2x-2y= 10 x-y/3 = 3 solve for x and y

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2x-2y=10......(1)

x-y/3=3........(2)

We will use the substitution method to solve the system:

From (2):

X= y/3 + 3

Now substitute in (1):

2x-2y = 10

2(y/3+3) -2y =10

2y/3 + 6 -2y=10

-4y/3 = 4

y= 4*3/-4 = -3

Then x= y/3 + 3= -3/3 + 3= 2

x= 2 and y= -3

2x-2y= 10 equation1

x-y/3 = 3 equation2 solve for x and y

I am going to solve this by elimination. Since both equations are in standard form already I am going to multiply equation2 by -2 to eliminate the x variable -this gives me

2x-2y=10

-2x+(2y)/3=6 then if we add the two equations we get-2y+(2y)/3=16 Now if we multiply this equation by 3 we can eliminate the fraction3(-2y+(2y)/3=16) => -6y+2y=48combine like terms to get-4y=48divide both sides of the equation by -4 to gety=-12Substitute -12 for y in original equation 1

2x-2(-12)=10simplify2x+24=10subtract 24 from both sides2x=-14divide both sides by 2x=-7Solution to the system of equations is the point (-7,-12)

Check by substituting x and y into both original equations

2(-7)-2(-12)= 10 (-7)-(-12)/3 = 3 perform arithmetic to see if both equations true

-14+24=10 -7+4=3

10=10 check 3=3 check

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