Two more algebra 1 problems before quiz tommorow?

Question

Sorry. It seems like there's another step for EVERY problem.

Systems of Equations using linear combinations

Need help on these two.

3x=13-y

2x-y=2

this is the first one.

second:

2x+y=6

3x-7y=9

Please help. I have no clue what I am doing and I am going to be quized tommorow.

Answer 1

First one:

Move the y to the other side:

3x + y =13

2x - y = 2

Add them together, y's cancel, solve for x and you get

x = 3.

Plug that back in to either of the equations and you get

y = 4.

Second one:

Multiply the first by 7. You get:

14x + 7y = 42

3x - 7y = 9

You add them together to cancel out the y's and you get

x = 3

Plug that back in and

y = 0

Answer 2

x=13 - y

....--------

......3

so

.....13 - y

2 (..-------..) - y = 2

........3

26 - 2y

(--------)...- y = 2

....3

26 - 2y - 3y

----------------..= 2

.......3

26 - 5y

---------....= 2

...3

26 - 5y = 2 (3)

26 - 5y = 6

-5y = -20

y = -4

so now

3x = 13 - (-4)

3x = 13 + 4

3x = 17

x = 17/3 [that's 17 over 3 ;-)]

(17/3, -4)

I'll set you up for the next one:

find y for the first equation:

2x + y = 6

y = 6 - 2x

Now plug it into the second equation.

3x - 7(6 - 2x) = 9

It's fun!