Question

solve the system of equations using elimination: $4x + 3y = -19$ and $3x + 3y = -15$.

Explanation:

Step1: Subtract the two equations

We have the system:

$$\begin{cases} 4x + 3y = -19 \\ 3x + 3y = -15 \end{cases}$$

Subtract the second equation from the first to eliminate \(y\):
\((4x + 3y)-(3x + 3y)=-19-(-15)\)
Simplify left side: \(4x + 3y - 3x - 3y=x\)
Simplify right side: \(-19 + 15=-4\)
So \(x = - 4\)

Step2: Substitute x into an equation

Substitute \(x=-4\) into the second equation \(3x + 3y=-15\):
\(3\times(-4)+3y=-15\)
Calculate \(3\times(-4)=-12\), so \(-12 + 3y=-15\)
Add 12 to both sides: \(3y=-15 + 12=-3\)
Divide both sides by 3: \(y=\frac{-3}{3}=-1\)

Answer:

The solution to the system is \(x = - 4\) and \(y=-1\) (or as an ordered pair \((-4,-1)\))