\left\{ \begin{array} { l } { 2 x + y = 20 } \\ { 6 x - 5 y = 12 } \end{array} \right.
Solve for x, y
x=7
y=6
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2x+y=20,6x-5y=12
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.
2x+y=20
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
2x=-y+20
Subtract y from both sides of the equation.
x=\frac{1}{2}\left(-y+20\right)
Divide both sides by 2.
x=-\frac{1}{2}y+10
Multiply \frac{1}{2} times -y+20.
6\left(-\frac{1}{2}y+10\right)-5y=12
Substitute -\frac{y}{2}+10 for x in the other equation, 6x-5y=12.
-3y+60-5y=12
Multiply 6 times -\frac{y}{2}+10.
-8y+60=12
Add -3y to -5y.
-8y=-48
Subtract 60 from both sides of the equation.
y=6
Divide both sides by -8.
x=-\frac{1}{2}\times 6+10
Substitute 6 for y in x=-\frac{1}{2}y+10. Because the resulting equation contains only one variable, you can solve for x directly.
x=-3+10
Multiply -\frac{1}{2} times 6.
x=7
Add 10 to -3.
x=7,y=6
The system is now solved.
2x+y=20,6x-5y=12
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}2&1\\6&-5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}20\\12\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}2&1\\6&-5\end{matrix}\right))\left(\begin{matrix}2&1\\6&-5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&1\\6&-5\end{matrix}\right))\left(\begin{matrix}20\\12\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&1\\6&-5\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}2&1\\6&-5\end{matrix}\right))\left(\begin{matrix}20\\12\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}2&1\\6&-5\end{matrix}\right))\left(\begin{matrix}20\\12\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{5}{2\left(-5\right)-6}&-\frac{1}{2\left(-5\right)-6}\\-\frac{6}{2\left(-5\right)-6}&\frac{2}{2\left(-5\right)-6}\end{matrix}\right)\left(\begin{matrix}20\\12\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}{16}&\frac{1}{16}\\\frac{3}{8}&-\frac{1}{8}\end{matrix}\right)\left(\begin{matrix}20\\12\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{16}\times 20+\frac{1}{16}\times 12\\\frac{3}{8}\times 20-\frac{1}{8}\times 12\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}7\\6\end{matrix}\right)
Do the arithmetic.
x=7,y=6
Extract the matrix elements x and y.
2x+y=20,6x-5y=12
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 2x+6y=6\times 20,2\times 6x+2\left(-5\right)y=2\times 12
To make 2x 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 2.
12x+6y=120,12x-10y=24
Simplify.
12x-12x+6y+10y=120-24
Subtract 12x-10y=24 from 12x+6y=120 by subtracting like terms on each side of the equal sign.
6y+10y=120-24
Add 12x to -12x. Terms 12x and -12x cancel out, leaving an equation with only one variable that can be solved.
16y=120-24
Add 6y to 10y.
16y=96
Add 120 to -24.
y=6
Divide both sides by 16.
6x-5\times 6=12
Substitute 6 for y in 6x-5y=12. Because the resulting equation contains only one variable, you can solve for x directly.
6x-30=12
Multiply -5 times 6.
6x=42
Add 30 to both sides of the equation.
x=7
Divide both sides by 6.
x=7,y=6
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}