\left\{ \begin{array} { l } { 2 x - 5 y = 25 } \\ { 6 x - 8 y = 54 } \end{array} \right.
Solve for x, y
x=5
y=-3
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2x-5y=25,6x-8y=54
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-5y=25
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
2x=5y+25
Add 5y to both sides of the equation.
x=\frac{1}{2}\left(5y+25\right)
Divide both sides by 2.
x=\frac{5}{2}y+\frac{25}{2}
Multiply \frac{1}{2} times 25+5y.
6\left(\frac{5}{2}y+\frac{25}{2}\right)-8y=54
Substitute \frac{25+5y}{2} for x in the other equation, 6x-8y=54.
15y+75-8y=54
Multiply 6 times \frac{25+5y}{2}.
7y+75=54
Add 15y to -8y.
7y=-21
Subtract 75 from both sides of the equation.
y=-3
Divide both sides by 7.
x=\frac{5}{2}\left(-3\right)+\frac{25}{2}
Substitute -3 for y in x=\frac{5}{2}y+\frac{25}{2}. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{-15+25}{2}
Multiply \frac{5}{2} times -3.
x=5
Add \frac{25}{2} to -\frac{15}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=5,y=-3
The system is now solved.
2x-5y=25,6x-8y=54
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}2&-5\\6&-8\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}25\\54\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}2&-5\\6&-8\end{matrix}\right))\left(\begin{matrix}2&-5\\6&-8\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-5\\6&-8\end{matrix}\right))\left(\begin{matrix}25\\54\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-5\\6&-8\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&-5\\6&-8\end{matrix}\right))\left(\begin{matrix}25\\54\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&-5\\6&-8\end{matrix}\right))\left(\begin{matrix}25\\54\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{8}{2\left(-8\right)-\left(-5\times 6\right)}&-\frac{-5}{2\left(-8\right)-\left(-5\times 6\right)}\\-\frac{6}{2\left(-8\right)-\left(-5\times 6\right)}&\frac{2}{2\left(-8\right)-\left(-5\times 6\right)}\end{matrix}\right)\left(\begin{matrix}25\\54\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{4}{7}&\frac{5}{14}\\-\frac{3}{7}&\frac{1}{7}\end{matrix}\right)\left(\begin{matrix}25\\54\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{7}\times 25+\frac{5}{14}\times 54\\-\frac{3}{7}\times 25+\frac{1}{7}\times 54\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}5\\-3\end{matrix}\right)
Do the arithmetic.
x=5,y=-3
Extract the matrix elements x and y.
2x-5y=25,6x-8y=54
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+6\left(-5\right)y=6\times 25,2\times 6x+2\left(-8\right)y=2\times 54
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-30y=150,12x-16y=108
Simplify.
12x-12x-30y+16y=150-108
Subtract 12x-16y=108 from 12x-30y=150 by subtracting like terms on each side of the equal sign.
-30y+16y=150-108
Add 12x to -12x. Terms 12x and -12x cancel out, leaving an equation with only one variable that can be solved.
-14y=150-108
Add -30y to 16y.
-14y=42
Add 150 to -108.
y=-3
Divide both sides by -14.
6x-8\left(-3\right)=54
Substitute -3 for y in 6x-8y=54. Because the resulting equation contains only one variable, you can solve for x directly.
6x+24=54
Multiply -8 times -3.
6x=30
Subtract 24 from both sides of the equation.
x=5
Divide both sides by 6.
x=5,y=-3
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}