Question

Solve the system of equations algebraically.

Answer 1

Answer:

Choice d is correct answer.

Step-by-step explanation:

We have given two equation.

3x-3y = 6                eq(1)

6x-7y = 8                eq(2)

We have to find the solution of given equations.

Multiplying by both sides of eq(1) by 2, we have

2(3x-3y) = 2(6)

6x-6y = 12                  eq(3)

Subtracting eq(3) to eq(2) , we have

6x-6y -(6x-7y) = 12-8

6x-6y-6x+7y = 4

Adding like terms, we have

6x-6x+7y-6y = 4

y = 4

Putting the value of y in eq(1), we have

3x-3(4) = 6

3x-12 = 6

Adding 12 to both sides of above equation, we have

3x-12+12= 6+12

3x = 18

Dividing by 3 to both sides of above equation , we have

3x/3 = 18/3

x = 6

Hence, the solution set is (6,4).

Answer 2

Answer:

The correct option is d.

Step-by-step explanation:

The given equations are

$3x-3y=6$               ....(1)

$6x-7y=8$                ..... (2)

Multiply equation (1) by  by 2, to make same coefficient of x.

$6x-6y=12$               ....(3)

Subtract equation (2) form equation (3).

$6x-6y-6x+7y=12-8$

$y=4$

Substitute this value in equation (1),

$3x-3(4)=6$

$3x-12=6$

$3x=18$

$x=6$

The solution of the given system of equation is (6,4).

Therefore the correct option is d.