# How do you solve 3x - 7y = 27 and - 5x + 4y = - 45 using substitution?

May 7, 2018

The solution is $\left(9 , 0\right)$.

#### Explanation:

Solve the system of equations:

$\text{Equation 1} :$ $3 x - 7 y = 27$

$\text{Equation 2} :$ $- 5 x + 4 y = - 45$

These are two linear equations in standard form. The solution to the system is the point they have in common, which is the point where they intersect.

Solve Equation 1 for $x$.

$3 x - 7 y = 27$

Add $7 y$ to both sides of the equation.

$3 x = 7 y + 27$

Divide both sides by $3$.

$x = \frac{7}{3} y + \frac{27}{3}$

$x = \frac{7}{3} y + 9$

Substitute $\frac{7}{3} y + 9$ for $x$ in Equation 2 and solve for $y$.

$- 5 x + 4 y = - 45$

$- 5 \left(\frac{7}{3} y + 9\right) + 4 y = - 45$

Expand.

$- \frac{35}{3} y - 45 + 4 y = - 45$

Simplify $- \frac{35}{3} y$ to $\frac{- 35 y}{3}$.

Add $45$ to both sides.

$\frac{- 35 y}{3} + 4 y = - 45 + 45$

$\frac{- 35 y}{3} + 4 y = 0$

Multiply $4 y$ by $\frac{3}{3}$ to get an equivalent fraction with $3$ as the denominator.

$\frac{- 35 y}{3} + 4 y \times \frac{3}{3} = 0$

$\frac{- 35 y}{3} + \frac{12 y}{3} = 0$

Simplify.

$\frac{- 23 y}{3} = 0$

Multiply both sides by $3$.

$- 23 y = 0 \times 3$

$- 23 y = 0$

Divide both sides by $- 23$.

$y = \frac{0}{- 23}$

$y = 0$

Substitute $0$ for $y$ in Equation 1 and solve for $x$.

$3 x - 7 y = 27$

$3 x - 7 \left(0\right) = 27$

$3 x = 27$

Divide both sides by $3$.

$x = \frac{27}{3}$

$x = 9$

The solution is $\left(9 , 0\right)$.

graph{(3x-7y-27)(-5x+4y+45)=0 [-10, 10, -5, 5]}