Question

Solve by substitute
{3x+2y=18
{3x+8y=36

Answer 1

\[3x+2y=18\\3x=18-2y\\x=\frac{18-2y}{3}\\x=\frac{18}3-\frac{2y}{3}\\x=6-\frac{2y}{3}\]plug this in for the second
\[3(6-\frac{2y}3)+8y=36\]
solve for y, then plug that in for y in the first eqn to find x.

Answer 2

So y=3?

Answer 3

\[3x+2y=18\]
\[2y=18-3x\]
divide by 2
\[y=9-1.5x\]

Now take the second equation and substitute all values of y
\[3x+8y=36\]
\[3x+8(9-1.5x)=36\]
\[3x+72-12x=36\]
Collect Like terms
\[72-36=12x-3x\]
\[36=9x\]
Simplify
\[x=4 \]

Go back to the equation where you found y
\[y=9-1.5x\]
You have x, so you can use it to find y
\[y=9-1.5(4)\]
\[y=9-6\]
\[y=3 \]

Answer 4

The easiest way to solve it is by using elimination, however, the best way to substitute is to isolate 3x in both equations:

3x = 18 - 2y
3x = 36 - 8y

3x = 3x
18 - 2y = 36 - 8y

8y - 2y = 36 - 18
6y = 18
y = 3