Question

elimination with addition and subtraction
question 3 of 15 (1 point) | question attempt: 1 of 3

solve the following system of equations.
$5x + 6y = -5$
$5x + 3y = 10$

$x = \square$
$y = \square$

Explanation:

Step1: Subtract the two equations

We have the system:

$$\begin{cases} 5x + 6y = -5 \\ 5x + 3y = 10 \end{cases}$$

Subtract the second equation from the first to eliminate \(x\):
\((5x + 6y) - (5x + 3y) = -5 - 10\)
Simplify: \(5x + 6y - 5x - 3y = -15\)
Which gives: \(3y = -15\)

Step2: Solve for \(y\)

Divide both sides of \(3y = -15\) by 3:
\(y = \frac{-15}{3} = -5\)

Step3: Substitute \(y\) into an equation

Substitute \(y = -5\) into the second equation \(5x + 3y = 10\):
\(5x + 3(-5) = 10\)
Simplify: \(5x - 15 = 10\)
Add 15 to both sides: \(5x = 10 + 15 = 25\)

Step4: Solve for \(x\)

Divide both sides of \(5x = 25\) by 5:
\(x = \frac{25}{5} = 5\)

Answer:

\(x = 5\), \(y = -5\)