Solve for x, y
x=-3
y=4
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23x-25y=-169,25x-23y=-167
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.
23x-25y=-169
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
23x=25y-169
Add 25y to both sides of the equation.
x=\frac{1}{23}\left(25y-169\right)
Divide both sides by 23.
x=\frac{25}{23}y-\frac{169}{23}
Multiply \frac{1}{23} times 25y-169.
25\left(\frac{25}{23}y-\frac{169}{23}\right)-23y=-167
Substitute \frac{25y-169}{23} for x in the other equation, 25x-23y=-167.
\frac{625}{23}y-\frac{4225}{23}-23y=-167
Multiply 25 times \frac{25y-169}{23}.
\frac{96}{23}y-\frac{4225}{23}=-167
Add \frac{625y}{23} to -23y.
\frac{96}{23}y=\frac{384}{23}
Add \frac{4225}{23} to both sides of the equation.
y=4
Divide both sides of the equation by \frac{96}{23}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=\frac{25}{23}\times 4-\frac{169}{23}
Substitute 4 for y in x=\frac{25}{23}y-\frac{169}{23}. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{100-169}{23}
Multiply \frac{25}{23} times 4.
x=-3
Add -\frac{169}{23} to \frac{100}{23} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=-3,y=4
The system is now solved.
23x-25y=-169,25x-23y=-167
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}23&-25\\25&-23\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-169\\-167\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}23&-25\\25&-23\end{matrix}\right))\left(\begin{matrix}23&-25\\25&-23\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}23&-25\\25&-23\end{matrix}\right))\left(\begin{matrix}-169\\-167\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}23&-25\\25&-23\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}23&-25\\25&-23\end{matrix}\right))\left(\begin{matrix}-169\\-167\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}23&-25\\25&-23\end{matrix}\right))\left(\begin{matrix}-169\\-167\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{23}{23\left(-23\right)-\left(-25\times 25\right)}&-\frac{-25}{23\left(-23\right)-\left(-25\times 25\right)}\\-\frac{25}{23\left(-23\right)-\left(-25\times 25\right)}&\frac{23}{23\left(-23\right)-\left(-25\times 25\right)}\end{matrix}\right)\left(\begin{matrix}-169\\-167\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{23}{96}&\frac{25}{96}\\-\frac{25}{96}&\frac{23}{96}\end{matrix}\right)\left(\begin{matrix}-169\\-167\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{23}{96}\left(-169\right)+\frac{25}{96}\left(-167\right)\\-\frac{25}{96}\left(-169\right)+\frac{23}{96}\left(-167\right)\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-3\\4\end{matrix}\right)
Do the arithmetic.
x=-3,y=4
Extract the matrix elements x and y.
23x-25y=-169,25x-23y=-167
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.
25\times 23x+25\left(-25\right)y=25\left(-169\right),23\times 25x+23\left(-23\right)y=23\left(-167\right)
To make 23x and 25x equal, multiply all terms on each side of the first equation by 25 and all terms on each side of the second by 23.
575x-625y=-4225,575x-529y=-3841
Simplify.
575x-575x-625y+529y=-4225+3841
Subtract 575x-529y=-3841 from 575x-625y=-4225 by subtracting like terms on each side of the equal sign.
-625y+529y=-4225+3841
Add 575x to -575x. Terms 575x and -575x cancel out, leaving an equation with only one variable that can be solved.
-96y=-4225+3841
Add -625y to 529y.
-96y=-384
Add -4225 to 3841.
y=4
Divide both sides by -96.
25x-23\times 4=-167
Substitute 4 for y in 25x-23y=-167. Because the resulting equation contains only one variable, you can solve for x directly.
25x-92=-167
Multiply -23 times 4.
25x=-75
Add 92 to both sides of the equation.
x=-3
Divide both sides by 25.
x=-3,y=4
The system is now solved.
Examples
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{ 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}