\left\{ \begin{array} { l } { 7 x - 8 y = 14 } \\ { 3 x - 2 y = 36 } \end{array} \right.
Solve for x, y
x=26
y=21
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7x-8y=14,3x-2y=36
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.
7x-8y=14
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
7x=8y+14
Add 8y to both sides of the equation.
x=\frac{1}{7}\left(8y+14\right)
Divide both sides by 7.
x=\frac{8}{7}y+2
Multiply \frac{1}{7} times 8y+14.
3\left(\frac{8}{7}y+2\right)-2y=36
Substitute \frac{8y}{7}+2 for x in the other equation, 3x-2y=36.
\frac{24}{7}y+6-2y=36
Multiply 3 times \frac{8y}{7}+2.
\frac{10}{7}y+6=36
Add \frac{24y}{7} to -2y.
\frac{10}{7}y=30
Subtract 6 from both sides of the equation.
y=21
Divide both sides of the equation by \frac{10}{7}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=\frac{8}{7}\times 21+2
Substitute 21 for y in x=\frac{8}{7}y+2. Because the resulting equation contains only one variable, you can solve for x directly.
x=24+2
Multiply \frac{8}{7} times 21.
x=26
Add 2 to 24.
x=26,y=21
The system is now solved.
7x-8y=14,3x-2y=36
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}7&-8\\3&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}14\\36\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}7&-8\\3&-2\end{matrix}\right))\left(\begin{matrix}7&-8\\3&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}7&-8\\3&-2\end{matrix}\right))\left(\begin{matrix}14\\36\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}7&-8\\3&-2\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}7&-8\\3&-2\end{matrix}\right))\left(\begin{matrix}14\\36\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}7&-8\\3&-2\end{matrix}\right))\left(\begin{matrix}14\\36\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{2}{7\left(-2\right)-\left(-8\times 3\right)}&-\frac{-8}{7\left(-2\right)-\left(-8\times 3\right)}\\-\frac{3}{7\left(-2\right)-\left(-8\times 3\right)}&\frac{7}{7\left(-2\right)-\left(-8\times 3\right)}\end{matrix}\right)\left(\begin{matrix}14\\36\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}{5}&\frac{4}{5}\\-\frac{3}{10}&\frac{7}{10}\end{matrix}\right)\left(\begin{matrix}14\\36\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{5}\times 14+\frac{4}{5}\times 36\\-\frac{3}{10}\times 14+\frac{7}{10}\times 36\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}26\\21\end{matrix}\right)
Do the arithmetic.
x=26,y=21
Extract the matrix elements x and y.
7x-8y=14,3x-2y=36
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.
3\times 7x+3\left(-8\right)y=3\times 14,7\times 3x+7\left(-2\right)y=7\times 36
To make 7x and 3x equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 7.
21x-24y=42,21x-14y=252
Simplify.
21x-21x-24y+14y=42-252
Subtract 21x-14y=252 from 21x-24y=42 by subtracting like terms on each side of the equal sign.
-24y+14y=42-252
Add 21x to -21x. Terms 21x and -21x cancel out, leaving an equation with only one variable that can be solved.
-10y=42-252
Add -24y to 14y.
-10y=-210
Add 42 to -252.
y=21
Divide both sides by -10.
3x-2\times 21=36
Substitute 21 for y in 3x-2y=36. Because the resulting equation contains only one variable, you can solve for x directly.
3x-42=36
Multiply -2 times 21.
3x=78
Add 42 to both sides of the equation.
x=26
Divide both sides by 3.
x=26,y=21
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}