Question

medium
solve each system of equations by substitution.
$y = 4x - 1$
$y = -2x - 1$
please do not write on these. they will be used in other classes.

Explanation:

Step1: Substitute \( y \) from first equation into second

We have the system:
\( y = 4x - 1 \)
\( y = -2x - 1 \)

Substitute \( y = 4x - 1 \) into \( y = -2x - 1 \):
\( 4x - 1 = -2x - 1 \)

Step2: Solve for \( x \)

Add \( 2x \) to both sides:
\( 4x + 2x - 1 = -1 \)
\( 6x - 1 = -1 \)

Add 1 to both sides:
\( 6x = -1 + 1 \)
\( 6x = 0 \)

Divide both sides by 6:
\( x = \frac{0}{6} = 0 \)

Step3: Solve for \( y \)

Substitute \( x = 0 \) into \( y = 4x - 1 \):
\( y = 4(0) - 1 = -1 \)

Answer:

The solution to the system is \( x = 0 \), \( y = -1 \) (or as an ordered pair \( (0, -1) \))