# Algebra Techniques: Solving Linear Systems by Elimination

The **substitution method** can be used to solve systems of equations in several variables, but if the equations happen to be linear, you can use a second technique known as **elimination**. With elimination, you multiply a pair of equations by two factors such that when you add the equations together, one of the variables cancels out.

After you eliminate one of the variables, you can either proceed with further elimination to solve for the other variable(s), or use substitution to solve for the remaining value(s). Or both!

## Step 1: How to Eliminate the First Variable

Take the linear system **3x - 4y = 1000** and **30x + y = 260** as our example. You can eliminate either the x or y first, it does not matter.

For example, to eliminate the x, multiply the first equation by -10 to obtain **-30x + 40y = -10000**. When you add this to the second equation, you get 0x + 41y = -9740, or 41y = -9740. Divide both sides by 41 to get** y = -237.560976**.

## Step 2a: Further Elimination

Now that you have the value for y, you can solve for x with either elimination or substitution. To find x by elimination, we must cancel out the y from the two equations.

To do this, multiply the second equation by 4 to obtain **120x + 4y = 1040**. Now add this to the first equation to obtain 123x + 0y = 2040, or **123x = 2040**. To solve for x, divide both sides by 123. This gives you **x = 16.585366.**

## Step 2b: Substitution

Alternatively, you can solve for x by plugging y = -237.560976 into one of the equations. For example, if you substitute this value into the second equation, you get

30x - 237.560976 = 260

30x = 497.560976

**x = 16.585366**

## Another Example

An event coordinator sets up 480 chairs in a rectangular pattern with x rows and y columns. She notices that if she decreases the number of rows by 2 and increases the number of columns by 3, she can seat 506 people. But if she increases the number of rows by 3 and decreases the number of columns by 2, she can only seat 486. What were the original number of rows and columns?

For this algebra word problem, we have the number of original rows = x and the number of original columns = y. This gives us the equations

xy = 480

(x - 2)(y + 3) = 506

(x + 3)(y - 2) = 486

The last pair of equations are equivalent to the pair

xy - 2y + 3x = 512

xy + 3y - 2x = 492

But since we know that xy = 480 from the first equation, we can simplify this system to

-2y + 3x = 512 - 480 = 32

3y - 2x = 492 - 480 = 12

So our reduced system of linear equations is

-2y + 3x = 32

3y - 2x = 12

Solving the first equation for y gives us y = 1.5x - 16. Plugging this into the second equation and solving for x gives us

3(1.5x - 16) - 2x = 12

4.5x - 48 - 2x = 12

2.5x = 60

x = 24.

Therefore, x = 24 and y = 480/24 = 20.

## What Others Are Reading

## Real-World Example

The cost of tickets for a show is $4 for students and members and $6 for the general public. If the organizers sold 155 tickets and brought in $728, how many members of the general public bought tickets?

Let S be the number of students and members, and let G be the number of general public audience members. The first relation between S and G is

S + G = 155

The second relation between S and G comes from the sales figures,

4S + 6G = 728

The first relation can be rewritten as 4S + 4G = 620, that is, you multiply each side by 4. If you subtract this equation from the second you get

(4S + 6G) - (4S + 4G) = 728 - 620

2G = 108

G = 54

Therefore, the organizers sold 54 tickets to people who were not students or members.

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## Comments 4 comments

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