\left\{ \begin{array} { l } { \frac { x - y } { 5 } - \frac { y } { 2 } = x - 1 } \\ { \frac { x } { 3 } + \frac { y + 2 } { 2 } = 1 } \end{array} \right.
Solve for x, y
x=3
y=-2
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2\left(x-y\right)-5y=10x-10
Consider the first equation. Multiply both sides of the equation by 10, the least common multiple of 5,2.
2x-2y-5y=10x-10
Use the distributive property to multiply 2 by x-y.
2x-7y=10x-10
Combine -2y and -5y to get -7y.
2x-7y-10x=-10
Subtract 10x from both sides.
-8x-7y=-10
Combine 2x and -10x to get -8x.
2x+3\left(y+2\right)=6
Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.
2x+3y+6=6
Use the distributive property to multiply 3 by y+2.
2x+3y=6-6
Subtract 6 from both sides.
2x+3y=0
Subtract 6 from 6 to get 0.
-8x-7y=-10,2x+3y=0
To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
-8x-7y=-10
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
-8x=7y-10
Add 7y to both sides of the equation.
x=-\frac{1}{8}\left(7y-10\right)
Divide both sides by -8.
x=-\frac{7}{8}y+\frac{5}{4}
Multiply -\frac{1}{8} times 7y-10.
2\left(-\frac{7}{8}y+\frac{5}{4}\right)+3y=0
Substitute -\frac{7y}{8}+\frac{5}{4} for x in the other equation, 2x+3y=0.
-\frac{7}{4}y+\frac{5}{2}+3y=0
Multiply 2 times -\frac{7y}{8}+\frac{5}{4}.
\frac{5}{4}y+\frac{5}{2}=0
Add -\frac{7y}{4} to 3y.
\frac{5}{4}y=-\frac{5}{2}
Subtract \frac{5}{2} from both sides of the equation.
y=-2
Divide both sides of the equation by \frac{5}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{7}{8}\left(-2\right)+\frac{5}{4}
Substitute -2 for y in x=-\frac{7}{8}y+\frac{5}{4}. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{7+5}{4}
Multiply -\frac{7}{8} times -2.
x=3
Add \frac{5}{4} to \frac{7}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=3,y=-2
The system is now solved.
2\left(x-y\right)-5y=10x-10
Consider the first equation. Multiply both sides of the equation by 10, the least common multiple of 5,2.
2x-2y-5y=10x-10
Use the distributive property to multiply 2 by x-y.
2x-7y=10x-10
Combine -2y and -5y to get -7y.
2x-7y-10x=-10
Subtract 10x from both sides.
-8x-7y=-10
Combine 2x and -10x to get -8x.
2x+3\left(y+2\right)=6
Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.
2x+3y+6=6
Use the distributive property to multiply 3 by y+2.
2x+3y=6-6
Subtract 6 from both sides.
2x+3y=0
Subtract 6 from 6 to get 0.
-8x-7y=-10,2x+3y=0
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}-8&-7\\2&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-10\\0\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}-8&-7\\2&3\end{matrix}\right))\left(\begin{matrix}-8&-7\\2&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-8&-7\\2&3\end{matrix}\right))\left(\begin{matrix}-10\\0\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}-8&-7\\2&3\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-8&-7\\2&3\end{matrix}\right))\left(\begin{matrix}-10\\0\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-8&-7\\2&3\end{matrix}\right))\left(\begin{matrix}-10\\0\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{-8\times 3-\left(-7\times 2\right)}&-\frac{-7}{-8\times 3-\left(-7\times 2\right)}\\-\frac{2}{-8\times 3-\left(-7\times 2\right)}&-\frac{8}{-8\times 3-\left(-7\times 2\right)}\end{matrix}\right)\left(\begin{matrix}-10\\0\end{matrix}\right)
For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{10}&-\frac{7}{10}\\\frac{1}{5}&\frac{4}{5}\end{matrix}\right)\left(\begin{matrix}-10\\0\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{10}\left(-10\right)\\\frac{1}{5}\left(-10\right)\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\\-2\end{matrix}\right)
Do the arithmetic.
x=3,y=-2
Extract the matrix elements x and y.
2\left(x-y\right)-5y=10x-10
Consider the first equation. Multiply both sides of the equation by 10, the least common multiple of 5,2.
2x-2y-5y=10x-10
Use the distributive property to multiply 2 by x-y.
2x-7y=10x-10
Combine -2y and -5y to get -7y.
2x-7y-10x=-10
Subtract 10x from both sides.
-8x-7y=-10
Combine 2x and -10x to get -8x.
2x+3\left(y+2\right)=6
Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.
2x+3y+6=6
Use the distributive property to multiply 3 by y+2.
2x+3y=6-6
Subtract 6 from both sides.
2x+3y=0
Subtract 6 from 6 to get 0.
-8x-7y=-10,2x+3y=0
In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.
2\left(-8\right)x+2\left(-7\right)y=2\left(-10\right),-8\times 2x-8\times 3y=0
To make -8x and 2x equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by -8.
-16x-14y=-20,-16x-24y=0
Simplify.
-16x+16x-14y+24y=-20
Subtract -16x-24y=0 from -16x-14y=-20 by subtracting like terms on each side of the equal sign.
-14y+24y=-20
Add -16x to 16x. Terms -16x and 16x cancel out, leaving an equation with only one variable that can be solved.
10y=-20
Add -14y to 24y.
y=-2
Divide both sides by 10.
2x+3\left(-2\right)=0
Substitute -2 for y in 2x+3y=0. Because the resulting equation contains only one variable, you can solve for x directly.
2x-6=0
Multiply 3 times -2.
2x=6
Add 6 to both sides of the equation.
x=3
Divide both sides by 2.
x=3,y=-2
The system is now solved.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}