Question

find the solution of the system of equations.
$9x + 2y = -38$
$9x + 10y = 26$

Explanation:

Step1: Subtract the first equation from the second equation to eliminate \(x\).

The two equations are:
\(9x + 2y = -38\) (Equation 1)
\(9x + 10y = 26\) (Equation 2)

Subtract Equation 1 from Equation 2:
\((9x + 10y) - (9x + 2y) = 26 - (-38)\)
Simplify the left side: \(9x + 10y - 9x - 2y = 8y\)
Simplify the right side: \(26 + 38 = 64\)
So we get \(8y = 64\)

Step2: Solve for \(y\).

Divide both sides of \(8y = 64\) by 8:
\(y = \frac{64}{8} = 8\)

Step3: Substitute \(y = 8\) into one of the original equations to solve for \(x\). Let's use Equation 1: \(9x + 2y = -38\)

Substitute \(y = 8\) into Equation 1:
\(9x + 2(8) = -38\)
Simplify: \(9x + 16 = -38\)
Subtract 16 from both sides: \(9x = -38 - 16 = -54\)
Divide both sides by 9: \(x = \frac{-54}{9} = -6\)

Answer:

The solution of the system of equations is \(x = -6\) and \(y = 8\), or in ordered pair form \((-6, 8)\)