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x+y=220,x-1.5y+180=320
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=220
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+220
Subtract y from both sides of the equation.
-y+220-1.5y+180=320
Substitute -y+220 for x in the other equation, x-1.5y+180=320.
-2.5y+220+180=320
Add -y to -\frac{3y}{2}.
-2.5y+400=320
Add 220 to 180.
-2.5y=-80
Subtract 400 from both sides of the equation.
y=32
Divide both sides of the equation by -2.5, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-32+220
Substitute 32 for y in x=-y+220. Because the resulting equation contains only one variable, you can solve for x directly.
x=188
Add 220 to -32.
x=188,y=32
The system is now solved.
x+y=220,x-1.5y+180=320
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\1&-1.5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}220\\140\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\1&-1.5\end{matrix}\right))\left(\begin{matrix}1&1\\1&-1.5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1&-1.5\end{matrix}\right))\left(\begin{matrix}220\\140\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\1&-1.5\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\\1&-1.5\end{matrix}\right))\left(\begin{matrix}220\\140\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\\1&-1.5\end{matrix}\right))\left(\begin{matrix}220\\140\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{1.5}{-1.5-1}&-\frac{1}{-1.5-1}\\-\frac{1}{-1.5-1}&\frac{1}{-1.5-1}\end{matrix}\right)\left(\begin{matrix}220\\140\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}0.6&0.4\\0.4&-0.4\end{matrix}\right)\left(\begin{matrix}220\\140\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0.6\times 220+0.4\times 140\\0.4\times 220-0.4\times 140\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}188\\32\end{matrix}\right)
Do the arithmetic.
x=188,y=32
Extract the matrix elements x and y.
x+y=220,x-1.5y+180=320
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.
x-x+y+1.5y-180=220-320
Subtract x-1.5y+180=320 from x+y=220 by subtracting like terms on each side of the equal sign.
y+1.5y-180=220-320
Add x to -x. Terms x and -x cancel out, leaving an equation with only one variable that can be solved.
2.5y-180=220-320
Add y to \frac{3y}{2}.
2.5y-180=-100
Add 220 to -320.
2.5y=80
Add 180 to both sides of the equation.
y=32
Divide both sides of the equation by 2.5, which is the same as multiplying both sides by the reciprocal of the fraction.
x-1.5\times 32+180=320
Substitute 32 for y in x-1.5y+180=320. Because the resulting equation contains only one variable, you can solve for x directly.
x-48+180=320
Multiply -1.5 times 32.
x+132=320
Add -48 to 180.
x=188
Subtract 132 from both sides of the equation.
x=188,y=32
The system is now solved.