How do you solve the system $2 x - 3 y = 12$ $2x-3y=12$ and $x = 4 y + 1$ $x=4y+1$ by substitution?

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Answer 1 · 2015-05-21T03:35:40

The answer is $x = 9$ $x=9$ and $y = 2$ $y=2$ .

Problem: Solve the system $2 x - 3 y = 12$ $2x-3y=12$ and $x = 4 y + 1$ $x=4y+1$ .

$x = 4 y + 1$ $x=4y+1$ is already solved for $x$ $x$ . Substitute it into the other equation: $2 x - 3 y = 12$ $2x-3y=12$ .

$2 (4 y + 1) - 3 y = 12$ $2(4y+1)-3y=12$

Distribute the $2$ $2$ .

$8 y + 2 - 3 y = 12$ $8y+2-3y=12$

Simplify.

$5 y + 2 = 12$ $5y+2=12$ =

$5 y = 10$ $5y=10$

Divide both sides by $5$ $5$ .

$y = 2$ $y=2$

Substitute the $2$ $2$ for the $y$ $y$ into one of the original equations.

$x = 4 \cdot 2 + 1 = 9$ $x=4*2+1=9$

$x = 9$ $x=9$

Substitute the values for $x$ $x$ and $y$ $y$ to check the answers.

$2 x - 3 y = 12$ $2x-3y=12$

$2 \cdot 9 - 3 \cdot 2 = 12$ $2*9-3*2=12$

$18 - 6 = 12$ $18-6=12$

$12 = 12$ $12=12$

$x = 4 y + 1$ $x=4y+1$

$9 = 4 \cdot 2 + 1$ $9=4*2+1$ =

$9 = 9$ $9=9$

Both equations check!

How do you solve the system 2x-3y=122x−3y=122x-3y=12 and x=4y+1x=4y+1x=4y+1 by substitution?