-71. solve these systems of equations using any method. check each solution, if possible. homework help
a. $2x + 3y = 9$
$-3x + 3y = -6$
b. $x = 8 - 2y$
$y - x = 4$
c. $y = -\frac{1}{2}x + 7$
$y = x - 8$
d. $9x + 10y = 14$
$7x + 5y = -3$
4-72. for each line below, make a table and graph. what do you notice? 4-72 hw etool homework help
a. $y = \frac{2}{3}x - 1$
b. $2x - 3y = 3$

Question

-71. solve these systems of equations using any method. check each solution, if possible. homework help
 a. $2x + 3y = 9$
 $-3x + 3y = -6$
 b. $x = 8 - 2y$
 $y - x = 4$
 c. $y = -\frac{1}{2}x + 7$
 $y = x - 8$
 d. $9x + 10y = 14$
 $7x + 5y = -3$
 4-72. for each line below, make a table and graph. what do you notice? 4-72 hw etool homework help
 a. $y = \frac{2}{3}x - 1$
 b. $2x - 3y = 3$

Accepted Answer

### Part 4-71a
# Explanation:
## Step1: Subtract the two equations
We have the system:
$$
\begin{cases}
2x + 3y = 9 \
-3x + 3y = -6
\end{cases}
$$
Subtract the second equation from the first to eliminate $y$:
$(2x + 3y) - (-3x + 3y) = 9 - (-6)$
Simplify: $2x + 3y + 3x - 3y = 9 + 6$
Which gives: $5x = 15$

## Step2: Solve for $x$
Divide both sides by 5: $x = \frac{15}{5} = 3$

## Step3: Substitute $x = 3$ into the first equation
Substitute $x = 3$ into $2x + 3y = 9$:
$2(3) + 3y = 9$
Simplify: $6 + 3y = 9$

## Step4: Solve for $y$
Subtract 6 from both sides: $3y = 9 - 6 = 3$
Divide by 3: $y = 1$

## Step5: Check the solution
Substitute $x = 3$ and $y = 1$ into the second equation $-3x + 3y$:
$-3(3) + 3(1) = -9 + 3 = -6$, which matches the right-hand side.

# Answer: $x = 3$, $y = 1$

### Part 4-71b
# Explanation:
## Step1: Substitute $x = 8 - 2y$ into $y - x = 4$
We have $x = 8 - 2y$ and $y - x = 4$. Substitute $x$ into the second equation:
$y - (8 - 2y) = 4$

## Step2: Simplify and solve for $y$
Simplify: $y - 8 + 2y = 4$
Combine like terms: $3y - 8 = 4$
Add 8 to both sides: $3y = 12$
Divide by 3: $y = 4$

## Step3: Substitute $y = 4$ into $x = 8 - 2y$
$x = 8 - 2(4) = 8 - 8 = 0$

## Step4: Check the solution
Substitute $x = 0$ and $y = 4$ into $y - x$: $4 - 0 = 4$, which matches.

# Answer: $x = 0$, $y = 4$

### Part 4-71c
# Explanation:
## Step1: Set the two equations equal
We have $y = -\frac{1}{2}x + 7$ and $y = x - 8$. Set them equal:
$x - 8 = -\frac{1}{2}x + 7$

## Step2: Solve for $x$
Add $\frac{1}{2}x$ to both sides: $x + \frac{1}{2}x - 8 = 7$
Combine like terms: $\frac{3}{2}x - 8 = 7$
Add 8 to both sides: $\frac{3}{2}x = 15$
Multiply both sides by $\frac{2}{3}$: $x = 15\times\frac{2}{3} = 10$

## Step3: Substitute $x = 10$ into $y = x - 8$
$y = 10 - 8 = 2$

## Step4: Check the solution
Substitute $x = 10$ and $y = 2$ into $y = -\frac{1}{2}x + 7$:
$2 = -\frac{1}{2}(10) + 7 = -5 + 7 = 2$, which matches.

# Answer: $x = 10$, $y = 2$

### Part 4-71d
# Explanation:
## Step1: Multiply the second equation by 2
We have:
$$
\begin{cases}
9x + 10y = 14 \
7x + 5y = -3
\end{cases}
$$
Multiply the second equation by 2: $14x + 10y = -6$

## Step2: Subtract the first equation from the new second equation
$(14x + 10y) - (9x + 10y) = -6 - 14$
Simplify: $5x = -20$

## Step3: Solve for $x$
Divide by 5: $x = -4$

## Step4: Substitute $x = -4$ into the first equation
$9(-4) + 10y = 14$
Simplify: $-36 + 10y = 14$

## Step5: Solve for $y$
Add 36 to both sides: $10y = 50$
Divide by 10: $y = 5$

## Step6: Check the solution
Substitute $x = -4$ and $y = 5$ into the second equation: $7(-4) + 5(5) = -28 + 25 = -3$, which matches.

# Answer: $x = -4$, $y = 5$

### Part 4-72a (Table and Graph for $y = \frac{2}{3}x - 1$)
# Explanation:
## Step1: Choose $x$ values
Let's choose $x = -3$, $0$, $3$ (to make calculation easy with the fraction).

## Step2: Calculate $y$ for each $x$
- When $x = -3$: $y = \frac{2}{3}(-3) - 1 = -2 - 1 = -3$
- When $x = 0$: $y = \frac{2}{3}(0) - 1 = -1$
- When $x = 3$: $y = \frac{2}{3}(3) - 1 = 2 - 1 = 1$

## Step3: Create the table
| $x$ | $y$ |
|-------|-------|
| -3    | -3    |
| 0     | -1    |
| 3     | 1     |

## Step4: Graph the points
Plot the points $(-3, -3)$, $(0, -1)$, $(3, 1)$ and draw a straight line through them.

## Step5: Notice
The line has a slope of $\frac{2}{3}$ (positive, rising from left to right) and a $y$-intercept at $(0, -1)$.

# Answer:
Table:
| $x$ | $y$ |
|-------|-------|
| -3    | -3    |
| 0     | -1    |
| 3     | 1     |

Graph: A straight line through the points $(-3, -3)$, $(0, -1)$, $(3, 1)$. Notice: Slope $\frac{2}{3}$, $y$-intercept $(0, -1)$.

### Part 4-72b (Table and Graph for $2x - 3y = 3$)
# Explanation:
## Step1: Rewrite in slope-intercept form
Solve for $y$: $2x - 3y = 3$ → $-3y = -2x + 3$ → $y = \frac{2}{3}x - 1$ (same as part a!)

## Step2: Choose $x$ values (same as part a for comparison)
$x = -3$, $0$, $3$

## Step3: Calculate $y$ for each $x$
- $x = -3$: $y = \frac{2}{3}(-3) - 1 = -3$
- $x = 0$: $y = -1$
- $x = 3$: $y = 1$

## Step4: Create the table
| $x$ | $y$ |
|-------|-------|
| -3    | -3    |
| 0     | -1    |
| 3     | 1     |

## Step5: Graph the points
Plot $(-3, -3)$, $(0, -1)$, $(3, 1)$ and draw a line.

## Step6: Notice
The line is identical to the one in part a! This means the two equations (from 4-72a and 4-72b) represent the same line (they are equivalent).

# Answer:
Table:
| $x$ | $y$ |
|-------|-------|
| -3    | -3    |
| 0     | -1    |
| 3     | 1     |

Graph: Same line as 4-72a. Notice: The two equations represent the same line (equivalent).

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QUESTION IMAGE

Question

Explanation:

Part 4-71a

Step1: Subtract the two equations

Step2: Solve for \(x\)

Step3: Substitute \(x = 3\) into the first equation

Step4: Solve for \(y\)

Step5: Check the solution

Step1: Substitute \(x = 8 - 2y\) into \(y - x = 4\)

Step2: Simplify and solve for \(y\)

Step3: Substitute \(y = 4\) into \(x = 8 - 2y\)

Step4: Check the solution

Step1: Set the two equations equal

Step2: Solve for \(x\)

Step3: Substitute \(x = 10\) into \(y = x - 8\)

Step4: Check the solution

Answer:

Part 4-71b