graphing** first.

## Solving Systems by Graphing

Remember that when you graph a line, you see all the different coordinates (or $ x/y$ combinations) that make the equation work. In systems, you have to make **both** equations work, so the intersection of the two lines shows the point that fits both equations (assuming the lines do in fact intersect; we’ll talk about that later). **The points of intersections satisfy both equations simultaneously. **

Put these equations into the $ y=mx+b$ ($ d=mj+b$) format, by solving for the $ d$ (which is like the $ y$):

$ \displaystyle \begin{array}{c}j+d=6;\text{ }\,\text{ }\,\text{solve for }d:\text{ }d=-j+6\text{ }\\25j+50d=200;\text{ }\,\,\text{solve for }d:\text{ }\\d=\displaystyle \frac{{200-25j}}{{50}}=-\displaystyle \frac{1}{2}j+4\end{array}$

Now graph both lines:

We can also use our graphing calculator to solve the systems of equations:

Note that with non-linear equations, there will most likely be more than one intersection; an example of how to get more than one solution via the Graphing Calculator can be found in the **Exponents and Radicals in Algebra** section. Also, there are some examples of systems of inequality here in the **Coordinate System and Graphing Lines** section.

## Solving Systems with Substitution

Substitution is the favorite way to solve for many students! It involves exactly what it says: substituting one variable in another equation so that you only have **one** variable in that equation. Here is the same problem:

You’re going to the mall with your friends and you have **$200** to spend from your recent birthday money. You discover a store that has all jeans for **$25** and all dresses for **$50**. You really, really want to take home **6** items of clothing because you “need” that many new things. How many pairs of jeans and how many dresses you can buy so you use the whole **$200** (tax not included)?

Below are our two equations, and let’s solve for “$ d$” in terms of “$ j$” in the first equation. Then, let’s substitute what we got for “$ d$” into the next equation. Even though it doesn’t matter which equation you start with, remember to always pick the “easiest” equation first (one that we can easily solve for a variable) to get a variable by itself.

We could buy **4 pairs of jeans and 2 dresses**. Note that we could have also solved for “$ j$” first; it really doesn’t matter. You’ll want to pick the variable that’s most easily solved for. Let’s try another substitution problem that’s a little bit different:

## Solving Systems with Linear Combination or Elimination

Probably the most useful way to solve systems is using linear combination, or linear elimination. The reason it’s most useful is that usually in real life we don’t have one variable in terms of another (in other words, a “$ y=$” situation).

The main purpose of the linear combination method is to add or subtract the equations so that one variable is eliminated. We can add, subtract, or multiply both sides of equations by the same numbers – let’s use real numbers as shown below. We are using the **Additive Property of Equality**, **Subtraction Property of Equality**, **Multiplicative Property of Equality**, and/or **Division Property of Equality** that we saw here in the **Types of Numbers and Algebraic Properties** section:

If we have a set of **2** equations with **2** unknowns, for example, we can manipulate them by adding, multiplying or subtracting (we usually prefer adding) so that we get one equation with one variable. Let’s use our previous problem:

We could buy **4 pairs of jeans and 2 dresses**.

## Types of equations

In the example above, we found **one unique solution** to the set of equations. Sometimes, however, for a set of equations, there are **no solutions** (when lines are parallel) or an **infinite number of solutions** or** infinitely many solutions** (when the two lines are actually the same line, and one is just a “multiple” of the other).

When there is **at least one solution**, the equations are **consistent equations**, since they have a solution. When there is only one solution, the system is called **independent**, since they cross at only one point. When equations have **infinite** solutions, they are the same equation, are **consistent**, and are called **dependent** or **coincident** (think of one just sitting on top of the other).

When equations have **no solutions**, they are called **inconsistent equations**, since** **we can never get a solution**. **

Here are graphs of inconsistent and dependent equations that were created on a graphing calculator:

## Systems with Three Equations

Let’s get a little more complicated with systems; in real life, we rarely just have two unknowns to solve for.

Let’s say at the same store, they also had pairs of shoes for **$20** and we managed to get **$60** more to spend! Now we have a new problem. To spend the even **$260**, how many pairs of jeans, dresses, and pairs of shoes should we get if want, for example, exactly **10** total items (Remember that jeans cost **$25** each and dresses cost **$50** each).

Let’s let $ j=$ the number of pair of jeans, $ d=$ the number of dresses, and $ s=$ the number of pairs of shoes we should buy. So far, we’ll have the following equations:

$ \displaystyle \begin{array}{c}j+d+s=10\text{ }\\25j+\text{ }50d+\,20s=260\end{array}$

We’ll need another equation, since for three variables, we need three equations (otherwise, we theoretically might have infinite ways to solve the problem). In this type of problem, you would also need something like this: **We want twice as many pairs of jeans as pairs of shoes**. Now, since we have the **same number of equations as variables**, we can potentially get one solution for the system of equations. Here are the three equations:

We’ll learn later how to put these in our calculator to easily solve using **matrices **(see the **Matrices and Solving Systems with Matrices** section). For now, we can use two equations at a time to eliminate a variable (using substitution and/or elimination), and keep doing this until we’ve solved for all variables. These can get really difficult to solve, but remember that in “real life”, there are computers to do all this work!

Remember again, that if we ever get to a point where we end up with something like this, it means there are an **infinite number of solutions**: $ 4=4$ (variables are gone and a number equals another number and they are the same). And if we up with something like this, it means there are **no solutions**: $ 5=2$ (variables are gone and two numbers are left and they don’t equal each other).

And another note: equations with three variables are represented by **planes**, not lines (you’ll learn about this in Geometry). If all the planes crossed in only one point, there is **one solution**, and if, for example, any two were parallel, we’d have **no solution**, and if, for example, two or three of them crossed in a line, we’d have an **infinite number of solutions. **

Let’s solve our system: $ \displaystyle \begin{array}{c}j+d+s=10\text{ }\\25j+\text{ }50d+20s=260\\j=2s\end{array}$:

We could buy **6 pairs of jeans, 1 dress, and 3 pairs of shoes**.

Here’s one more example of a three-variable system of equations, where we’ll only use linear elimination:

$ \displaystyle \begin{align}5x-6y-\,7z\,&=\,7\\6x-4y+10z&=\,-34\\2x+4y-\,3z\,&=\,29\end{align}$

I know – this is really difficult stuff! But if you do it step-by-step and keep using the equations you need with the right variables, you can do it. Think of it like a puzzle – you may not know exactly where you’re going, but do what you can in baby steps, and you’ll get there (sort of like life sometimes, right?!). And we’ll learn much easier ways to do these types of problems.

## Algebra Word Problems with Systems

Let’s do more word problems; you’ll notice that many of these are the same type that we did earlier in the **Algebra Word Problems** section, but now we can use more than one variable. This will actually make the problems easier! Again, when doing these word problems:

**If you’re wondering what the variables (or unknowns) should be when working on a word problem, look at what the problem is asking. These are usually (but not always) what your variables are!****If you’re not sure how to set up the equations, use regular numbers (simple ones!) and see what you’re doing. Then put the variables back in!**

Here are some problems:

**Investment Word Problem**

**Mixture Word Problems**

Here’s a **mixture word problem**. With mixture problems, remember if the problem calls for a **pure solution or concentrate**, use **100% **(if the percentage is that solution) or **0% **(if the percentage is another solution).

Here’s another mixture problem:

**Distance Word Problem:**

Here’s a distance word problem using systems**; distance problems **have to do with an object’s speed, time, and distance. Note that, as well as the distance word problem here in the **Algebra Word Problems** section, there’s an example of a **Parametric Distance Problem** here in the **Parametric Equations** section.

**Which Plumber Problem**

Many word problems have to do with an initial charge or setup charge, and a charge or rate per time period. In these cases, the **initial charge will be the **$ \boldsymbol {y}$**-intercept**, and the **rate will be the slope**. Here is an example:

**Geometry Word Problem:**

Many times, we’ll have a geometry problem as an algebra word problem; these might involve perimeter, area, or sometimes angle measurements (so don’t forget these things!). Let’s do one involving angle measurements.

See – these are getting easier! Here’s one that’s a little tricky though:

**Work Problem****: **

Let’s do a “**work problem**” that is typically seen when studying **Rational Equations **(fraction with variables in them) and can be found here in the **Rational Functions, E****quations and**** Inequalities** section.

Note that there’s also a simpler version of this problem here in the **Direct, Inverse, Joint and Combined Variation** section.

**Three Variable Word Problem:**

Let’s do one more with three equations and three unknowns:

**The “Candy” Problem**

Sometimes we get lucky and can solve a system of equations where we have more unknowns (variables) then equations. (Actually, I think it’s not so much luck, but having good problem writers!) Here’s one like that:

There are more **Systems Word Problems** in the **Matrices and Solving Systems with Matrices** section, **Linear Programming **section, and **Right Triangle Trigonometry** section.

**Understand these problems, and practice, practice, practice!**

**For Practice**: Use the **Mathway** widget below to try a **Systems of Equations** problem. Click on **Submit** (the blue arrow to the right of the problem) and click on **Solve by Substitution** or **Solve by Addition/Elimination **to see the answer**.**

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on **Tap to view steps**, or **Click Here**, you can register at **Mathway** for a **free trial**, and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

On to **Scatter Plots, Correlation, and Regression **– you are ready!