\left\{ \begin{array} { l } { 20 x - 20 y = 150 } \\ { 4 x + 4 y = 150 } \end{array} \right.
Solve for x, y
x = \frac{45}{2} = 22\frac{1}{2} = 22.5
y=15
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20x-20y=150,4x+4y=150
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.
20x-20y=150
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
20x=20y+150
Add 20y to both sides of the equation.
x=\frac{1}{20}\left(20y+150\right)
Divide both sides by 20.
x=y+\frac{15}{2}
Multiply \frac{1}{20} times 20y+150.
4\left(y+\frac{15}{2}\right)+4y=150
Substitute y+\frac{15}{2} for x in the other equation, 4x+4y=150.
4y+30+4y=150
Multiply 4 times y+\frac{15}{2}.
8y+30=150
Add 4y to 4y.
8y=120
Subtract 30 from both sides of the equation.
y=15
Divide both sides by 8.
x=15+\frac{15}{2}
Substitute 15 for y in x=y+\frac{15}{2}. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{45}{2}
Add \frac{15}{2} to 15.
x=\frac{45}{2},y=15
The system is now solved.
20x-20y=150,4x+4y=150
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}20&-20\\4&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}150\\150\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}20&-20\\4&4\end{matrix}\right))\left(\begin{matrix}20&-20\\4&4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&-20\\4&4\end{matrix}\right))\left(\begin{matrix}150\\150\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}20&-20\\4&4\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}20&-20\\4&4\end{matrix}\right))\left(\begin{matrix}150\\150\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}20&-20\\4&4\end{matrix}\right))\left(\begin{matrix}150\\150\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{4}{20\times 4-\left(-20\times 4\right)}&-\frac{-20}{20\times 4-\left(-20\times 4\right)}\\-\frac{4}{20\times 4-\left(-20\times 4\right)}&\frac{20}{20\times 4-\left(-20\times 4\right)}\end{matrix}\right)\left(\begin{matrix}150\\150\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{1}{40}&\frac{1}{8}\\-\frac{1}{40}&\frac{1}{8}\end{matrix}\right)\left(\begin{matrix}150\\150\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{40}\times 150+\frac{1}{8}\times 150\\-\frac{1}{40}\times 150+\frac{1}{8}\times 150\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{45}{2}\\15\end{matrix}\right)
Do the arithmetic.
x=\frac{45}{2},y=15
Extract the matrix elements x and y.
20x-20y=150,4x+4y=150
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.
4\times 20x+4\left(-20\right)y=4\times 150,20\times 4x+20\times 4y=20\times 150
To make 20x and 4x equal, multiply all terms on each side of the first equation by 4 and all terms on each side of the second by 20.
80x-80y=600,80x+80y=3000
Simplify.
80x-80x-80y-80y=600-3000
Subtract 80x+80y=3000 from 80x-80y=600 by subtracting like terms on each side of the equal sign.
-80y-80y=600-3000
Add 80x to -80x. Terms 80x and -80x cancel out, leaving an equation with only one variable that can be solved.
-160y=600-3000
Add -80y to -80y.
-160y=-2400
Add 600 to -3000.
y=15
Divide both sides by -160.
4x+4\times 15=150
Substitute 15 for y in 4x+4y=150. Because the resulting equation contains only one variable, you can solve for x directly.
4x+60=150
Multiply 4 times 15.
4x=90
Subtract 60 from both sides of the equation.
x=\frac{45}{2}
Divide both sides by 4.
x=\frac{45}{2},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}