Question

solve each system of equations by substitution.

1. ( y = 6x - 2 )

( y = 4x + 10 )

1. ( x = -2y + 3 )

( x = -5y - 9 )

Explanation:

Problem 3: Solve the system \(

$$\begin{cases} y = 6x - 2 \\ y = 4x + 10 \end{cases}$$

\) by substitution

Step 1: Substitute \( y \) from first equation into second

Since both equations equal \( y \), set them equal:
\( 6x - 2 = 4x + 10 \)

Step 2: Solve for \( x \)

Subtract \( 4x \) from both sides:
\( 6x - 4x - 2 = 10 \)
\( 2x - 2 = 10 \)

Add 2 to both sides:
\( 2x = 10 + 2 \)
\( 2x = 12 \)

Divide by 2:
\( x = \frac{12}{2} = 6 \)

Step 3: Substitute \( x = 6 \) into \( y = 6x - 2 \)

\( y = 6(6) - 2 \)
\( y = 36 - 2 = 34 \)

Problem 4: Solve the system \(

$$\begin{cases} x = -2y + 3 \\ x = -5y - 9 \end{cases}$$

\) by substitution

Step 1: Substitute \( x \) from first equation into second

Set \( -2y + 3 = -5y - 9 \)

Step 2: Solve for \( y \)

Add \( 5y \) to both sides:
\( -2y + 5y + 3 = -9 \)
\( 3y + 3 = -9 \)

Subtract 3 from both sides:
\( 3y = -9 - 3 \)
\( 3y = -12 \)

Divide by 3:
\( y = \frac{-12}{3} = -4 \)

Step 3: Substitute \( y = -4 \) into \( x = -2y + 3 \)

\( x = -2(-4) + 3 \)
\( x = 8 + 3 = 11 \)

Answer:

s:

• Problem 3: \( x = 6 \), \( y = 34 \) (or the ordered pair \( (6, 34) \))
• Problem 4: \( x = 11 \), \( y = -4 \) (or the ordered pair \( (11, -4) \))