Question

find the solution of the system of equations.\

$$\begin{cases}5x + 4y = -19\\\\5x - 3y = -12\\end{cases}$$

\
show your work here\
x =\
y =

Explanation:

Step1: Subtract the two equations

We have the system:

$$\begin{cases} 5x + 4y = -19 \quad (1)\\ 5x - 3y = -12 \quad (2) \end{cases}$$

Subtract equation (2) from equation (1) to eliminate \(x\):
$$(5x + 4y)-(5x - 3y)=-19-(-12)$$
Simplify the left - hand side: \(5x + 4y - 5x+3y = 7y\)
Simplify the right - hand side: \(-19 + 12=-7\)
So we get the equation \(7y=-7\)

Step2: Solve for \(y\)

From \(7y=-7\), divide both sides by 7:
\(y=\frac{-7}{7}=-1\)

Step3: Substitute \(y = - 1\) into one of the original equations

Substitute \(y=-1\) into equation (2): \(5x-3\times(-1)=-12\)
Simplify the left - hand side: \(5x + 3=-12\)
Subtract 3 from both sides: \(5x=-12 - 3=-15\)
Divide both sides by 5: \(x=\frac{-15}{5}=-3\)

Answer:

\(x = - 3\), \(y=-1\)