Solve for x, y
x=8
y=15
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10x+6y=170,6x+10y=198
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.
10x+6y=170
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
10x=-6y+170
Subtract 6y from both sides of the equation.
x=\frac{1}{10}\left(-6y+170\right)
Divide both sides by 10.
x=-\frac{3}{5}y+17
Multiply \frac{1}{10} times -6y+170.
6\left(-\frac{3}{5}y+17\right)+10y=198
Substitute -\frac{3y}{5}+17 for x in the other equation, 6x+10y=198.
-\frac{18}{5}y+102+10y=198
Multiply 6 times -\frac{3y}{5}+17.
\frac{32}{5}y+102=198
Add -\frac{18y}{5} to 10y.
\frac{32}{5}y=96
Subtract 102 from both sides of the equation.
y=15
Divide both sides of the equation by \frac{32}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{3}{5}\times 15+17
Substitute 15 for y in x=-\frac{3}{5}y+17. Because the resulting equation contains only one variable, you can solve for x directly.
x=-9+17
Multiply -\frac{3}{5} times 15.
x=8
Add 17 to -9.
x=8,y=15
The system is now solved.
10x+6y=170,6x+10y=198
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}10&6\\6&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}170\\198\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}10&6\\6&10\end{matrix}\right))\left(\begin{matrix}10&6\\6&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&6\\6&10\end{matrix}\right))\left(\begin{matrix}170\\198\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}10&6\\6&10\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}10&6\\6&10\end{matrix}\right))\left(\begin{matrix}170\\198\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}10&6\\6&10\end{matrix}\right))\left(\begin{matrix}170\\198\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{10}{10\times 10-6\times 6}&-\frac{6}{10\times 10-6\times 6}\\-\frac{6}{10\times 10-6\times 6}&\frac{10}{10\times 10-6\times 6}\end{matrix}\right)\left(\begin{matrix}170\\198\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{5}{32}&-\frac{3}{32}\\-\frac{3}{32}&\frac{5}{32}\end{matrix}\right)\left(\begin{matrix}170\\198\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{32}\times 170-\frac{3}{32}\times 198\\-\frac{3}{32}\times 170+\frac{5}{32}\times 198\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}8\\15\end{matrix}\right)
Do the arithmetic.
x=8,y=15
Extract the matrix elements x and y.
10x+6y=170,6x+10y=198
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.
6\times 10x+6\times 6y=6\times 170,10\times 6x+10\times 10y=10\times 198
To make 10x and 6x equal, multiply all terms on each side of the first equation by 6 and all terms on each side of the second by 10.
60x+36y=1020,60x+100y=1980
Simplify.
60x-60x+36y-100y=1020-1980
Subtract 60x+100y=1980 from 60x+36y=1020 by subtracting like terms on each side of the equal sign.
36y-100y=1020-1980
Add 60x to -60x. Terms 60x and -60x cancel out, leaving an equation with only one variable that can be solved.
-64y=1020-1980
Add 36y to -100y.
-64y=-960
Add 1020 to -1980.
y=15
Divide both sides by -64.
6x+10\times 15=198
Substitute 15 for y in 6x+10y=198. Because the resulting equation contains only one variable, you can solve for x directly.
6x+150=198
Multiply 10 times 15.
6x=48
Subtract 150 from both sides of the equation.
x=8
Divide both sides by 6.
x=8,y=15
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}