\left\{ \begin{array} { l } { 35 x - 27 y = 43 } \\ { 27 x - 35 y = 19 } \end{array} \right.
Solve for x, y
x=2
y=1
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35x-27y=43,27x-35y=19
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.
35x-27y=43
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
35x=27y+43
Add 27y to both sides of the equation.
x=\frac{1}{35}\left(27y+43\right)
Divide both sides by 35.
x=\frac{27}{35}y+\frac{43}{35}
Multiply \frac{1}{35} times 27y+43.
27\left(\frac{27}{35}y+\frac{43}{35}\right)-35y=19
Substitute \frac{27y+43}{35} for x in the other equation, 27x-35y=19.
\frac{729}{35}y+\frac{1161}{35}-35y=19
Multiply 27 times \frac{27y+43}{35}.
-\frac{496}{35}y+\frac{1161}{35}=19
Add \frac{729y}{35} to -35y.
-\frac{496}{35}y=-\frac{496}{35}
Subtract \frac{1161}{35} from both sides of the equation.
y=1
Divide both sides of the equation by -\frac{496}{35}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=\frac{27+43}{35}
Substitute 1 for y in x=\frac{27}{35}y+\frac{43}{35}. Because the resulting equation contains only one variable, you can solve for x directly.
x=2
Add \frac{43}{35} to \frac{27}{35} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=2,y=1
The system is now solved.
35x-27y=43,27x-35y=19
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}35&-27\\27&-35\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}43\\19\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}35&-27\\27&-35\end{matrix}\right))\left(\begin{matrix}35&-27\\27&-35\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}35&-27\\27&-35\end{matrix}\right))\left(\begin{matrix}43\\19\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}35&-27\\27&-35\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}35&-27\\27&-35\end{matrix}\right))\left(\begin{matrix}43\\19\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}35&-27\\27&-35\end{matrix}\right))\left(\begin{matrix}43\\19\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{35}{35\left(-35\right)-\left(-27\times 27\right)}&-\frac{-27}{35\left(-35\right)-\left(-27\times 27\right)}\\-\frac{27}{35\left(-35\right)-\left(-27\times 27\right)}&\frac{35}{35\left(-35\right)-\left(-27\times 27\right)}\end{matrix}\right)\left(\begin{matrix}43\\19\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{35}{496}&-\frac{27}{496}\\\frac{27}{496}&-\frac{35}{496}\end{matrix}\right)\left(\begin{matrix}43\\19\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{35}{496}\times 43-\frac{27}{496}\times 19\\\frac{27}{496}\times 43-\frac{35}{496}\times 19\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2\\1\end{matrix}\right)
Do the arithmetic.
x=2,y=1
Extract the matrix elements x and y.
35x-27y=43,27x-35y=19
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.
27\times 35x+27\left(-27\right)y=27\times 43,35\times 27x+35\left(-35\right)y=35\times 19
To make 35x and 27x equal, multiply all terms on each side of the first equation by 27 and all terms on each side of the second by 35.
945x-729y=1161,945x-1225y=665
Simplify.
945x-945x-729y+1225y=1161-665
Subtract 945x-1225y=665 from 945x-729y=1161 by subtracting like terms on each side of the equal sign.
-729y+1225y=1161-665
Add 945x to -945x. Terms 945x and -945x cancel out, leaving an equation with only one variable that can be solved.
496y=1161-665
Add -729y to 1225y.
496y=496
Add 1161 to -665.
y=1
Divide both sides by 496.
27x-35=19
Substitute 1 for y in 27x-35y=19. Because the resulting equation contains only one variable, you can solve for x directly.
27x=54
Add 35 to both sides of the equation.
x=2
Divide both sides by 27.
x=2,y=1
The system is now solved.
Examples
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{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
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Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
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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}