Answer:
Next she should;
C. Add 6·x + 3·y = -24 to 6·x - 2·y = -4
Explanation:
To show that given a a system of two equations in (having) two unknowns, when one of the equations is replaced by the sum the equation being replaced and the multiple of the other equation gives a system with the same solution as the previous system, we proceed as follows;
The given system of equations are;
6·x - 2·y = -4
2·x + y = -8
Gabriela multiplies the second equation by 3, to get;
3 × (2·x + y) = 3 ×-8 = -24
6·x + 3·y = -24
She should add the equation above to the first equation to get;
Add 6·x + 3·y = -24 to 6·x - 2·y = -4
(6·x + 3·y) + (6·x - 2·y) = -24 + -4
12·x + y = -24
The new system becomes;
12·x + y = -28
2·x + y = -8
Subtracting the second equation from the first equation in the new system gives;
12·x + y - (2·x + y) = -28 - (-8) = -28 + 8 = -20
10·x + y - y = 10·x + 0 = -20
x = -20/10 = -2
From the second equation, we get;
y = -8 - 2·x = -8 - 2×(-2) = -4
y = -4
We verify if the solution is the same as the first system as follows;
6·x - 2·y = -4
2·x + y = -8
Therefore
6·(-2) - 2·(-4) = -4 (Correct)
2·(-2) + (-4) = -8 (Correct)
Therefore, the next step is to add 6·x + 3·y = -24 to 6·x - 2·y = -4