Question

lo4: solve each system using elimination.

1. $2x + 4y = -14$

$x - 2y = 9$

1. $-5x - 5y = -30$

$10x + 3y = -3$

1. $x - 3y = -13$

$2x + 15y = 16$

lo5: solve each system using elimination.

1. $-4x - 4y = 0$

$7x + 5y = 10$

1. $-2x + 3y = -5$

$-x + 4y = 5$

Explanation:

Problem 11: Solve \(

$$\begin{cases} 2x + 4y = -14 \\ x - 2y = 9 \end{cases}$$

\) using elimination

Step 1: Prepare to eliminate \( y \)

Multiply the second equation by 2 to make the coefficients of \( y \) opposites.
\( 2(x - 2y) = 2(9) \)
\( 2x - 4y = 18 \)

Step 2: Add the two equations

Add the first equation \( 2x + 4y = -14 \) and the new second equation \( 2x - 4y = 18 \).
\( (2x + 4y) + (2x - 4y) = -14 + 18 \)
\( 4x = 4 \)

Step 3: Solve for \( x \)

Divide both sides by 4.
\( x = \frac{4}{4} = 1 \)

Step 4: Substitute \( x = 1 \) into the second original equation

\( 1 - 2y = 9 \)
Subtract 1 from both sides: \( -2y = 9 - 1 = 8 \)
Divide by -2: \( y = \frac{8}{-2} = -4 \)

Step 1: Prepare to eliminate \( x \)

Multiply the first equation by 2 to make the coefficients of \( x \) opposites.
\( 2(-5x - 5y) = 2(-30) \)
\( -10x - 10y = -60 \)

Step 2: Add the two equations

Add the new first equation \( -10x - 10y = -60 \) and the second equation \( 10x + 3y = -3 \).
\( (-10x - 10y) + (10x + 3y) = -60 + (-3) \)
\( -7y = -63 \)

Step 3: Solve for \( y \)

Divide both sides by -7.
\( y = \frac{-63}{-7} = 9 \)

Step 4: Substitute \( y = 9 \) into the first original equation

\( -5x - 5(9) = -30 \)
\( -5x - 45 = -30 \)
Add 45 to both sides: \( -5x = -30 + 45 = 15 \)
Divide by -5: \( x = \frac{15}{-5} = -3 \)

Step 1: Prepare to eliminate \( x \)

Multiply the first equation by -2 to make the coefficients of \( x \) opposites.
\( -2(x - 3y) = -2(-13) \)
\( -2x + 6y = 26 \)

Step 2: Add the two equations

Add the new first equation \( -2x + 6y = 26 \) and the second equation \( 2x + 15y = 16 \).
\( (-2x + 6y) + (2x + 15y) = 26 + 16 \)
\( 21y = 42 \)

Step 3: Solve for \( y \)

Divide both sides by 21.
\( y = \frac{42}{21} = 2 \)

Step 4: Substitute \( y = 2 \) into the first original equation

\( x - 3(2) = -13 \)
\( x - 6 = -13 \)
Add 6 to both sides: \( x = -13 + 6 = -7 \)

Answer:

\( x = 1, y = -4 \)

Problem 12: Solve \(

$$\begin{cases} -5x - 5y = -30 \\ 10x + 3y = -3 \end{cases}$$

\) using elimination