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...

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