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unit 3, lesson 6 – ready, set, go ready solve each system of equations.…

Question

unit 3, lesson 6 – ready, set, go
ready
solve each system of equations.

  1. $y = 3x$

$y = -2x - 15$

  1. $3x + y = 21$

$y = -2x - 15$

  1. $3x + 2y = 7$

$x - 2y = -3$

  1. $5x - y = 13$

$-2x + y = -1$

  1. $3x + y = 21$

$-3x + 5y = -3$

  1. $3x + 2y = 7$

$x + y = 2$

Explanation:

Response
Problem 1:

Step1: Substitute \( y = 3x \) into \( y=-2x - 15 \)

\( 3x=-2x - 15 \)

Step2: Solve for \( x \)

\( 3x + 2x=-15 \)
\( 5x=-15 \)
\( x = \frac{-15}{5}=-3 \)

Step3: Substitute \( x = -3 \) into \( y = 3x \)

\( y = 3\times(-3)=-9 \)

Step1: Substitute \( y=-2x - 15 \) into \( 3x + y=21 \)

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

Step2: Simplify and solve for \( x \)

\( 3x-2x-15 = 21 \)
\( x-15 = 21 \)
\( x=21 + 15=36 \)

Step3: Substitute \( x = 36 \) into \( y=-2x - 15 \)

\( y=-2\times36-15=-72 - 15=-87 \)

Step1: Add the two equations \( 3x + 2y=7 \) and \( x-2y=-3 \)

\( (3x + 2y)+(x - 2y)=7+(-3) \)

Step2: Simplify and solve for \( x \)

\( 4x=4 \)
\( x = 1 \)

Step3: Substitute \( x = 1 \) into \( x-2y=-3 \)

\( 1-2y=-3 \)
\( -2y=-3 - 1=-4 \)
\( y=\frac{-4}{-2}=2 \)

Answer:

\( x=-3, y = -9 \)

Problem 2: