Question

Solve the system of equations using elimination 3c - 8d =7 and c + 2d = -7

Answer 1

Given.

$(3c-8d=7)=(c+2d=-7)$

Multiply 4 to the second equation.

$4(c+2d=-7)$

$3c+8d=-28$

Compare the like terms.

$4c+3c-8d+8d=7-28$

Solve.

$7c=-21$

Divide 7 to both sides

$\frac{7c}{7}=\frac{-21}{7}$

$c=-3$

Substitute (-3) for c into the original second equation.

$c+2d=-7$

$-3+2d=-7$

Add (3) to both sides.

$(-3+2d=-7)+3$

$2d=-4$

Divide (2) to both sides.

$\frac{2d}{2}=\frac{-4}{2}$

$d=-2$

Check your work.

$3(-3)-8(-2)=7$

$-9+16=7$

$7=7$

$c+2d=-7$

$-3+2(-2)=-7$

$-3-4=-7$

$-7=-7$

Therefore your answer is.

$c=-3\\d=-2$

Answer 2

Answer:

d=-2 c=-3

Step-by-step explanation:

3c-8d=7

4(c+2d=-7)

3c-8d=7

4c+8d=-28

4c+3c-8d+8d=7-28

7c=-21

c=-3

c+2d=-7

-3+2d=-7

2d=-4

d=-2