Solve for x, y
x=70
y=65
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x+y=135,55x+25y=5475
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=135
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+135
Subtract y from both sides of the equation.
55\left(-y+135\right)+25y=5475
Substitute -y+135 for x in the other equation, 55x+25y=5475.
-55y+7425+25y=5475
Multiply 55 times -y+135.
-30y+7425=5475
Add -55y to 25y.
-30y=-1950
Subtract 7425 from both sides of the equation.
y=65
Divide both sides by -30.
x=-65+135
Substitute 65 for y in x=-y+135. Because the resulting equation contains only one variable, you can solve for x directly.
x=70
Add 135 to -65.
x=70,y=65
The system is now solved.
x+y=135,55x+25y=5475
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\55&25\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}135\\5475\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\55&25\end{matrix}\right))\left(\begin{matrix}1&1\\55&25\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\55&25\end{matrix}\right))\left(\begin{matrix}135\\5475\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\55&25\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\\55&25\end{matrix}\right))\left(\begin{matrix}135\\5475\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\\55&25\end{matrix}\right))\left(\begin{matrix}135\\5475\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{25}{25-55}&-\frac{1}{25-55}\\-\frac{55}{25-55}&\frac{1}{25-55}\end{matrix}\right)\left(\begin{matrix}135\\5475\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{5}{6}&\frac{1}{30}\\\frac{11}{6}&-\frac{1}{30}\end{matrix}\right)\left(\begin{matrix}135\\5475\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{5}{6}\times 135+\frac{1}{30}\times 5475\\\frac{11}{6}\times 135-\frac{1}{30}\times 5475\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}70\\65\end{matrix}\right)
Do the arithmetic.
x=70,y=65
Extract the matrix elements x and y.
x+y=135,55x+25y=5475
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.
55x+55y=55\times 135,55x+25y=5475
To make x and 55x equal, multiply all terms on each side of the first equation by 55 and all terms on each side of the second by 1.
55x+55y=7425,55x+25y=5475
Simplify.
55x-55x+55y-25y=7425-5475
Subtract 55x+25y=5475 from 55x+55y=7425 by subtracting like terms on each side of the equal sign.
55y-25y=7425-5475
Add 55x to -55x. Terms 55x and -55x cancel out, leaving an equation with only one variable that can be solved.
30y=7425-5475
Add 55y to -25y.
30y=1950
Add 7425 to -5475.
y=65
Divide both sides by 30.
55x+25\times 65=5475
Substitute 65 for y in 55x+25y=5475. Because the resulting equation contains only one variable, you can solve for x directly.
55x+1625=5475
Multiply 25 times 65.
55x=3850
Subtract 1625 from both sides of the equation.
x=70
Divide both sides by 55.
x=70,y=65
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}