algebra 1 homework help.?

solving systems of linear equations1.y=2x+13x-4y=12.5x-6y=23x=6-3y3.x+y=12x-3y=84.x+2y=32x+y=55.y-3x=53(x+1)=y-26.2y=1-4x2x+y=0

1)y = 2x + 13x - 4(2x + 1) = 1y = 2x + 1-5x = 5y = 2(-1) + 1x = -1x = -1y = -12)5(6 - 3y) - 6y = 23x = 6 - 3y30 - 15y - 6y = 23x = 6 - 3yy = 1/3x = 6 - 3(1/3) y = 1/3x = 53)x + y = 12x - 3y = 8y = 1 - x2x - 3(1-x) = 8y = 1 - x5x = 11y = 1 - 11/5x = 11/5y = -6/5x = 11/54)x = 3 - 2y2(3-2y) + y = 5x = 3 - 2y6 - 4y + y = 5x = 3 - 2y-3y = -1x = 3 - 2(1/3)y = 1/3x = 7/3y = 1/35)y - 3x = 53(x+1) = y - 2y = 5 + 3x3x + 3 = 5 + 3x - 2y = 5 + 3x0x = 0Indeterminated6)2y = 1 - 4x2x + y = 0y = 1/2 - 2x2x + 1/2 - 2x = 0y = 1/2 - 2x0x = -1/2Impossible!

For number one just plug in the first equation into the second. You know what y is in terms of x so just shove it in about solve for x, then plug x in to solve for y. So 3x-2x-1=1, then x-1=1, so x=2. Then you can put that value into either equation and you find that y=5. So your solution is (2,5).Just do that for the others.

I'm working on a "take home final" from my college Algb. class. I can't take math and speak it!!!!! I'm going to be calling on all my Math coworkers!!!! Good luck!

starting w/ the first one you can use substitution:3x-4(2x+1)=13x-8x-4=1-5x=5x=-1once u figure either of the variables u plug it to figure out the other one