\left\{ \begin{array} { l } { 78 x + 40 y = 1280 } \\ { 120 x + 80 y = 2800 } \end{array} \right.
Solve for x, y
x = -\frac{20}{3} = -6\frac{2}{3} \approx -6.666666667
y=45
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78x+40y=1280,120x+80y=2800
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.
78x+40y=1280
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
78x=-40y+1280
Subtract 40y from both sides of the equation.
x=\frac{1}{78}\left(-40y+1280\right)
Divide both sides by 78.
x=-\frac{20}{39}y+\frac{640}{39}
Multiply \frac{1}{78} times -40y+1280.
120\left(-\frac{20}{39}y+\frac{640}{39}\right)+80y=2800
Substitute \frac{-20y+640}{39} for x in the other equation, 120x+80y=2800.
-\frac{800}{13}y+\frac{25600}{13}+80y=2800
Multiply 120 times \frac{-20y+640}{39}.
\frac{240}{13}y+\frac{25600}{13}=2800
Add -\frac{800y}{13} to 80y.
\frac{240}{13}y=\frac{10800}{13}
Subtract \frac{25600}{13} from both sides of the equation.
y=45
Divide both sides of the equation by \frac{240}{13}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{20}{39}\times 45+\frac{640}{39}
Substitute 45 for y in x=-\frac{20}{39}y+\frac{640}{39}. Because the resulting equation contains only one variable, you can solve for x directly.
x=-\frac{300}{13}+\frac{640}{39}
Multiply -\frac{20}{39} times 45.
x=-\frac{20}{3}
Add \frac{640}{39} to -\frac{300}{13} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=-\frac{20}{3},y=45
The system is now solved.
78x+40y=1280,120x+80y=2800
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}78&40\\120&80\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1280\\2800\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}78&40\\120&80\end{matrix}\right))\left(\begin{matrix}78&40\\120&80\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}78&40\\120&80\end{matrix}\right))\left(\begin{matrix}1280\\2800\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}78&40\\120&80\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}78&40\\120&80\end{matrix}\right))\left(\begin{matrix}1280\\2800\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}78&40\\120&80\end{matrix}\right))\left(\begin{matrix}1280\\2800\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{80}{78\times 80-40\times 120}&-\frac{40}{78\times 80-40\times 120}\\-\frac{120}{78\times 80-40\times 120}&\frac{78}{78\times 80-40\times 120}\end{matrix}\right)\left(\begin{matrix}1280\\2800\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}{18}&-\frac{1}{36}\\-\frac{1}{12}&\frac{13}{240}\end{matrix}\right)\left(\begin{matrix}1280\\2800\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{18}\times 1280-\frac{1}{36}\times 2800\\-\frac{1}{12}\times 1280+\frac{13}{240}\times 2800\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{20}{3}\\45\end{matrix}\right)
Do the arithmetic.
x=-\frac{20}{3},y=45
Extract the matrix elements x and y.
78x+40y=1280,120x+80y=2800
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.
120\times 78x+120\times 40y=120\times 1280,78\times 120x+78\times 80y=78\times 2800
To make 78x and 120x equal, multiply all terms on each side of the first equation by 120 and all terms on each side of the second by 78.
9360x+4800y=153600,9360x+6240y=218400
Simplify.
9360x-9360x+4800y-6240y=153600-218400
Subtract 9360x+6240y=218400 from 9360x+4800y=153600 by subtracting like terms on each side of the equal sign.
4800y-6240y=153600-218400
Add 9360x to -9360x. Terms 9360x and -9360x cancel out, leaving an equation with only one variable that can be solved.
-1440y=153600-218400
Add 4800y to -6240y.
-1440y=-64800
Add 153600 to -218400.
y=45
Divide both sides by -1440.
120x+80\times 45=2800
Substitute 45 for y in 120x+80y=2800. Because the resulting equation contains only one variable, you can solve for x directly.
120x+3600=2800
Multiply 80 times 45.
120x=-800
Subtract 3600 from both sides of the equation.
x=-\frac{20}{3}
Divide both sides by 120.
x=-\frac{20}{3},y=45
The system is now solved.
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