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x+207y=3400,x+150y=3250
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+207y=3400
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-207y+3400
Subtract 207y from both sides of the equation.
-207y+3400+150y=3250
Substitute -207y+3400 for x in the other equation, x+150y=3250.
-57y+3400=3250
Add -207y to 150y.
-57y=-150
Subtract 3400 from both sides of the equation.
y=\frac{50}{19}
Divide both sides by -57.
x=-207\times \frac{50}{19}+3400
Substitute \frac{50}{19} for y in x=-207y+3400. Because the resulting equation contains only one variable, you can solve for x directly.
x=-\frac{10350}{19}+3400
Multiply -207 times \frac{50}{19}.
x=\frac{54250}{19}
Add 3400 to -\frac{10350}{19}.
x=\frac{54250}{19},y=\frac{50}{19}
The system is now solved.
x+207y=3400,x+150y=3250
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&207\\1&150\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3400\\3250\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&207\\1&150\end{matrix}\right))\left(\begin{matrix}1&207\\1&150\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&207\\1&150\end{matrix}\right))\left(\begin{matrix}3400\\3250\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&207\\1&150\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&207\\1&150\end{matrix}\right))\left(\begin{matrix}3400\\3250\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&207\\1&150\end{matrix}\right))\left(\begin{matrix}3400\\3250\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{150}{150-207}&-\frac{207}{150-207}\\-\frac{1}{150-207}&\frac{1}{150-207}\end{matrix}\right)\left(\begin{matrix}3400\\3250\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{50}{19}&\frac{69}{19}\\\frac{1}{57}&-\frac{1}{57}\end{matrix}\right)\left(\begin{matrix}3400\\3250\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{50}{19}\times 3400+\frac{69}{19}\times 3250\\\frac{1}{57}\times 3400-\frac{1}{57}\times 3250\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{54250}{19}\\\frac{50}{19}\end{matrix}\right)
Do the arithmetic.
x=\frac{54250}{19},y=\frac{50}{19}
Extract the matrix elements x and y.
x+207y=3400,x+150y=3250
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+207y-150y=3400-3250
Subtract x+150y=3250 from x+207y=3400 by subtracting like terms on each side of the equal sign.
207y-150y=3400-3250
Add x to -x. Terms x and -x cancel out, leaving an equation with only one variable that can be solved.
57y=3400-3250
Add 207y to -150y.
57y=150
Add 3400 to -3250.
y=\frac{50}{19}
Divide both sides by 57.
x+150\times \frac{50}{19}=3250
Substitute \frac{50}{19} for y in x+150y=3250. Because the resulting equation contains only one variable, you can solve for x directly.
x+\frac{7500}{19}=3250
Multiply 150 times \frac{50}{19}.
x=\frac{54250}{19}
Subtract \frac{7500}{19} from both sides of the equation.
x=\frac{54250}{19},y=\frac{50}{19}
The system is now solved.