HELP I NEED HELP ASAP HELP I NEED HELP ASAP
HELP I NEED HELP ASAP
HELP I NEED HELP ASAP
HELP I NEED HELP ASAP
HELP I NEED HELP ASAP
HELP I NEED HELP ASAP
HELP I NEED HELP ASAP

Next she should;

C. Add 6·x + 3·y = -24 to 6·x - 2·y = -4

Step-by-step 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

c

Step-by-step explanation:

a28=58

step-by-step explanation:

a1=a1+(n-1)d

a28= 4+ (28-1)2

a28=4+(27)(2)

a28=4+54

a28=58