How do you solve 3x+ 4y=27 and -8x+ y=33 using substitution?

Apr 6, 2018

The solution is $\left(- 3 , 9\right)$.

Explanation:

Solve the system:

$\text{Equation 1} :$ $3 x + 4 y = 27$

$\text{Equation 2} :$ $- 8 x + y = 33$

Both equations are linear equations in standard form. The solution to the system is the point that the two lines have in common. The $x$ and $y$ coordinates will be determined by substitution.

Solve Equation 1 for $x$.

$3 x + 4 y = 27$

$3 x = 27 - 4 y$

Divide both sides by $3$.

$x = \frac{27}{3} - \frac{4 y}{3}$

Simplify.

$x = 9 - \frac{4 y}{3}$

Substitute $9 - 4 \frac{y}{3}$ for $x$ in Equation 2. Solve for $y$.

$- 8 x + y = 33$

$- 8 \left(9 - \frac{4 y}{3}\right) + y = 33$

$- 72 + \frac{32 y}{3} + y = 33$

Multiply $y$ by $\frac{3}{3}$ to create an equivalent fraction with $3$ in the denominator.

$- 72 + \frac{32 y}{3} + y \times \frac{3}{3} = 33$

$- 72 + \frac{32 y}{3} + \frac{3 y}{3} = 33$

$- 72 + \frac{35 y}{3} = 33$

Add $72$ to both sides.

$\frac{35 y}{3} = 33 + 72$

Simplify.

$\frac{35 y}{3} = 105$

Multiply both sides by $3$.

$35 y = 105 \times 3$

$35 y = 315$

Divide both sides by $35$

$y = \frac{315}{35}$

$y = 9$

Substitute $9$ for $y$ in Equation 1. Solve for $x$.

$3 x + 4 \left(9\right) = 27$

$3 x + 36 = 27$

$3 x = 27 - 36$

$3 x = - 9$

$x = \frac{- 9}{3}$

$x = - 3$

The solution is the point of intersection for the two lines, which is $\left(- 3 , 9\right)$.

graph{(3x+4y-27)(y-8x-33)=0 [-14.24, 11.07, 1.37, 14.03]}