Solve for x, y
x = -\frac{50}{7} = -7\frac{1}{7} \approx -7.142857143
y = \frac{155}{7} = 22\frac{1}{7} \approx 22.142857143
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x+y=15,5x+75y=1625
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.
x+y=15
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+15
Subtract y from both sides of the equation.
5\left(-y+15\right)+75y=1625
Substitute -y+15 for x in the other equation, 5x+75y=1625.
-5y+75+75y=1625
Multiply 5 times -y+15.
70y+75=1625
Add -5y to 75y.
70y=1550
Subtract 75 from both sides of the equation.
y=\frac{155}{7}
Divide both sides by 70.
x=-\frac{155}{7}+15
Substitute \frac{155}{7} for y in x=-y+15. Because the resulting equation contains only one variable, you can solve for x directly.
x=-\frac{50}{7}
Add 15 to -\frac{155}{7}.
x=-\frac{50}{7},y=\frac{155}{7}
The system is now solved.
x+y=15,5x+75y=1625
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\5&75\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}15\\1625\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\5&75\end{matrix}\right))\left(\begin{matrix}1&1\\5&75\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\5&75\end{matrix}\right))\left(\begin{matrix}15\\1625\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\5&75\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}1&1\\5&75\end{matrix}\right))\left(\begin{matrix}15\\1625\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}1&1\\5&75\end{matrix}\right))\left(\begin{matrix}15\\1625\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{75}{75-5}&-\frac{1}{75-5}\\-\frac{5}{75-5}&\frac{1}{75-5}\end{matrix}\right)\left(\begin{matrix}15\\1625\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{15}{14}&-\frac{1}{70}\\-\frac{1}{14}&\frac{1}{70}\end{matrix}\right)\left(\begin{matrix}15\\1625\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{15}{14}\times 15-\frac{1}{70}\times 1625\\-\frac{1}{14}\times 15+\frac{1}{70}\times 1625\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{50}{7}\\\frac{155}{7}\end{matrix}\right)
Do the arithmetic.
x=-\frac{50}{7},y=\frac{155}{7}
Extract the matrix elements x and y.
x+y=15,5x+75y=1625
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+5y=5\times 15,5x+75y=1625
To make x 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 1.
5x+5y=75,5x+75y=1625
Simplify.
5x-5x+5y-75y=75-1625
Subtract 5x+75y=1625 from 5x+5y=75 by subtracting like terms on each side of the equal sign.
5y-75y=75-1625
Add 5x to -5x. Terms 5x and -5x cancel out, leaving an equation with only one variable that can be solved.
-70y=75-1625
Add 5y to -75y.
-70y=-1550
Add 75 to -1625.
y=\frac{155}{7}
Divide both sides by -70.
5x+75\times \frac{155}{7}=1625
Substitute \frac{155}{7} for y in 5x+75y=1625. Because the resulting equation contains only one variable, you can solve for x directly.
5x+\frac{11625}{7}=1625
Multiply 75 times \frac{155}{7}.
5x=-\frac{250}{7}
Subtract \frac{11625}{7} from both sides of the equation.
x=-\frac{50}{7}
Divide both sides by 5.
x=-\frac{50}{7},y=\frac{155}{7}
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}