Question

Solve the following system of equations using the elimination method.
3x - 4y, x + 3y = -10

A. (-2, -4)
B. (-4, -2)
C. (-16, -2)
D. (-16, 2)

Answer 1

There is a missing equals sign in the equation "3x - 4y".

Assuming the equation is meant to be "3x - 4y = -10", let's solve the system of equations using the elimination method.

Given equations:
1) 3x - 4y = -10
2) x + 3y = -10

Multiply equation 2) by 3 to match the coefficients of y:
3(x + 3y) = 3(-10)
3x + 9y = -30

Now, we have the following equations:
1) 3x - 4y = -10
3) 3x + 9y = -30

Multiply equation 1) by 3 to match the coefficients of x:
3(3x - 4y) = 3(-10)
9x - 12y = -30

Now, we have the following equations:
4) 9x - 12y = -30
3) 3x + 9y = -30

Subtract equation 3) from equation 4) to eliminate x:
(9x - 12y) - (3x + 9y) = (-30) - (-30)
9x - 12y - 3x - 9y = -30 + 30
6x - 21y = 0
Simplify:
2x - 7y = 0 -------- (Equation 5)

Solve equations 2) and 5) as a new system:
2) x + 3y = -10
5) 2x - 7y = 0

Multiply equation 2) by 2 to match the coefficients of x:
2(x + 3y) = 2(-10)
2x + 6y = -20

Now, we have the following equations:
6) 2x + 6y = -20
5) 2x - 7y = 0

Add equation 5) to equation 6) to eliminate x:
(2x - 7y) + (2x + 6y) = 0 + (-20)
2x - 7y + 2x + 6y = -20
4x - y = -20
Add y to both sides of the equation:
4x - y + y = -20 + y
4x = -20 + y
Divide by 4:
x = (-20 + y)/4

Now, substitute the value of x into equation 2):
(-20 + y)/4 + 3y = -10
Multiply through by 4 to clear the fraction:
-20 + y + 12y = -40
Combine like terms:
13y - 20 = -40
Add 20 to both sides:
13y = -20 - 40
13y = -60
Divide by 13:
y = -60/13

Substitute y = -60/13 back into the equation for x:
x = (-20 + y)/4
x = (-20 - 60/13)/4
x = (-260 + 60)/52
x = -200/52
x = -50/13

The solution to the system of equations is (x, y) = (-50/13, -60/13).

None of the answer choices match the solution, so none of the given options is correct.