Question

unit 5 lesson 1
set
solve each system of equations using the algebraic method of substitution. check your so
both equations.

1. \\(\

$$\begin{cases} y = 2x - 13 \\\\ y = x - 5 \\end{cases}$$

\\)

1. \\(\

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

\\)

1. \\(\

$$\begin{cases} y = x + 1 \\\\ x + 2y = 8 \\end{cases}$$

\\)

1. \\(\

$$\begin{cases} x = 9 - 2y \\\\ 3x + 5y = 20 \\end{cases}$$

\\)

1. \\(\

$$\begin{cases} y = 2x - 4 \\\\ 3y + 21x = 15 \\end{cases}$$

\\)

1. \\(\

$$\begin{cases} x = -1 - 2y \\\\ 3x + 5y = -1 \\end{cases}$$

\\)

Answer:

Let's solve each system of equations one by one using the substitution method.

Problem 9

\[

$$\begin{cases} y = 2x - 13 \\ y = x - 6 \end{cases}$$

\]

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

Since both equations are solved for \( y \), we can set them equal to each other:
\( 2x - 13 = x - 6 \)

Step 2: Solve for \( x \)

Subtract \( x \) from both sides:
\( 2x - x - 13 = x - x - 6 \)
\( x - 13 = -6 \)

Add 13 to both sides:
\( x - 13 + 13 = -6 + 13 \)
\( x = 7 \)

Step 3: Substitute \( x = 7 \) into \( y = x - 6 \) to find \( y \)

\( y = 7 - 6 \)
\( y = 1 \)

Step 4: Check the solution in both equations

• For \( y = 2x - 13 \): \( 1 = 2(7) - 13 = 14 - 13 = 1 \) (True)
• For \( y = x - 6 \): \( 1 = 7 - 6 = 1 \) (True)

So the solution is \( (7, 1) \).

Problem 10

\[

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

\]

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

Set \( -2x + 9 = x - 3 \)

Step 2: Solve for \( x \)

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

Add 3 to both sides:
\( 9 + 3 = 3x - 3 + 3 \)
\( 12 = 3x \)

Divide both sides by 3:
\( x = 4 \)

Step 3: Substitute \( x = 4 \) into \( y = x - 3 \) to find \( y \)

\( y = 4 - 3 \)
\( y = 1 \)

Step 4: Check the solution in both equations

• For \( y = -2x + 9 \): \( 1 = -2(4) + 9 = -8 + 9 = 1 \) (True)
• For \( y = x - 3 \): \( 1 = 4 - 3 = 1 \) (True)

So the solution is \( (4, 1) \).

Problem 11

\[

$$\begin{cases} y = x + 1 \\ x + 2y = 8 \end{cases}$$

\]

Step 1: Substitute \( y = x + 1 \) into \( x + 2y = 8 \)

\( x + 2(x + 1) = 8 \)

Step 2: Solve for \( x \)

Expand the left side:
\( x + 2x + 2 = 8 \)
\( 3x + 2 = 8 \)

Subtract 2 from both sides:
\( 3x + 2 - 2 = 8 - 2 \)
\( 3x = 6 \)

Divide both sides by 3:
\( x = 2 \)

Step 3: Substitute \( x = 2 \) into \( y = x + 1 \) to find \( y \)

\( y = 2 + 1 \)
\( y = 3 \)

Step 4: Check the solution in both equations

• For \( y = x + 1 \): \( 3 = 2 + 1 = 3 \) (True)
• For \( x + 2y = 8 \): \( 2 + 2(3) = 2 + 6 = 8 \) (True)

So the solution is \( (2, 3) \).

Problem 12

\[

$$\begin{cases} x = 9 - 2y \\ 3x + 5y = 20 \end{cases}$$

\]

Step 1: Substitute \( x = 9 - 2y \) into \( 3x + 5y = 20 \)

\( 3(9 - 2y) + 5y = 20 \)

Step 2: Solve for \( y \)

Expand the left side:
\( 27 - 6y + 5y = 20 \)
\( 27 - y = 20 \)

Subtract 27 from both sides:
\( -y = 20 - 27 \)
\( -y = -7 \)

Multiply both sides by -1:
\( y = 7 \)

Step 3: Substitute \( y = 7 \) into \( x = 9 - 2y \) to find \( x \)

\( x = 9 - 2(7) \)
\( x = 9 - 14 \)
\( x = -5 \)

Step 4: Check the solution in both equations

• For \( x = 9 - 2y \): \( -5 = 9 - 2(7) = 9 - 14 = -5 \) (True)
• For \( 3x + 5y = 20 \): \( 3(-5) + 5(7) = -15 + 35 = 20 \) (True)

So the solution is \( (-5, 7) \).

Problem 13

\[

$$\begin{cases} y = 2x - 4 \\ 3y + 21x = 15 \end{cases}$$

\]

Step 1: Substitute \( y = 2x - 4 \) into \( 3y + 21x = 15 \)

\( 3(2x - 4) + 21x = 15 \)

Step 2: Solve for \( x \)

Expand the left side:
\( 6x - 12 + 21x = 15 \)
\( 27x - 12 = 15 \)

Add 12 to both sides:
\( 27x = 15 + 12 \)
\( 27x = 27 \)

Divide both sides by 27:
\( x = 1 \)

Step 3: Substitute \( x = 1 \) into \( y = 2x - 4 \) to find \( y \)

\( y = 2(1) - 4 \)
\( y = 2 - 4 \)
\( y = -2 \)

Step 4: Check the solution in both equations

• For \( y = 2x - 4 \): \( -2 = 2(1) - 4 = 2 - 4 = -2 \) (True)
• For \( 3y + 21x = 15 \): \( 3(-2) + 21(1) = -6 + 21 = 15 \) (True)

So the solution is \( (1, -2) \).

Problem 14

\[

$$\begin{cases} x = -1 - 2y \\ 3x + 5y = -1 \end{cases}$$

\]

Step 1: Substitute \( x = -1 - 2y \) into \( 3x + 5y = -1 \)

\( 3(-1 - 2y) + 5y = -1 \)

Step 2: Solve for \( y \)

Expand the left side:
\( -3 - 6y + 5y = -1 \)
\( -3 - y = -1 \)

Add 3 to both sides:
\( -y = -1 + 3 \)
\( -y = 2 \)

Multiply both sides by -1:
\( y = -2 \)

Step 3: Substitute \( y = -2 \) into \( x = -1 - 2y \) to find \( x \)

\( x = -1 - 2(-2) \)
\( x = -1 + 4 \)
\( x = 3 \)

Step 4: Check the solution in both equations

• For \( x = -1 - 2y \): \( 3 = -1 - 2(-2) = -1 + 4 = 3 \) (True)
• For \( 3x + 5y = -1 \): \( 3(3) + 5(-2) = 9 - 10 = -1 \) (True)

So the solution is \( (3, -2) \).

Final Answers

• Problem 9: \( \boldsymbol{(7, 1)} \)
• Problem 10: \( \boldsymbol{(4, 1)} \)
• Problem 11: \( \boldsymbol{(2, 3)} \)
• Problem 12: \( \boldsymbol{(-5, 7)} \)
• Problem 13: \( \boldsymbol{(1, -2)} \)
• Problem 14: \( \boldsymbol{(3, -2)} \)