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07272007  #1 
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algebra 1 homework help.?
solving systems of linear equations1.y=2x+13x4y=12.5x6y=23x=63y3.x+y=12x3y=84.x+2y=32x+y=55.y3x=53(x+1)=y26.2y=14x2x+y=0

07272007  #2 
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1)y = 2x + 13x  4(2x + 1) = 1y = 2x + 15x = 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(1x) = 8y = 1  x5x = 11y = 1  11/5x = 11/5y = 6/5x = 11/54)x = 3  2y2(32y) + y = 5x = 3  2y6  4y + y = 5x = 3  2y3y = 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!

07272007  #3 
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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 3x2x1=1, then x1=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.

07272007  #4 
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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!

07272007  #5 
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starting w/ the first one you can use substitution:3x4(2x+1)=13x8x4=15x=5x=1once u figure either of the variables u plug it to figure out the other one

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