1. Solve the system of two linear equations.

Question

-3x - 4y = -23
3x - 5y = 5

2. How many solutions does the system of equations have?
8x - 6y = 48
8x + 6y = 48

3. Solve the system of two equations using elimination with multiplication.
4x - 6y = -48
-2x + 4y = 30

4. Solve the system of two equations.
3x - 4y = 40
5x - 5y = 55

5. How many solutions does the system have?
12x - 15y = 18
4x - 5y = 6
……

Answer 1

1. x = 4, y = 3

2. Infinite solutions (The two equations are the same)

3. x = -6, y = -4

4. x = -135, y = -101.25

5. One solution

Answer 2

Number one is wrong

Answer 3

I apologize for that mistake. Let's re-calculate the first problem:

1. Solve the system of two linear equations.
-3x - 4y = -23
3x - 5y = 5

Adding the two equations together to eliminate x:
-3x - 4y + 3x - 5y = -23 + 5
-y = -18
y = 18

Substitute y back into the first equation:
-3x - 4(18) = -23
-3x - 72 = -23
-3x = 49
x = -49/3

Therefore, the solution to the system of equations is x = -49/3, y = 18.

Answer 4

Try again

Answer 5

I apologize for the mistake. Let's re-calculate the first problem again:

1. Solve the system of two linear equations.
-3x - 4y = -23
3x - 5y = 5

From the second equation, rearrange it to solve for x:
3x = 5 + 5y
x = (5 + 5y) / 3

Substitute x into the first equation:
-3((5 + 5y) / 3) - 4y = -23
-5 - 5y - 4y = -23
-9y = -18
y = 2

Now, substitute y back into the second equation:
3x - 5(2) = 5
3x - 10 = 5
3x = 15
x = 5

Therefore, the solution to the system of equations is x = 5, y = 2.

Answer 6

Number 3 is wrong

Answer 7

Let's re-calculate the third problem using elimination with multiplication:

3. Solve the system of two equations using elimination with multiplication.
4x - 6y = -48
-2x + 4y = 30

Multiply the second equation by 2 to eliminate x:
8x - 12y = -96
-2x + 4y = 30

Now, add the two equations:
8x - 12y + (-2x + 4y) = -96 + 30
6x - 8y = -66

Divide by 2 to simplify:
3x - 4y = -33

Now we have a new equation:
3x - 4y = -33
4x - 6y = -48

Multiply the first equation by 2:
6x - 8y = -66
4x - 6y = -48

Subtract the second equation from the first:
2x = 18
x = 9

Substitute x back into the first equation:
4(9) - 6y = -48
36 - 6y = -48
-6y = -84
y = 14

Therefore, the solution to the system of equations is x = 9, y = 14.

Answer 8

Wrong

Answer 9

Let's re-calculate the third problem using elimination with multiplication:

3. Solve the system of two equations using elimination with multiplication.
4x - 6y = -48
-2x + 4y = 30

First, simplify the equations to help in elimination:
-4x + 6y = 48
-2x + 4y = 30

Now, multiply the second equation by 2 to get the same coefficient for x (4x and -4x):
-4x + 8y = 60

Now, we have the following equations:
-4x + 6y = 48
-4x + 8y = 60

Subtract the first equation from the second equation to eliminate x:
2y = 12
y = 6

Substitute y back into one of the original equations:
4x - 6(6) = -48
4x - 36 = -48
4x = -12
x = -3

Therefore, the solution to the system of equations is x = -3, y = 6.

Answer 10

Number 4 is wrong

Answer 11

Let's re-calculate the fourth problem:

4. Solve the system of two equations.
3x - 4y = 40
5x - 5y = 55

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

First, let's multiply the first equation by 5 and the second equation by 4 to eliminate y:

15x - 20y = 200
20x - 20y = 220

Now, subtract the first equation from the second equation:
5x = 20
x = 4

Substitute x back into the first equation:
3(4) - 4y = 40
12 - 4y = 40
-4y = 28
y = -7

Therefore, the solution to the system of equations is x = 4, y = -7.