\left\{ \begin{array} { l } { 4 x + 7 y = 100 } \\ { 5 x + 9 y = 90 } \end{array} \right.
Solve for x, y
x=270
y=-140
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4x+7y=100,5x+9y=90
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.
4x+7y=100
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
4x=-7y+100
Subtract 7y from both sides of the equation.
x=\frac{1}{4}\left(-7y+100\right)
Divide both sides by 4.
x=-\frac{7}{4}y+25
Multiply \frac{1}{4} times -7y+100.
5\left(-\frac{7}{4}y+25\right)+9y=90
Substitute -\frac{7y}{4}+25 for x in the other equation, 5x+9y=90.
-\frac{35}{4}y+125+9y=90
Multiply 5 times -\frac{7y}{4}+25.
\frac{1}{4}y+125=90
Add -\frac{35y}{4} to 9y.
\frac{1}{4}y=-35
Subtract 125 from both sides of the equation.
y=-140
Multiply both sides by 4.
x=-\frac{7}{4}\left(-140\right)+25
Substitute -140 for y in x=-\frac{7}{4}y+25. Because the resulting equation contains only one variable, you can solve for x directly.
x=245+25
Multiply -\frac{7}{4} times -140.
x=270
Add 25 to 245.
x=270,y=-140
The system is now solved.
4x+7y=100,5x+9y=90
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}4&7\\5&9\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}100\\90\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}4&7\\5&9\end{matrix}\right))\left(\begin{matrix}4&7\\5&9\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&7\\5&9\end{matrix}\right))\left(\begin{matrix}100\\90\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&7\\5&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}4&7\\5&9\end{matrix}\right))\left(\begin{matrix}100\\90\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}4&7\\5&9\end{matrix}\right))\left(\begin{matrix}100\\90\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}{4\times 9-7\times 5}&-\frac{7}{4\times 9-7\times 5}\\-\frac{5}{4\times 9-7\times 5}&\frac{4}{4\times 9-7\times 5}\end{matrix}\right)\left(\begin{matrix}100\\90\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}9&-7\\-5&4\end{matrix}\right)\left(\begin{matrix}100\\90\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}9\times 100-7\times 90\\-5\times 100+4\times 90\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}270\\-140\end{matrix}\right)
Do the arithmetic.
x=270,y=-140
Extract the matrix elements x and y.
4x+7y=100,5x+9y=90
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.
5\times 4x+5\times 7y=5\times 100,4\times 5x+4\times 9y=4\times 90
To make 4x and 5x equal, multiply all terms on each side of the first equation by 5 and all terms on each side of the second by 4.
20x+35y=500,20x+36y=360
Simplify.
20x-20x+35y-36y=500-360
Subtract 20x+36y=360 from 20x+35y=500 by subtracting like terms on each side of the equal sign.
35y-36y=500-360
Add 20x to -20x. Terms 20x and -20x cancel out, leaving an equation with only one variable that can be solved.
-y=500-360
Add 35y to -36y.
-y=140
Add 500 to -360.
y=-140
Divide both sides by -1.
5x+9\left(-140\right)=90
Substitute -140 for y in 5x+9y=90. Because the resulting equation contains only one variable, you can solve for x directly.
5x-1260=90
Multiply 9 times -140.
5x=1350
Add 1260 to both sides of the equation.
x=270
Divide both sides by 5.
x=270,y=-140
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}