## Pre-Algebra Tutorial

*Multiple Linear Equations*

#### Intro

This particular tutorial will demonstrate how one can find multiple linear equations. If you believe you need a better understanding of single linear equations, please view my 3 part tutorial (Graph proportional relationship). We will first use an example before graphing. You will be given 2 equations: 1) **x+2y=1 ** 2) **2x+4y=2**

Before we begin, I must note that the number in front of a variable such as x or y is known as the coefficient. In equation 1, y has a coefficient of 2. In equation 2, x has a coefficient of 2 and y has a coefficient of 4.

Example 1: **Confirm the equations**

Use the first equation…

**x+2y=1 ** We must first get x by itself by subtracting 2y from each side…

**x=1-2y ** Now we know what x equals!

To solve the actual equation, we have to substitute this into the second equation…

**2x+4y=2** Now substitute…

**2(1-2y)+4y=2** And now we confirm if the equation is true…

**2-4y+4y=2 **

**2=2 True**

Now, there are some other tools we can use to help solve different equations-

Example 2: **Solve for x**

Equation 1) **3x-4y=10** Equation 2) **2x-4y=4**

We can actually take the second equation and subtract from the first to eliminate y:

**3x-2x-4y-(-4y)=10-4 **

Remember to put like terms next to each other to make the equation cleaner:

**x=6**

Note: **-4y-(-4y)** is the same as **-4y+4y**, therefore they will eliminate each other

Sometimes there are equations that aren’t quite as simple-

Example 3: **Solve for y**

Equation 1) **2x+4y=22** Equation 2) **4x+y=16**

Since neither of the equations have variables with the same coefficient, we can actually use multiplication to manipulate the equation in a more useful format:

**2(2x+4y)=(22)2** Multiply each side by 2 to get…

**4x+8y=44** Now each equation has 4x and we can subtract equation 2 from equation 1…

**4x-4x+8y-y=44-16** Notice we are eliminating x and solving for y…

**7y=28** Now simplify

**y=4 **

To review, example 1 is solving through substitution and examples 2 and 3 use elimination. They are both useful depending on the situation and your preference.

Example 4: **Solve for both x and y**

Equation 1) **2x-3y=12** Equation 2) **3x+4y=1**

Notice that only multiplying one equation will not work. Therefore we will multiply equation 1 by 3 and then multiply equation 2 by 2…

Step 1: 3(2x-3y)=3(12) –> **6x-9y=36** new equation 1

Step 2: 2(3x+4y)=2(1) –> **6x+8y=2** new equation 2

Subtract equation 2 from equation 1 to eliminate x and solve for y

Step 3: 6x-6x-9y-8y=36-2 –> -17y=34 –> **y=-2**

We have found the value for y. Take this value and use it in any equation to solve for x…

Step 4: 2x-3y=12 –> 2x-3(-2)=12 –> 2x+6=12 –> 2x=6 –> **x=3**

Not done yet! We have to confirm our answers by substituting in both the values we have found for x and y into one of the equations…

Step 5: 2x-3y=12 –> 2(3)-3(-2)=12 –> 6+6=12 –> **12=12 TRUE**

Finally, we can use this visual to see what these 2 equations really represent

When we find the x and y for the equations, we are actually finding the point at which they intersect. From there, we can substitute any coordinates in to find slope, y-intercepts, or confirm if a point is on the line.

Slope of 2y+x=8 using coordinates (2,3) and (8,0)

m=(0-3)/(8-2) –> m=(-3)/6 –> **m=-2**

#### Sample Problem

1. Solve for x and y using your preferred method and confirm it

2. Find the slope of each equation

Equation 1) **3y+7x=45**

Equation 2) **y=x-5**

#### Solution

1. Step 1: Notice that equation 2 is in a great format and can be placed directly into equation 1

3y+7x=45 –> 3(x-5)+7x=45 –> 3x-15+7x=45 –> 10x+60 –> **x=6**

Step 2: Now that we know x, lets find y…

y=x-5 –> y=(6)-5 –> **y=1**

Step 3: Confirm by substituting

3y+7x=45 –> 3(1)+7(6)=45 –> 3+42=45 –> **45=45 TRUE**

y=x-5 –> (1)=(6)-5 –> **1=1 TRUE**

2. Step 1: We need to find other coordinates for each equation to use in the slope formula

Equation 1) **3y+7x=45** –> Lets use x=0 –> 3y+7(0)=45 –> y=15

Use the points (6,1) and (0,15) to find the slope

m=(15-1)/(0-6) –> m=14/-6 –> **m=-7/3**

Equation 2) **y=x-5** –> lets use x=5 –> y=(5)-5 –> y=0

Use the points (6,1) and (5,0) to find the slope

m=(0-1)/(5-6) –> m=-1/-1 –> **m=1**

We can confirm that one line is positive slope and the other is negative slope. We can also confirm that they intersect at (6,1).

# About The Author

Math, Financing, And Accounting |

I am a recent graduate from Kent State University with an Associate of Applied Business and a Bachelor of Business Administration. During my studies, I excelled at math, financing, and accounting and was offered tutoring jobs at the university. Unfortunately, I had to decline the jobs at the time du... |