To solve the system of equations, we can use either substitution or elimination method. Here, I will use the elimination method.

The system of equations is:

3x + 4y = 38
5x - 5y = -30

First, we can multiply the first equation by 5 and the second equation by 3 to make the coefficients of x the same in both equations:

15x + 20y = 190
15x - 15y = -90

Now, subtract the second equation from the first:

15x + 20y - (15x - 15y) = 190 - (-90)
15x + 20y - 15x + 15y = 190 + 90
35y = 280

Divide by 35 to solve for y:

y = 280 / 35
y = 8

Substitute y = 8 into the first equation to solve for x:

3x + 4*8 = 38
3x + 32 = 38
3x = 38 - 32
3x = 6
x = 6 / 3
x = 2

So, the solution to the system of equations is (2, 8), which corresponds to option C.

Question

To solve the system of equations, we can use either substitution or elimination method. Here, I will use the elimination method.

The system of equations is:

3x + 4y = 38
5x - 5y = -30

First, we can multiply the first equation by 5 and the second equation by 3 to make the coefficients of x the same in both equations:

15x + 20y = 190
15x - 15y = -90

Now, subtract the second equation from the first:

15x + 20y - (15x - 15y) = 190 - (-90)
15x + 20y - 15x + 15y = 190 + 90
35y = 280

Divide by 35 to solve for y:

y = 280 / 35
y = 8

Substitute y = 8 into the first equation to solve for x:

3x + 4*8 = 38
3x + 32 = 38
3x = 38 - 32
3x = 6
x = 6 / 3
x = 2

So, the solution to the system of equations is (2, 8), which corresponds to option C.

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