Question

solve the following system of equations.
$4x - 3y = -17$
$9x + 8y = 6$
$x = \square$
$y = \square$

Explanation:

Step1: Eliminate a variable (e.g., y)

Multiply the first equation \(4x - 3y = -17\) by 8 and the second equation \(9x + 8y = 6\) by 3 to make the coefficients of y opposite.
First equation: \(8\times(4x - 3y)=8\times(-17)\) → \(32x - 24y = -136\)
Second equation: \(3\times(9x + 8y)=3\times6\) → \(27x + 24y = 18\)

Step2: Add the two new equations

Add \(32x - 24y = -136\) and \(27x + 24y = 18\) to eliminate y:
\((32x + 27x)+(-24y + 24y)=-136 + 18\)
\(59x = -118\)

Step3: Solve for x

Divide both sides by 59:
\(x=\frac{-118}{59}=-2\)

Step4: Substitute x into one of the original equations (e.g., the second one)

Substitute \(x = -2\) into \(9x + 8y = 6\):
\(9\times(-2)+8y = 6\)
\(-18 + 8y = 6\)

Step5: Solve for y

Add 18 to both sides: \(8y = 6 + 18 = 24\)
Divide by 8: \(y=\frac{24}{8}=3\)

Answer:

\(x = -2\)
\(y = 3\)