i have another algebra problem: solve linear systems by elimination?

I don't remember it perfectly but i'm pretty sure it's like this:

6x - 8y =36 From that you get that:

- 6x - y =15 -7y=21, so you divide by -7

and you get that y=-3.

0 - 7y =21

To solve x replace y in any equation: 6x +3 =15 (- -3 is +3)

6x=12, divide by 6 and you get that x=2 so the answer is (2,-3)

Second one:

2x - y = -11

+ -2x - 13 = y

0 -y-13= -11+y

-y - 13=-11 +y -Add y to both sides to get:

-13= -11 =2y -Add 11 to both and you get:

-2 =2y -Divide by 2

-1=y

Replace any equation:

2x - y = - 11

2x +1= -11 -As with before - -1 is +1. You substract one to get:

2x= -12 -Divide by 2 and you get that:

x= -6

Making the solution (-6,-1)

right here thake 2x-y=32 and subtract y making it 2x=32+y Then take y-5x=13 and subtract y making it 5x=13-y the y's cancel then you definately upload the relax issues togeather so ur answer could desire to be7x=40 5 then you definately divide so the respond is x=6.40 two then you definately in simple terms bypass on from there wish i helped!;-)

try mymaths.co.uk take the lessons and you can also choose what level