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x+y=20,150x+250y=4500
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=20
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+20
Subtract y from both sides of the equation.
150\left(-y+20\right)+250y=4500
Substitute -y+20 for x in the other equation, 150x+250y=4500.
-150y+3000+250y=4500
Multiply 150 times -y+20.
100y+3000=4500
Add -150y to 250y.
100y=1500
Subtract 3000 from both sides of the equation.
y=15
Divide both sides by 100.
x=-15+20
Substitute 15 for y in x=-y+20. Because the resulting equation contains only one variable, you can solve for x directly.
x=5
Add 20 to -15.
x=5,y=15
The system is now solved.
x+y=20,150x+250y=4500
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\150&250\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}20\\4500\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\150&250\end{matrix}\right))\left(\begin{matrix}1&1\\150&250\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\150&250\end{matrix}\right))\left(\begin{matrix}20\\4500\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\150&250\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\\150&250\end{matrix}\right))\left(\begin{matrix}20\\4500\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\\150&250\end{matrix}\right))\left(\begin{matrix}20\\4500\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{250}{250-150}&-\frac{1}{250-150}\\-\frac{150}{250-150}&\frac{1}{250-150}\end{matrix}\right)\left(\begin{matrix}20\\4500\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}{2}&-\frac{1}{100}\\-\frac{3}{2}&\frac{1}{100}\end{matrix}\right)\left(\begin{matrix}20\\4500\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{2}\times 20-\frac{1}{100}\times 4500\\-\frac{3}{2}\times 20+\frac{1}{100}\times 4500\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}5\\15\end{matrix}\right)
Do the arithmetic.
x=5,y=15
Extract the matrix elements x and y.
x+y=20,150x+250y=4500
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.
150x+150y=150\times 20,150x+250y=4500
To make x and 150x equal, multiply all terms on each side of the first equation by 150 and all terms on each side of the second by 1.
150x+150y=3000,150x+250y=4500
Simplify.
150x-150x+150y-250y=3000-4500
Subtract 150x+250y=4500 from 150x+150y=3000 by subtracting like terms on each side of the equal sign.
150y-250y=3000-4500
Add 150x to -150x. Terms 150x and -150x cancel out, leaving an equation with only one variable that can be solved.
-100y=3000-4500
Add 150y to -250y.
-100y=-1500
Add 3000 to -4500.
y=15
Divide both sides by -100.
150x+250\times 15=4500
Substitute 15 for y in 150x+250y=4500. Because the resulting equation contains only one variable, you can solve for x directly.
150x+3750=4500
Multiply 250 times 15.
150x=750
Subtract 3750 from both sides of the equation.
x=5
Divide both sides by 150.
x=5,y=15
The system is now solved.