how would you solver this system of equations.....?

Question

i solver it by elimination and for some reason got the wrong answer i don't know why. how would you solve this?

720 = 22x + 28y
30 = x + y

Answer 1

Multiply the second equation by 22:

660 = 22x + 22y

Subtract that from the first equation:

60 = 6y

y = 60/6

y = 10

Multiply the second equation by 28:

840 = 28x + 28y

Subtract the first equation from that:

120 = 6x

x = 120/6

x = 20



See also http://jeff.aaron.ca/cgi-bin/equations

Answer 2

First let's simplify the first equation.



720 = 22x + 28y



Divide both sides by 2.



360 = 11x + 14 y



11 is prime, so unfortunately the simplification ends here.



360 = 11x + 14 y



30 = x + y



Isolate x above. 30 - y = x. Plug that in in the other equation:



360 = 11x + 14 y



360 = 11 ( 30 - y ) + 14 y



360 = 330 - 11 y + 14 y

360 = 330 + 3 y



30 = 3 y

y = 10



Since x = 30 - y, x = 30 - 10, or 20.

Answer 3

I would use substitution.

x=30-y

720=22(30-y)+28y

720=660-22y+28y

60=6y

y=10

substituting back in gives x=20

the final answer is (20, 10)

Answer 4

11x + 14y = 360

- 11x - 11y = - 330----ADD



3y = 30

y = 10



30 = x + 10

x = 20



x = 20 , y = 10

Answer 5

720=22(30-y)+28y

720=660-22y+28y

720=660+6y

60=6y

y=10



x=30-10=20

x=20