Question

Olve the system of equations below.

3x+4y=10
6x-2y=40

(6,-2)

(2,6)

(2,-6)

(-2,-6)

Answer 1

The correct solutions are (6, -2).

Explanation:

For the first equation, rearrange to make x the subject.

3x + 4y = 10

`3x = 10 -4y`

Divide the whole equation by 3 to isolate x:

`3x = 10 -4y\\3x / 3 = (10)/(3) - (4)/(3)y\\x = (10)/(3) - (4)/(3)y\\`

Now substitute this into the second equation:

`6x - 2y = 40\\6((10)/(3) - (4)/(3)y) - 2y = 40\\6((10)/(3)) + 6((4)/(3)) - 2y = 40`

`20 - 8y - 2y = 40\\20 - 10y = 40`

Subtract 20 from both sides:

20 - 10y = 40

20 - 10y - 20 = 40 - 20

-10y = 20

Divide both sides by 2:

-10y ÷ 10 = 20 ÷ 10

-y = 2 ∴ y = -2

Plug this value back into the first equation:

3x + 4y = 10

3x + 4(-2) = 10

3x + (-8) = 10

3x - 8 = 10

Add 8 to both sides:

3x - 8 + 8 = 10 + 8

3x = 18

Divide both sides by 3:

3x ÷ 3 = 18 ÷ 3

x = 6

Therefore, the correct solutions are (6, -2).

Hope this helps!