Question

6.5a · solving systems of equations by substitution
solve each system of equation by substitution.

1. ( x = y + 7 )

( x + y = 9 )

1. ( 2x - y = 4 )

( y = 2x - 4 )

1. ( x = -y + 5 )

( x - 4y = 10 )

1. ( y = -6x + 4 )

( 2x + 2y = -12 )

1. ( 5x - 3y = -11 )

( x = 2y + 2 )

1. ( y = 3x - 5 )

( 4x - y = 3 )

Explanation:

Problem 1

Step1: Substitute \( x = y + 7 \) into \( x + y = 9 \)

\( (y + 7) + y = 9 \)

Step2: Simplify and solve for \( y \)

\( 2y + 7 = 9 \)
\( 2y = 9 - 7 \)
\( 2y = 2 \)
\( y = 1 \)

Step3: Substitute \( y = 1 \) into \( x = y + 7 \)

\( x = 1 + 7 \)
\( x = 8 \)

Step1: Substitute \( y = 2x - 4 \) into \( 2x - y = 4 \)

\( 2x - (2x - 4) = 4 \)

Step2: Simplify

\( 2x - 2x + 4 = 4 \)
\( 4 = 4 \) (This is an identity, so there are infinitely many solutions. The system is dependent.)

Step1: Substitute \( x = -y + 5 \) into \( x - 4y = 10 \)

\( (-y + 5) - 4y = 10 \)

Step2: Simplify and solve for \( y \)

\( -5y + 5 = 10 \)
\( -5y = 10 - 5 \)
\( -5y = 5 \)
\( y = -1 \)

Step3: Substitute \( y = -1 \) into \( x = -y + 5 \)

\( x = -(-1) + 5 \)
\( x = 1 + 5 \)
\( x = 6 \)

Answer:

\( x = 8 \), \( y = 1 \)

Problem 2