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**Solving Systems Of Linear Equations By Addition**: At a Glance

- Topics At a Glance
- Systems of Equations
**Systems of Linear Equations**- Solving Systems of Linear Equations by Graphing
- Solving Systems of Linear Equations by Substitution
**Solving Systems of Linear Equations by Addition**- Solving Linear Systems
- More Vocabulary
- Word Problems and Lines
- Solving Word Problems
- Word Problems with Two Lines
- More About Word Problems
- Translating a Word Problem into a System of Equations
- Solving Word Problems with Systems of Equations
- In the Real World
- I Like Abstract Stuff; Why Should I Care?
- How to Solve a Math Problem

When we have two equations, we can add them together to get a new equation. We don't even need to ask for their consent. If

*a* = *b*

and

*c* = *d*,

we can add the left-hand sides of the equations and the right-hand sides of the equations to get

*a* + *c* = *b* + *d*.

When we do this, we say we're **adding the equations**. Come to think of it, that term may be too self-explanatory for bolding. Oh well, too late now.

We can also more complicated equations. If

3*x* + 2*y* = 7

and

*y* = 3*x* – 2,

then

3*x* + 2*y* + *y* = 7 + 3*x* – 2.

We'll use this concept to solve systems of linear equations.

Solve the system of linear equations

We add the equations to find that

*y* + *x* – *y* = 3*x* + 2 + 4.

This simplifies to *x* = 3*x* + 6.

All of a sudden the variable *y* is gone. We don't know where it went, but hopefully it's in a better place. We now have an equation we can solve for *x*. Doing this yields *x* = -3. Now we can put -3 in for *x* in the first equation to find *y*:

3(-3) + 2 = -7.

The solution to the system appears to be (-3, -7). Let's check this answer in the original equations. We want to check it before we wreck it.

For the first equation, when *x* = -3 and *y* = -7, the left-hand side is -7 and the right-hand side is 3(-3) + 2, which also happens to be -7. These values work in the first equation.

For the second equation, when *x* = -3 and *y* = -7, the left-hand side of the equation is *x* -*y* = (-3) – (-4), which is 4, the same as the right-hand side of the equation.

These values work in both the equations, so the solution really is (-3, -7).

The previous example worked because the the coefficients of *y* in the two different equations were additive inverses of each other. The first equation had a positive *y*, and the second equation had a negative *y*.

When we added the equations, we **eliminated** the variable *y*. Now *y* sleeps with the fishes.

This example was one of the kindest ones we could have given you, because the coefficients of *y* in the two equations were already additive inverses. In general, we need to do much more legwork to get coefficients that are additive inverses. Once we're done with our legwork, we can focus on our glutes.

Solve the system of equations

The coefficients of *x* aren't additive inverses, but we can be sneaky and mold them into what we want them to be. Take the first equation and multiply each side by -2. This gives us

-2(*x* + 5*y* ) = -2(11),

which simplifies to

-2*x* -10*y* = -22.

Write this next to the second equation:

Now the coefficients of *x* are additive inverses of each other, so we can work our system-solving magic. Nothing up our sleeves here...

Add the equations to eliminate the *x* and find

-10*y*-3*y* = -18,

then solve to get

.

Now we can stick that value in for *y* in the first original equation to find

and solve to get

.

We think the solution to the system of equations is .

As always, we're going to cover our hineys and double-check this in the original equations. For the first equation, when and the left-hand side is

which is 11, exactly as it should be.

For the second equation, when and the left-hand side is

This agrees with the right-hand side of the second equation, so we've got it. Man, being right never gets old.

When we solve a system by addition, we could also say we're solving it by **elimination**. We eliminate one variable, find the value of the other, and then find the value of the variable we eliminated. It's like the NFL playoffs. These variables are one and done.

The general steps we're using here are similar to the ones we used with the substitution method.

- Eliminate a variable to get an equation in one variable.

- Solve the equation from (1).

- Use one of the original equations to find the value of the other variable.

- Check your answer in both original equations.

So far, the first step ("Eliminate a variable'') hasn't been too difficult. We've only needed to multiply one equation in the system by a number in order to eliminate a variable. Now it's time to up the ante. Of course, we don't mean "ante" in the poker sense. We can't condone gambling, unless it's on the stock market.

We frequently need to multiply *both* equations by something in order to eliminate a variable.

Eliminate the variable *x* from the system of equations to get an equation in *y*.

The coefficients on the *x* are 3 and 2, which have a common multiple of 6. If we multiply the first equation by 2 and the second equation by 3, we're almost there:

Now the coefficients on *x* in the two equations are the same, which is close, but no cigar. Again, we are not referring to a literal cigar. We can't condone smoking.

We want our coefficients to be additive inverses, and we can multiply either equation by -1. In other words, we need to flip all the signs in one equation. We could flip a coin to determine which one we should choose, but how do we decide which coin to flip?

We'll make things easy on ourselves and go chronologically. Emphasis on "logically." If we pick the first equation, we have

Now we add the equations to find -7*y* = 7, and the *x* has been eliminated.

There's one other thing that can happen: our linear system can contain fractions. When this happens, we just get rid of the fractions first and then solve the system as we've been doing. If they don't go quietly, don't be afraid to get a bit rough. Show those fractions who's boss.

Solve the system of equations.

First, we get rid of the fractions. Multiply the first equation through by 6 and the second equation through by 5 to find

We know what to do from here. In this case, we can eliminate either variable without too much trouble, assuming they don't get frisky. Let's eliminate *y*, since its coefficients already have opposite signs. Thank you, *y*, for cooperating. Perhaps the judge will reward it during sentencing.

Multiply the second equation by 2, so now we're looking at the system of equations

We add these equations to find 33*x* = 26,

which means, unfortunately, that

Use the first original equation to find *y*. We plug in our value of *x* into

,

which gives us

.

This simplifies to

,

which means

and so

Now we think the solution to the system is , but we need to check our answer in the other equation, which was

Let's plug in the values we found for *x* and *y* and see if they work. If they don't, we'll try to keep from throwing a hissy fit. Look at things through a clean pair of perspectacles.

which is indeed 2. These values work, so we found the solution to the system of equations.

Example 1

If we want to solve this system by eliminating the variable |

Example 2

For the following system of equations, eliminate the variable |

Exercise 1

We can solve the following system by eliminating the variable *x*. However, we first need to multiply one equation in each system by a number. Which equation, and what number?

Exercise 2

We can solve the following system by eliminating the variable *x*. However, we first need to multiply one equation in each system by a number. Which equation, and what number?

Exercise 3

We can solve the following system by eliminating the variable *x*. However, we first need to multiply one equation in each system by a number. Which equation, and what number?

Exercise 4

Use addition to solve the following system of equations. Warning: there may be fractions.

Exercise 5

Use addition to solve the following system of equations. Warning: there may be fractions.

Exercise 6

Use addition to solve the following system of equations. Warning: there may be fractions.

Exercise 7

Use addition to solve the following system of equations. Warning: there may be fractions.

Exercise 8

Use addition to solve the following system of equations. Warning: there may be fractions.

Exercise 9

Use addition, i.e. elimination, to solve the following system of equations. Remember, it's possible for a system of equations to have no solutions or infinitely many solutions. If it has infinitely many solutions, you don't need to write them all down. We'll let you off the hook with that one.

Exercise 10

Use addition, i.e. elimination, to solve the following system of equations. Remember, it's possible for a system of equations to have no solutions or infinitely many solutions. If it has infinitely many solutions, you don't need to write them all down. We'll let you off the hook with that one.

Exercise 11

Use addition, i.e. elimination, to solve the following system of equations. Remember, it's possible for a system of equations to have no solutions or infinitely many solutions. If it has infinitely many solutions, you don't need to write them all down. We'll let you off the hook with that one.

Exercise 12

Exercise 13