QUESTION IMAGE
Question
solve the following system of equations.
$5x + 6y = -5$
$5x + 3y = 10$
$x = \square$
$y = \square$
Step1: Subtract the two equations
We have the system:
$$\begin{cases}
5x + 6y = -5 \quad (1)\\
5x + 3y = 10 \quad (2)
\end{cases}$$
Subtract equation (2) from equation (1) to eliminate \(x\):
$$(5x + 6y)-(5x + 3y)=-5 - 10$$
Simplify the left side: \(5x+6y - 5x - 3y=3y\)
Simplify the right side: \(-15\)
So we get \(3y=-15\)
Step2: Solve for \(y\)
Divide both sides of \(3y = - 15\) by 3:
$$y=\frac{-15}{3}=-5$$
Step3: Substitute \(y\) into equation (2)
Substitute \(y = - 5\) into \(5x+3y = 10\):
$$5x+3\times(-5)=10$$
Simplify: \(5x-15 = 10\)
Step4: Solve for \(x\)
Add 15 to both sides: \(5x=10 + 15=25\)
Divide both sides by 5: \(x=\frac{25}{5}=5\)
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\(x = 5\), \(y=-5\)