To solve the system of equations, we can use either the substitution method or the elimination method. Let's solve it using the elimination method:
Multiply the first equation by 4 to make the coefficients of y the same in both equations:
12x + 4y = 8
Subtract the second equation from the modified first equation:
(12x + 4y) - (7x - 4y) = 8 - 30
12x + 4y - 7x + 4y = -22
5x + 8y = -22
Now you have a system of equations:
5x + 8y = -22 (Equation 1)
7x - 4y = 30 (Equation 2)
Multiply Equation 1 by 7 and Equation 2 by 5 to make the coefficients of x the same in both equations:
35x + 56y = -154 (Equation 3)
35x - 20y = 150 (Equation 4)
Now subtract Equation 4 from Equation 3:
(35x + 56y) - (35x - 20y) = -154 - 150
35x + 56y - 35x + 20y = -304
76y = -304
y = -4
Now substitute the value of y back into either Equation 1 or Equation 2. Let's use Equation 1:
5x + 8(-4) = -22
5x - 32 = -22
5x = 10
x = 2
Therefore, the solution to the system of equations is (2, -4).
Solve the system of equations.
3x+y=2
7x−4y=30(1 point)
Responses
(2, −4)
left parenthesis 2 comma negative 4 right parenthesis
(−2, 8)
left parenthesis negative 2 comma 8 right parenthesis
(12, 12)
left parenthesis start fraction 1 over 2 end fraction comma start fraction 1 over 2 end fraction right parenthesis
(2, −2)
1 answer
1 answer