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3x+2=12
Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 2,3.
3x=12-2
Subtract 2 from both sides.
3x=10
Subtract 2 from 12 to get 10.
x=\frac{10}{3}
Divide both sides by 3.
3\times \frac{10}{3}-2\times 2y=2
Consider the second equation. Insert the known values of variables into the equation.
10-2\times 2y=2
Multiply 3 and \frac{10}{3} to get 10.
10-4y=2
Multiply 2 and 2 to get 4.
-4y=2-10
Subtract 10 from both sides.
-4y=-8
Subtract 10 from 2 to get -8.
y=\frac{-8}{-4}
Divide both sides by -4.
y=2
Divide -8 by -4 to get 2.
x=\frac{10}{3} y=2
The system is now solved.