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x+y=373,x-4y=100
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=373
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+373
Subtract y from both sides of the equation.
-y+373-4y=100
Substitute -y+373 for x in the other equation, x-4y=100.
-5y+373=100
Add -y to -4y.
-5y=-273
Subtract 373 from both sides of the equation.
y=\frac{273}{5}
Divide both sides by -5.
x=-\frac{273}{5}+373
Substitute \frac{273}{5} for y in x=-y+373. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{1592}{5}
Add 373 to -\frac{273}{5}.
x=\frac{1592}{5},y=\frac{273}{5}
The system is now solved.
x+y=373,x-4y=100
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\1&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}373\\100\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\1&-4\end{matrix}\right))\left(\begin{matrix}1&1\\1&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1&-4\end{matrix}\right))\left(\begin{matrix}373\\100\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\1&-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}1&1\\1&-4\end{matrix}\right))\left(\begin{matrix}373\\100\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&-4\end{matrix}\right))\left(\begin{matrix}373\\100\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}{-4-1}&-\frac{1}{-4-1}\\-\frac{1}{-4-1}&\frac{1}{-4-1}\end{matrix}\right)\left(\begin{matrix}373\\100\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{4}{5}&\frac{1}{5}\\\frac{1}{5}&-\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}373\\100\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{5}\times 373+\frac{1}{5}\times 100\\\frac{1}{5}\times 373-\frac{1}{5}\times 100\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1592}{5}\\\frac{273}{5}\end{matrix}\right)
Do the arithmetic.
x=\frac{1592}{5},y=\frac{273}{5}
Extract the matrix elements x and y.
x+y=373,x-4y=100
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+4y=373-100
Subtract x-4y=100 from x+y=373 by subtracting like terms on each side of the equal sign.
y+4y=373-100
Add x to -x. Terms x and -x cancel out, leaving an equation with only one variable that can be solved.
5y=373-100
Add y to 4y.
5y=273
Add 373 to -100.
y=\frac{273}{5}
Divide both sides by 5.
x-4\times \frac{273}{5}=100
Substitute \frac{273}{5} for y in x-4y=100. Because the resulting equation contains only one variable, you can solve for x directly.
x-\frac{1092}{5}=100
Multiply -4 times \frac{273}{5}.
x=\frac{1592}{5}
Add \frac{1092}{5} to both sides of the equation.
x=\frac{1592}{5},y=\frac{273}{5}
The system is now solved.