Skip to main content
Solve for x, y
Tick mark Image
Graph

Similar Problems from Web Search

Share

x+y=27a,0.08x+0.13y=32.9
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=27a
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+27a
Subtract y from both sides of the equation.
0.08\left(-y+27a\right)+0.13y=32.9
Substitute -y+27a for x in the other equation, 0.08x+0.13y=32.9.
-0.08y+\frac{54a}{25}+0.13y=32.9
Multiply 0.08 times -y+27a.
0.05y+\frac{54a}{25}=32.9
Add -\frac{2y}{25} to \frac{13y}{100}.
0.05y=-\frac{54a}{25}+\frac{329}{10}
Subtract \frac{54a}{25} from both sides of the equation.
y=-\frac{216a}{5}+658
Multiply both sides by 20.
x=-\left(-\frac{216a}{5}+658\right)+27a
Substitute 658-\frac{216a}{5} for y in x=-y+27a. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{216a}{5}-658+27a
Multiply -1 times 658-\frac{216a}{5}.
x=\frac{351a}{5}-658
Add 27a to -658+\frac{216a}{5}.
x=\frac{351a}{5}-658,y=-\frac{216a}{5}+658
The system is now solved.
x+y=27a,0.08x+0.13y=32.9
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\0.08&0.13\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}27a\\32.9\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\0.08&0.13\end{matrix}\right))\left(\begin{matrix}1&1\\0.08&0.13\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\0.08&0.13\end{matrix}\right))\left(\begin{matrix}27a\\32.9\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\0.08&0.13\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\\0.08&0.13\end{matrix}\right))\left(\begin{matrix}27a\\32.9\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\\0.08&0.13\end{matrix}\right))\left(\begin{matrix}27a\\32.9\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{0.13}{0.13-0.08}&-\frac{1}{0.13-0.08}\\-\frac{0.08}{0.13-0.08}&\frac{1}{0.13-0.08}\end{matrix}\right)\left(\begin{matrix}27a\\32.9\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}2.6&-20\\-1.6&20\end{matrix}\right)\left(\begin{matrix}27a\\32.9\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2.6\times 27a-20\times 32.9\\-1.6\times 27a+20\times 32.9\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{351a}{5}-658\\-\frac{216a}{5}+658\end{matrix}\right)
Do the arithmetic.
x=\frac{351a}{5}-658,y=-\frac{216a}{5}+658
Extract the matrix elements x and y.
x+y=27a,0.08x+0.13y=32.9
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.
0.08x+0.08y=0.08\times 27a,0.08x+0.13y=32.9
To make x and \frac{2x}{25} equal, multiply all terms on each side of the first equation by 0.08 and all terms on each side of the second by 1.
0.08x+0.08y=\frac{54a}{25},0.08x+0.13y=32.9
Simplify.
0.08x-0.08x+0.08y-0.13y=\frac{54a}{25}-32.9
Subtract 0.08x+0.13y=32.9 from 0.08x+0.08y=\frac{54a}{25} by subtracting like terms on each side of the equal sign.
0.08y-0.13y=\frac{54a}{25}-32.9
Add \frac{2x}{25} to -\frac{2x}{25}. Terms \frac{2x}{25} and -\frac{2x}{25} cancel out, leaving an equation with only one variable that can be solved.
-0.05y=\frac{54a}{25}-32.9
Add \frac{2y}{25} to -\frac{13y}{100}.
-0.05y=\frac{54a}{25}-\frac{329}{10}
Add \frac{54a}{25} to -32.9.
y=-\frac{216a}{5}+658
Multiply both sides by -20.
0.08x+0.13\left(-\frac{216a}{5}+658\right)=32.9
Substitute 658-\frac{216a}{5} for y in 0.08x+0.13y=32.9. Because the resulting equation contains only one variable, you can solve for x directly.
0.08x-\frac{702a}{125}+\frac{4277}{50}=32.9
Multiply 0.13 times 658-\frac{216a}{5}.
0.08x=\frac{702a}{125}-\frac{1316}{25}
Subtract \frac{4277}{50}-\frac{702a}{125} from both sides of the equation.
x=\frac{351a}{5}-658
Divide both sides of the equation by 0.08, which is the same as multiplying both sides by the reciprocal of the fraction.
x=\frac{351a}{5}-658,y=-\frac{216a}{5}+658
The system is now solved.