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3y=10-2
Consider the second equation. Subtract 2 from both sides.
3y=8
Subtract 2 from 10 to get 8.
y=\frac{8}{3}
Divide both sides by 3.
2x+5\times \frac{8}{3}=9
Consider the first equation. Insert the known values of variables into the equation.
2x+\frac{40}{3}=9
Multiply 5 and \frac{8}{3} to get \frac{40}{3}.
2x=9-\frac{40}{3}
Subtract \frac{40}{3} from both sides.
2x=-\frac{13}{3}
Subtract \frac{40}{3} from 9 to get -\frac{13}{3}.
x=\frac{-\frac{13}{3}}{2}
Divide both sides by 2.
x=\frac{-13}{3\times 2}
Express \frac{-\frac{13}{3}}{2} as a single fraction.
x=\frac{-13}{6}
Multiply 3 and 2 to get 6.
x=-\frac{13}{6}
Fraction \frac{-13}{6} can be rewritten as -\frac{13}{6} by extracting the negative sign.
x=-\frac{13}{6} y=\frac{8}{3}
The system is now solved.