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x+2y=143,2x+y=121
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+2y=143
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-2y+143
Subtract 2y from both sides of the equation.
2\left(-2y+143\right)+y=121
Substitute -2y+143 for x in the other equation, 2x+y=121.
-4y+286+y=121
Multiply 2 times -2y+143.
-3y+286=121
Add -4y to y.
-3y=-165
Subtract 286 from both sides of the equation.
y=55
Divide both sides by -3.
x=-2\times 55+143
Substitute 55 for y in x=-2y+143. Because the resulting equation contains only one variable, you can solve for x directly.
x=-110+143
Multiply -2 times 55.
x=33
Add 143 to -110.
x=33,y=55
The system is now solved.
x+2y=143,2x+y=121
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&2\\2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}143\\121\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&2\\2&1\end{matrix}\right))\left(\begin{matrix}1&2\\2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&2\\2&1\end{matrix}\right))\left(\begin{matrix}143\\121\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&2\\2&1\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&2\\2&1\end{matrix}\right))\left(\begin{matrix}143\\121\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&2\\2&1\end{matrix}\right))\left(\begin{matrix}143\\121\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}{1-2\times 2}&-\frac{2}{1-2\times 2}\\-\frac{2}{1-2\times 2}&\frac{1}{1-2\times 2}\end{matrix}\right)\left(\begin{matrix}143\\121\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}{3}&\frac{2}{3}\\\frac{2}{3}&-\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}143\\121\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{3}\times 143+\frac{2}{3}\times 121\\\frac{2}{3}\times 143-\frac{1}{3}\times 121\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}33\\55\end{matrix}\right)
Do the arithmetic.
x=33,y=55
Extract the matrix elements x and y.
x+2y=143,2x+y=121
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.
2x+2\times 2y=2\times 143,2x+y=121
To make x and 2x equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by 1.
2x+4y=286,2x+y=121
Simplify.
2x-2x+4y-y=286-121
Subtract 2x+y=121 from 2x+4y=286 by subtracting like terms on each side of the equal sign.
4y-y=286-121
Add 2x to -2x. Terms 2x and -2x cancel out, leaving an equation with only one variable that can be solved.
3y=286-121
Add 4y to -y.
3y=165
Add 286 to -121.
y=55
Divide both sides by 3.
2x+55=121
Substitute 55 for y in 2x+y=121. Because the resulting equation contains only one variable, you can solve for x directly.
2x=66
Subtract 55 from both sides of the equation.
x=33
Divide both sides by 2.
x=33,y=55
The system is now solved.