Question

solve the following system of equations. -3x + 5y = 19; 9x + 2y = -23; x = \square; y = \square

Explanation:

Step1: Eliminate x by multiplying first equation

Multiply the first equation \(-3x + 5y = 19\) by \(3\) to get \(-9x + 15y = 57\).

Step2: Add the two equations

Add the new equation \(-9x + 15y = 57\) and the second equation \(9x + 2y = -23\). So, \((-9x + 9x)+(15y + 2y)=57 + (-23)\), which simplifies to \(17y = 34\).

Step3: Solve for y

Divide both sides of \(17y = 34\) by \(17\): \(y=\frac{34}{17}=2\).

Step4: Substitute y into first equation

Substitute \(y = 2\) into \(-3x + 5y = 19\): \(-3x+5\times2 = 19\), so \(-3x + 10 = 19\).

Step5: Solve for x

Subtract \(10\) from both sides: \(-3x=19 - 10 = 9\). Then divide by \(-3\): \(x=\frac{9}{-3}=-3\).

Answer:

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