Question

kuta software - infinite algebra 1
solving systems of equations by substitution
solve each system by substitution.

1. ( y = 6x - 11 )

( -2x - 3y = -7 )

1. ( 2x - 3y = -1 )

( y = x - 1 )

1. ( y = -3x + 5 )

( 5x - 4y = -3 )

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

( y = -5x - 17 )

1. ( y = -2 )

( 4x - 3y = 18 )

1. ( y = 5x - 7 )

( -3x - 2y = -12 )

1. ( -4x + y = 6 )

( -5x - y = 21 )

1. ( -7x - 2y = -13 )

( x - 2y = 11 )

1. ( -5x + y = -2 )

( -3x + 6y = -12 )

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

( 3x - 8y = 24 )

Explanation:

Let's solve the first system of equations (problem 1) as an example:

We have the system:

1. \( y = 6x - 11 \)
2. \( -2x - 3y = -7 \)

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

We know \( y = 6x - 11 \), so replace \( y \) in \( -2x - 3y = -7 \) with \( 6x - 11 \).

$$ -2x - 3(6x - 11) = -7 $$

Step 2: Simplify and solve for \( x \).

First, distribute the -3:

$$ -2x - 18x + 33 = -7 $$

Combine like terms:

$$ -20x + 33 = -7 $$

Subtract 33 from both sides:

$$ -20x = -7 - 33 $$

$$ -20x = -40 $$

Divide both sides by -20:

$$ x = \frac{-40}{-20} = 2 $$

Step 3: Substitute \( x = 2 \) back into the first equation to find \( y \).

Using \( y = 6x - 11 \):

$$ y = 6(2) - 11 $$

$$ y = 12 - 11 = 1 $$

Answer:

The solution to the system is \( x = 2 \), \( y = 1 \) (or the ordered pair \( (2, 1) \)).