\left\{ \begin{array} { l } { 5 x - y = 110 } \\ { 9 y - x = 110 } \end{array} \right.
Solve for x, y
x=25
y=15
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5x-y=110,-x+9y=110
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.
5x-y=110
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
5x=y+110
Add y to both sides of the equation.
x=\frac{1}{5}\left(y+110\right)
Divide both sides by 5.
x=\frac{1}{5}y+22
Multiply \frac{1}{5} times y+110.
-\left(\frac{1}{5}y+22\right)+9y=110
Substitute \frac{y}{5}+22 for x in the other equation, -x+9y=110.
-\frac{1}{5}y-22+9y=110
Multiply -1 times \frac{y}{5}+22.
\frac{44}{5}y-22=110
Add -\frac{y}{5} to 9y.
\frac{44}{5}y=132
Add 22 to both sides of the equation.
y=15
Divide both sides of the equation by \frac{44}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=\frac{1}{5}\times 15+22
Substitute 15 for y in x=\frac{1}{5}y+22. Because the resulting equation contains only one variable, you can solve for x directly.
x=3+22
Multiply \frac{1}{5} times 15.
x=25
Add 22 to 3.
x=25,y=15
The system is now solved.
5x-y=110,-x+9y=110
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}5&-1\\-1&9\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}110\\110\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}5&-1\\-1&9\end{matrix}\right))\left(\begin{matrix}5&-1\\-1&9\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-1\\-1&9\end{matrix}\right))\left(\begin{matrix}110\\110\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&-1\\-1&9\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}5&-1\\-1&9\end{matrix}\right))\left(\begin{matrix}110\\110\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}5&-1\\-1&9\end{matrix}\right))\left(\begin{matrix}110\\110\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{9}{5\times 9-\left(-\left(-1\right)\right)}&-\frac{-1}{5\times 9-\left(-\left(-1\right)\right)}\\-\frac{-1}{5\times 9-\left(-\left(-1\right)\right)}&\frac{5}{5\times 9-\left(-\left(-1\right)\right)}\end{matrix}\right)\left(\begin{matrix}110\\110\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{9}{44}&\frac{1}{44}\\\frac{1}{44}&\frac{5}{44}\end{matrix}\right)\left(\begin{matrix}110\\110\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{9}{44}\times 110+\frac{1}{44}\times 110\\\frac{1}{44}\times 110+\frac{5}{44}\times 110\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}25\\15\end{matrix}\right)
Do the arithmetic.
x=25,y=15
Extract the matrix elements x and y.
5x-y=110,-x+9y=110
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.
-5x-\left(-y\right)=-110,5\left(-1\right)x+5\times 9y=5\times 110
To make 5x and -x equal, multiply all terms on each side of the first equation by -1 and all terms on each side of the second by 5.
-5x+y=-110,-5x+45y=550
Simplify.
-5x+5x+y-45y=-110-550
Subtract -5x+45y=550 from -5x+y=-110 by subtracting like terms on each side of the equal sign.
y-45y=-110-550
Add -5x to 5x. Terms -5x and 5x cancel out, leaving an equation with only one variable that can be solved.
-44y=-110-550
Add y to -45y.
-44y=-660
Add -110 to -550.
y=15
Divide both sides by -44.
-x+9\times 15=110
Substitute 15 for y in -x+9y=110. Because the resulting equation contains only one variable, you can solve for x directly.
-x+135=110
Multiply 9 times 15.
-x=-25
Subtract 135 from both sides of the equation.
x=25
Divide both sides by -1.
x=25,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}