Solve the system of equations.

To solve this system of equations, we can use the method of substitution.

First, we can rearrange the second equation to solve for y:

-6x + y = 32

y = 6x + 32

Next, we can substitute this expression for y into the first equation:

4x - 8y = -36

4x - 8(6x + 32) = -36

Simplifying this equation, we get:

4x - 48x - 256 = -36

-44x = 220

x = -5

Now that we know the value of x, we can substitute it back into either of the original equations to solve for y. Using the second equation:

-6x + y = 32

-6(-5) + y = 32

y = 2

Therefore, the solution to the system of equations is:

x = -5

y = 2

6x−3y=−6

9x−3y=−18

We can use the method of elimination to solve this system of equations.

First, we'll write the equations in standard format:

6x - 3y = -6

9x - 3y = -18

Notice that both equations have a common term of -3y. To eliminate this term, we can subtract the first equation from the second:

(9x - 3y) - (6x - 3y) = -18 - (-6)

3x = -12

x = -4

Now that we know the value of x, we can substitute it back into either equation to solve for y. Using the first equation:

6x - 3y = -6

6(-4) - 3y = -6

-24 - 3y = -6

y = 6

Therefore, the solution to the system of equations is:

x = -4

y = 6

5x–4y=–44

–x+3y=22

We can use the method of substitution to solve this system of equations:

From the second equation, we can solve for x in terms of y:

-x + 3y = 22

x = 3y - 22

Now, we can substitute this expression for x into the first equation:

5x - 4y = -44

5(3y - 22) - 4y = -44

15y - 110 - 4y = -44

11y = 66

y = 6

We've now found the value of y, so we can substitute this back into either equation to solve for x. Using the second equation:

-x + 3y = 22

-x + 3(6) = 22

-x + 18 = 22

x = -4

Therefore, the solution to the system of equations is:

x = -4

y = 6

−6x+2y=−32

4x+3y=−9

We can use the method of elimination to solve this system of equations:

First, we'll multiply the first equation by -2 to eliminate the y term:

-2(-6x + 2y = -32)

12x - 4y = 64

Now we can add this equation to the second equation to eliminate the x term:

12x - 4y = 64

4x + 3y = -9

----------------------

16x - y = 55

Now we can solve for y by subtracting 16x from both sides:

16x - y = 55

-y = 55 - 16x

y = -55 + 16x

We can substitute this expression for y back into either of the original equations to solve for x. Using the first equation:

-6x + 2y = -32

-6x + 2(-55 + 16x) = -32

-6x - 110 + 32x = -32

26x = 78

x = 3

Now we can substitute x = 3 into our expression for y:

y = -55 + 16x

y = -55 + 16(3)

y = -7

Therefore, the solution to the system of equations is:

x = 3

y = -7

The Math Club sold slices of pizza, p, at a profit of $3 each and cans of soda, s, at $2 each, to raise money for a trip. They sold 200 items and made a profit of $550. Write the pair of linear equations that model this situation.(1 point)

Equation 1, per item: = 200
Equation 2, total: = 550

Let's use the following variables:

p = number of pizza slices sold

s = number of cans of soda sold

From the problem, we know the following:

- The Math Club sold 200 items in total:

p + s = 200

- The profit from selling each slice of pizza is $3, and they sold p slices:

3p

- The profit from selling each can of soda is $2, and they sold s cans:

2s

- The total profit made was $550:

3p + 2s = 550

Therefore, the pair of linear equations that model this situation are:

p + s = 200

3p + 2s = 550

The Lakewood baseball team is selling T-shirts for a fundraiser. The shirts cost $100 for the printing design and setup, plus $10 per shirt. The team is going to sell the shirts for $15 each. How many shirts do they need to sell to break even?(1 point)

Let's use the following variables:

x = number of shirts sold

Using the information given in the problem, we can write the cost function and revenue function:

Cost function:

C(x) = 100 + 10x

This is because the cost of making the shirts includes a fixed cost of $100, plus $10 for each shirt produced.

Revenue function:

R(x) = 15x

This is because the team will sell each shirt for $15, and x represents the number of shirts sold.

To find the number of shirts the team needs to sell to break even, we need to set the cost equal to the revenue and solve for x:

C(x) = R(x)

100 + 10x = 15x

100 = 5x

x = 20

Therefore, the team needs to sell 20 shirts to break even.

The Kesling Middle School PTA is planning a carnival to raise money for the school’s art department. They estimate that the event will be very popular and that they will have 500 people attend. They plan to charge adults $10 and children $5 for admission. The PTA wants to earn $3,500 from admission charges. How many adults and how many children need to attend for the PTA to reach their goal of $3,500?(1 point)

Let's use the following variables:

a = number of adults attending

c = number of children attending

Using the information given in the problem, we can write the following system of equations:

- The total number of people attending is 500:

a + c = 500

- The PTA wants to earn $3,500 from admission:

10a + 5c = 3500

We have two equations and two variables, so we can solve for a and c.

To do this, we can start by isolating one of the variables in the first equation:

a + c = 500

a = 500 - c

Now we can substitute this expression for a into the second equation:

10a + 5c = 3500

10(500 - c) + 5c = 3500

5000 - 10c + 5c = 3500

-5c = -1500

c = 300

Now we know that 300 children will attend. We can substitute this value back into one of the equations to solve for a:

a + c = 500

a + 300 = 500

a = 200

Therefore, 200 adults and 300 children need to attend for the PTA to reach their goal of $3,500.

Solve the following system of linear equations by graphing. Graph on your own piece of paper. In your submitted answer, describe what the graph looks like and what this tells you about the solution to the system of linear equations.

y=x+3

−4x+4y=28

To graph the system of linear equations, we can start by putting each equation into slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept.

y = x + 3

This equation is already in slope-intercept form, with a slope of 1 and a y-intercept of 3. To graph it, we can start at the y-intercept (0, 3), and then use the slope of 1 to find another point on the line. One point we can use is (1, 4):

```
|
4 |- *
| \
3 |- \
| *
2 |- \
| \
1 |- *
| \
0 |-________________\______________
| 0 | 1 | 2 | 3 | 4 |
```

Next, we'll put the second equation into slope-intercept form:

-4x + 4y = 28

4y = 4x + 28

y = x + 7

This equation also has a slope of 1, but a y-intercept of 7. To graph it, we can start at the y-intercept (0, 7), and then use the slope of 1 to find another point on the line. One point we can use is (1, 8):

```
|
| *
| \
| *
| \
| *
| \
7 |- *
| \
| *
| \
| *
|________________*\_____________
| 0 | 1 | 2 | 3 | 4 |
```

Now we can see that the two lines intersect at the point (2, 5). This is the solution to the system of linear equations.

The graph looks like two lines that intersect at a single point. This tells us that there is exactly one solution to the system of linear equations.