Question

solve the following system of equations.
$9x - 8y = -25$
$-3x - 4y = -5$
$x = \square$
$y = \square$

Explanation:

Step1: Multiply the second equation by 2

We multiply the equation \(-3x - 4y = -5\) by 2 to make the coefficients of \(y\) in both equations opposites (or easier to eliminate).
\(2\times(-3x - 4y)=2\times(-5)\)
\(-6x - 8y = -10\)

Step2: Subtract the new second equation from the first equation

Now we have the first equation \(9x - 8y = -25\) and the new second equation \(-6x - 8y = -10\). Subtract the second from the first:
\((9x - 8y)-(-6x - 8y)=-25-(-10)\)
\(9x - 8y + 6x + 8y=-25 + 10\)
Combine like terms: \(15x=-15\)

Step3: Solve for \(x\)

Divide both sides of \(15x = -15\) by 15:
\(x=\frac{-15}{15}=-1\)

Step4: Substitute \(x = -1\) into one of the original equations to solve for \(y\)

Let's use the second original equation \(-3x - 4y = -5\). Substitute \(x=-1\):
\(-3\times(-1)-4y=-5\)
\(3 - 4y=-5\)
Subtract 3 from both sides: \(-4y=-5 - 3=-8\)
Divide both sides by -4: \(y=\frac{-8}{-4}=2\)

Answer:

\(x = -1\)
\(y = 2\)