How do you solve the system 2x-3y=12 and x=4y+12x−3y=12andx=4y+1?

1 Answer
Camilleon
Jun 2, 2018

x = 9x=9
y = 2y=2

Explanation:

2x - 3y = 122x−3y=12
x = 4y + 1x=4y+1

Solving by Substitution

First, we're going to use an equation for the value of a variable in order to plug it into the opposite equation of the system. Because x = 4y + 1x=4y+1 is already an equation for the value of a variable, we'll be using it. In the other equation of the system, plug in xx's value where xx is. So:

2(4y + 1) - 3y = 122(4y+1)−3y=12

Next, you'll be distributing. What this means is that you'll be multiplying the outside number, 22, by the terms in the parentheses, 4y4y and 11. So:

2 * 4y = 8y2⋅4y=8y
2 * 1 = 22⋅1=2

Re-write your equation.

8y + 2 - 3y = 128y+2−3y=12

Combine like terms. 8y - 3y = 5y8y−3y=5y, so:

5y + 2 = 125y+2=12

This is a two-step equation. To solve it, subtract 2 from both sides to isolate for yy. You should now have:

5y = 105y=10

Divide by 55 to isolate for yy:

y = 2y=2

Plug the value of yy back into the equation for the value of xx:

x = 4y + 1x=4y+1
x = 4(2) + 1x=4(2)+1
x = 8 + 1x=8+1
x = 9x=9

To truly prove that xx is 9 and yy is 2:

2x - 3y = 122x−3y=12
2(9) - 3(2) = 122(9)−3(2)=12
18 - 6 = 1218−6=12
12 = 1212=12

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