Solve by substitution (two systems of equations)

[Guest]

Can someone tell me if these are correct by substitution please? Thanks. Andrea

1) 5x + y = 10
. . .3x – 3y = 6

Divide the second equation by 3, and solve for x:
. . .x – y = 2
. . .x = y + 2

Substitute into the first equation, and solve for y:
. . .5(y + 2) + y = 10
. . .5y + 10 + y = 10
. . .6y = 0
. . .y = 0

Back-substitute to solve for x:
. . .3x – 3(0) = 6
. . .3x = 6
. . .x = 2

2) y = 13x + 14
. . .x = y – 1

Substitute the first equation into the second in place of y:
. . .x = 13x + 14 – 1
. . .x – 13x = 13
. . .-12x = 13
. . .x = -13/12

Back-substitute to solve for y:
. . .y = 13(-13/12) + 14
. . .y = (156 – 13 + 168) / 12
. . .y = 311 / 12
. . .y = 25 and 11 / 12ths

 

[stapel]

You can check your answers by plugging them back into the original systems:

1) Checking (x, y) = (2, 0):

. . . . .5x + y = 10
. . . . .5(2) + (0) ?=? 10
. . . . .10 + 0 ?=? 10
. . . . .10 = 10

. . . . .3x - 3y = 6
. . . . .3(2) - 3(0) ?=? 6
. . . . .6 - 0 ?=? 6
. . . . .6 = 6

The solution "checks" in each of the two equations, so the solution is valid. Good work!

2) Checking (x, y) = (-13/12, 311/12):

. . . . .y = 13x + 14
. . . . .311/12 ?=? 13(-13/12) + 168/12
. . . . .311/12 ?=? -169/12 + 168/12
. . . . .311/12 ?=? -1/12

The solution does not "check", so there must be an error somewhere.

Plug the first equation into the second in place of "y":

. . . . .x = (13x + 14) - 1
. . . . .x = 13x + 13
. . . . .-13 = 12x
. . . . .-13/12 = x

So this still looks good. Continuing:

. . . . .y = 13x + 14
. . . . .y = 13(-13/12) + 14
. . . . .y = -169/12 + 168/12

...and so forth. Remember to check your new solution in both equations.

Eliz.

P.S. Thank you for showing your work!!

 

[Guest]

I redid my math in the second problem and see where I screwed up. I show x=-13/12 and y=-1/12 and when you plug them into each of the original equations they come out true.

Thanks for looking at these for me.
Andrea