Question

How to solve system of equations?
8q-15r=-40
4q+2r=56

Answer 1

q=10 and r=8

Explanation:

8q-15r=-40 ...........equation(i)

4q+2r=56................eqn(ii)

Now,

Multiplying eqn (ii) by 2 then we get,

8q+4r=112..........eqn(iii)

Here, Subtracting eqn (i) and (iii)

8q-15r=-40

8q+4r=112

(-) (-) (-)

---------------------

-19r=-152

r= -152/-19

r= 8

Again, Putting the value of r in eqn (ii)

4q+2r=56

4q+2×8=56

4q=56-16

q=40/4

q=10

Answer 2

q=10, r=8

Explanation:

You have to first cancel one of the variables.

8q-15r=-40

4q+2r=56

To do this, multiply the second equation by -2.

8q-15r=-40

-2(4q+2r)=(56)(-2)

Multiply it out.

8q-15r=-40

-8q-4r=-112

Combine like terms.

-19r = -152

Divide by -19 to get the value of r.

r=8

Substitute r back into one of the equations, it does not matter which one. I'll do both.

8q-15(8)=-40

8q-120=-40

8q=80

q=10

4q+2(8)=56

4q+16=56

4q=40

q=10

You should get the same value for q for both.

In the instance where the variables are x and y, which they usually are, you would write your answer as an ordered pair, (x, y).

(10, 8)