Solve for x, y
x = -\frac{513}{83} = -6\frac{15}{83} \approx -6.180722892
y = \frac{819}{83} = 9\frac{72}{83} \approx 9.86746988
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-13x-20y=-117,14x-4y=-126
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.
-13x-20y=-117
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
-13x=20y-117
Add 20y to both sides of the equation.
x=-\frac{1}{13}\left(20y-117\right)
Divide both sides by -13.
x=-\frac{20}{13}y+9
Multiply -\frac{1}{13} times 20y-117.
14\left(-\frac{20}{13}y+9\right)-4y=-126
Substitute -\frac{20y}{13}+9 for x in the other equation, 14x-4y=-126.
-\frac{280}{13}y+126-4y=-126
Multiply 14 times -\frac{20y}{13}+9.
-\frac{332}{13}y+126=-126
Add -\frac{280y}{13} to -4y.
-\frac{332}{13}y=-252
Subtract 126 from both sides of the equation.
y=\frac{819}{83}
Divide both sides of the equation by -\frac{332}{13}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{20}{13}\times \frac{819}{83}+9
Substitute \frac{819}{83} for y in x=-\frac{20}{13}y+9. Because the resulting equation contains only one variable, you can solve for x directly.
x=-\frac{1260}{83}+9
Multiply -\frac{20}{13} times \frac{819}{83} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=-\frac{513}{83}
Add 9 to -\frac{1260}{83}.
x=-\frac{513}{83},y=\frac{819}{83}
The system is now solved.
-13x-20y=-117,14x-4y=-126
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}-13&-20\\14&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-117\\-126\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}-13&-20\\14&-4\end{matrix}\right))\left(\begin{matrix}-13&-20\\14&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-13&-20\\14&-4\end{matrix}\right))\left(\begin{matrix}-117\\-126\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}-13&-20\\14&-4\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}-13&-20\\14&-4\end{matrix}\right))\left(\begin{matrix}-117\\-126\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}-13&-20\\14&-4\end{matrix}\right))\left(\begin{matrix}-117\\-126\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{4}{-13\left(-4\right)-\left(-20\times 14\right)}&-\frac{-20}{-13\left(-4\right)-\left(-20\times 14\right)}\\-\frac{14}{-13\left(-4\right)-\left(-20\times 14\right)}&-\frac{13}{-13\left(-4\right)-\left(-20\times 14\right)}\end{matrix}\right)\left(\begin{matrix}-117\\-126\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{1}{83}&\frac{5}{83}\\-\frac{7}{166}&-\frac{13}{332}\end{matrix}\right)\left(\begin{matrix}-117\\-126\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{83}\left(-117\right)+\frac{5}{83}\left(-126\right)\\-\frac{7}{166}\left(-117\right)-\frac{13}{332}\left(-126\right)\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{513}{83}\\\frac{819}{83}\end{matrix}\right)
Do the arithmetic.
x=-\frac{513}{83},y=\frac{819}{83}
Extract the matrix elements x and y.
-13x-20y=-117,14x-4y=-126
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.
14\left(-13\right)x+14\left(-20\right)y=14\left(-117\right),-13\times 14x-13\left(-4\right)y=-13\left(-126\right)
To make -13x and 14x equal, multiply all terms on each side of the first equation by 14 and all terms on each side of the second by -13.
-182x-280y=-1638,-182x+52y=1638
Simplify.
-182x+182x-280y-52y=-1638-1638
Subtract -182x+52y=1638 from -182x-280y=-1638 by subtracting like terms on each side of the equal sign.
-280y-52y=-1638-1638
Add -182x to 182x. Terms -182x and 182x cancel out, leaving an equation with only one variable that can be solved.
-332y=-1638-1638
Add -280y to -52y.
-332y=-3276
Add -1638 to -1638.
y=\frac{819}{83}
Divide both sides by -332.
14x-4\times \frac{819}{83}=-126
Substitute \frac{819}{83} for y in 14x-4y=-126. Because the resulting equation contains only one variable, you can solve for x directly.
14x-\frac{3276}{83}=-126
Multiply -4 times \frac{819}{83}.
14x=-\frac{7182}{83}
Add \frac{3276}{83} to both sides of the equation.
x=-\frac{513}{83}
Divide both sides by 14.
x=-\frac{513}{83},y=\frac{819}{83}
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}