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Mathematics 11 Online
OpenStudy (anonymous):

Please help me find the vaule of X and Y!! 5x-2y=-15 3x+8y=37

OpenStudy (anonymous):

step1; multiply equation 1 with 4 step2; add the two equations

OpenStudy (anonymous):

?

OpenStudy (johnweldon1993):

when you do as @Kanwar245 has said you get 20x - 8y = -60 and 3x + 8y = 37 when he said add the 2 equations...look what happens when you do that...the -8y and 8y cancel....all you have left is the x's so that's how you solve for x 23x = -23 x = -1 if you wanted to solve for y first what you want to do ...is multiply the whole top equation by -3 and then multiply the whole bottom equation by 5 this will give you -15x +6y = 45 15x+40y=185 now when you add the 2 equations ...the x's cancel out...and all you have is a y 46y = 230 y = 5 you have your x and your y :)

OpenStudy (anonymous):

thank you SO much! I have another equation to ask, if thats okay? What's the value of X and Y if the equation is: -x+y=1 x+y=11

OpenStudy (johnweldon1993):

no problem :) and heres a substitution question we have -x + y = 1 x + y = 11 solve for y in the first equation -x + y = 1 add x to both sides...makes this become y = x + 1 now you have what "y" equals right? plus that into your second equation where you see a y x + y = 11 x + (x + 1) = 11 subtract 1 from both sides x + x = 10 x+x = 2x right? 2x = 10 divide both sides by 2 x = 5 so now we have x...plug that into the 1st equation wherever you see an "x" -x + y = 1 -(5) + y = 1 -5 + y = 1 add 5 to both sides y = 6 plug BOTH numbers into the second equation to check x + y = 11 (5) + (6) = 11 5 + 6 = 11 11 = 11? yes it does so x = 5 and y = 6 :)

OpenStudy (johnweldon1993):

and also you can plug both numbers back into the first equation as well -x + y = 1 -(5) + (6) = 1 1 = 1...yes again....so we have the correct solution :)

OpenStudy (anonymous):

you're awesome and totally a lifesaver! thank you thank you thank you (:

OpenStudy (johnweldon1993):

lol anytime :) feel free to fan me and message me if you need help with anything else :)

OpenStudy (anonymous):

will do (:

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