# SOLUTION: I don't know which category this problem would fall under but could someone help me out with a step by step explanation please. Solve the system by addition -5x+2y=10 x-3y=11

Algebra ->  Algebra  -> Coordinate Systems and Linear Equations -> SOLUTION: I don't know which category this problem would fall under but could someone help me out with a step by step explanation please. Solve the system by addition -5x+2y=10 x-3y=11      Log On

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 Click here to see ALL problems on Linear-systems Question 81762: I don't know which category this problem would fall under but could someone help me out with a step by step explanation please. Solve the system by addition -5x+2y=10 x-3y=11Answer by rapaljer(4667)   (Show Source): You can put this solution on YOUR website!In order to make one of the variables subtract out -5x+2y=10 x-3y=11 you might want to multiply both sides of the second equation times 5. Do you see that this would give you a "+5x" that will subtract out the "-5x" in the second equation? -5x+2y=10 5(x-3y)=5(11) -5x+2y=10 5x-15y=55 -13y=65 y=-5 Substitute y=-5 into either equation, the first one will do: -5x+2y=10 -5x+2(-5)=10 -5x -10=10 -5x -10+10 =10+10 -5x=20 x=-4 Now, check in the OTHER equation, that is, the second equation: x-3y=11 -4-3(-5) = 11 -4+15=11 It checks!! For more problems just like this IN COLOR, see my website by clicking on my tutor name "rapaljer" anywhere in algebra.com, and look for "MATH IN LIVING COLOR", then click either "Basic Algebra" or "Intermediate Algebra". Look for "Systems of Equations", which is in Chapter 4 of Basic Algebra or Chapter 5 of Intermediate Algebra. I hope this will be helpful to you. R^2 at SCC