# How do you solve 3x+6y=9 and 2x+8y=10 using substitution?

Mar 30, 2017

x=1; y=1

#### Explanation:

$\left[\text{Eqn 1"] 3x+6y=9; ["Eqn 2}\right] 2 x + 8 y = 10$

Multiply [Eqn 1] by 2; [Eqn 2] by 3, so $x$ can be in the same multiple:

$\left[\text{Eqn 3}\right] 6 x + 12 y = 18$

$\left[\text{Eqn 4}\right] 6 x + 24 y = 30$

Hence, we can find $y$ by $\text{[Eqn 4]"-"[Eqn 3]}$:
$6 x - 6 x + 24 y - 12 y = 30 - 18$
$12 y = 12$
$y = 1$

Now, just substitute $y$ into either [Eqn 1] or [2]:
For example, I will subst. into [Eqn 1].

$3 x + 6 \left(1\right) = 9$
$3 x = 9 - 6 = 3$
$x = 1$

Hence, the solution:
x=1; y=1