use substitution to solve each system of equations.
1. $\begin{cases} y = -x + 4 \\ y = 3x \end{cases}$
2. $\begin{cases} y = 2x - 10 \\ 2y = x - 8 \end{cases}$
3. $\begin{cases} x - 2y = 12 \\ y = 3x + 14 \end{cases}$
4. $\begin{cases} x = 2y - 6 \\ y = 3x - 7 \end{cases}$
5. $\begin{cases} 6x - 4y = 18 \\ x = -6y - 7 \end{cases}$
6. $\begin{cases} y = -217x + 5/7 \\ 3x + 10y = 6 \end{cases}$

Question

Accepted Answer

### Problem 1
# Explanation:
## Step1: Substitute $ y = 3x $ into $ y=-x + 4 $
$ 3x=-x + 4 $
## Step2: Solve for $ x $
Add $ x $ to both sides: $ 3x+x=-x+x + 4 $ → $ 4x = 4 $
Divide by 4: $ x=\frac{4}{4}=1 $
## Step3: Substitute $ x = 1 $ into $ y = 3x $
$ y=3\times1 = 3 $
# Answer:
$ x = 1 $, $ y = 3 $

### Problem 2
# Explanation:
## Step1: Substitute $ y = 2x-10 $ into $ 2y=x - 8 $
$ 2(2x-10)=x - 8 $
## Step2: Expand and solve for $ x $
$ 4x-20=x - 8 $
Subtract $ x $ from both sides: $ 4x-x-20=x-x - 8 $ → $ 3x-20=-8 $
Add 20 to both sides: $ 3x-20 + 20=-8 + 20 $ → $ 3x = 12 $
Divide by 3: $ x=\frac{12}{3}=4 $
## Step3: Substitute $ x = 4 $ into $ y = 2x-10 $
$ y=2\times4-10=8 - 10=-2 $
# Answer:
$ x = 4 $, $ y=-2 $

### Problem 3
# Explanation:
## Step1: Substitute $ y = 3x + 14 $ into $ x-2y = 12 $
$ x-2(3x + 14)=12 $
## Step2: Expand and solve for $ x $
$ x-6x-28 = 12 $
Combine like terms: $ -5x-28 = 12 $
Add 28 to both sides: $ -5x-28 + 28=12 + 28 $ → $ -5x = 40 $
Divide by -5: $ x=\frac{40}{-5}=-8 $
## Step3: Substitute $ x=-8 $ into $ y = 3x + 14 $
$ y=3\times(-8)+14=-24 + 14=-10 $
# Answer:
$ x=-8 $, $ y=-10 $

### Problem 4
# Explanation:
## Step1: Substitute $ y = 3x-7 $ into $ x = 2y-6 $
$ x=2(3x-7)-6 $
## Step2: Expand and solve for $ x $
$ x = 6x-14-6 $
$ x = 6x-20 $
Subtract $ 6x $ from both sides: $ x-6x=6x-6x-20 $ → $ -5x=-20 $
Divide by -5: $ x=\frac{-20}{-5}=4 $
## Step3: Substitute $ x = 4 $ into $ y = 3x-7 $
$ y=3\times4-7=12 - 7 = 5 $
# Answer:
$ x = 4 $, $ y = 5 $

### Problem 5
# Explanation:
## Step1: Substitute $ x=-6y-7 $ into $ 6x-4y = 18 $
$ 6(-6y-7)-4y = 18 $
## Step2: Expand and solve for $ y $
$ -36y-42-4y = 18 $
Combine like terms: $ -40y-42 = 18 $
Add 42 to both sides: $ -40y-42 + 42=18 + 42 $ → $ -40y = 60 $
Divide by -40: $ y=\frac{60}{-40}=-\frac{3}{2} $
## Step3: Substitute $ y = -\frac{3}{2} $ into $ x=-6y-7 $
$ x=-6\times(-\frac{3}{2})-7=9 - 7 = 2 $
# Answer:
$ x = 2 $, $ y=-\frac{3}{2} $

### Problem 6
# Explanation:
## Step1: Substitute $ Y=-217x + 5/7 $ (assuming $ Y $ is $ y $) into $ 3x + 10y = 6 $
$ 3x+10(-217x+\frac{5}{7})=6 $
## Step2: Expand and solve for $ x $
$ 3x-2170x+\frac{50}{7}=6 $
Combine like terms: $ -2167x+\frac{50}{7}=6 $
Subtract $ \frac{50}{7} $ from both sides: $ -2167x=6-\frac{50}{7}=\frac{42 - 50}{7}=-\frac{8}{7} $
Divide by -2167: $ x=\frac{-\frac{8}{7}}{-2167}=\frac{8}{15169} $
## Step3: Substitute $ x=\frac{8}{15169} $ into $ y=-217x+\frac{5}{7} $
$ y=-217\times\frac{8}{15169}+\frac{5}{7}=-\frac{1736}{15169}+\frac{5\times2167}{15169}=-\frac{1736}{15169}+\frac{10835}{15169}=\frac{9099}{15169}\approx0.6 $ (or keep as fraction $ \frac{9099}{15169} $)
# Answer:
$ x=\frac{8}{15169} $, $ y=\frac{9099}{15169} $ (or approximate values)

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QUESTION IMAGE

Question

Explanation:

Problem 1

Step1: Substitute \( y = 3x \) into \( y=-x + 4 \)

Step2: Solve for \( x \)

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

Step1: Substitute \( y = 2x-10 \) into \( 2y=x - 8 \)

Step2: Expand and solve for \( x \)

Step3: Substitute \( x = 4 \) into \( y = 2x-10 \)

Step1: Substitute \( y = 3x + 14 \) into \( x-2y = 12 \)

Step2: Expand and solve for \( x \)

Step3: Substitute \( x=-8 \) into \( y = 3x + 14 \)

Answer:

Problem 2