Question

substitution - practice
examples 1 - 3
use substitution to solve each system of equations.

1. $y = 5x + 1$

$4x + y = 10$

1. $y = 4x + 5$

$2x + y = 17$

1. $y = 3x - 34$

$y = 2x - 5$

1. $y = 3x - 2$

$y = 2x - 5$

1. $2x + y = 3$

$4x + 4y = 8$

1. $3x + 4y = -3$

$x + 2y = -1$
money harvey has some $1 bills and some $5 bills. in all, he has 6 bills worth $22. let $x$ be the number of $1 bills and let $y$ be the number of $5 bills. write a system of equations to represent the information, and use substitution to determine how many bills of each denomination harvey has.

Explanation:

Problem 1: \(

$$\begin{cases} y = 5x + 1 \\ 4x + y = 10 \end{cases}$$

\)

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

Since \( y = 5x + 1 \), substitute \( y \) in \( 4x + y = 10 \):
\( 4x + (5x + 1) = 10 \)

Step 2: Solve for \( x \)

Simplify: \( 4x + 5x + 1 = 10 \)
\( 9x + 1 = 10 \)
Subtract 1: \( 9x = 9 \)
Divide by 9: \( x = 1 \)

Step 3: Solve for \( y \)

Substitute \( x = 1 \) into \( y = 5x + 1 \):
\( y = 5(1) + 1 = 6 \)

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

Substitute \( y = 4x + 5 \) into \( 2x + y = 17 \):
\( 2x + (4x + 5) = 17 \)

Step 2: Solve for \( x \)

Simplify: \( 2x + 4x + 5 = 17 \)
\( 6x + 5 = 17 \)
Subtract 5: \( 6x = 12 \)
Divide by 6: \( x = 2 \)

Step 3: Solve for \( y \)

Substitute \( x = 2 \) into \( y = 4x + 5 \):
\( y = 4(2) + 5 = 13 \)

Step 1: Set the two equations equal

Since both equal \( y \), set \( 3x - 34 = 2x - 5 \)

Step 2: Solve for \( x \)

Subtract \( 2x \): \( x - 34 = -5 \)
Add 34: \( x = 29 \)

Step 3: Solve for \( y \)

Substitute \( x = 29 \) into \( y = 2x - 5 \):
\( y = 2(29) - 5 = 53 \)

Answer:

\( x = 1, y = 6 \)

Problem 2: \(

$$\begin{cases} y = 4x + 5 \\ 2x + y = 17 \end{cases}$$

\)