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50x+y=200,60x+y=260
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.
50x+y=200
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
50x=-y+200
Subtract y from both sides of the equation.
x=\frac{1}{50}\left(-y+200\right)
Divide both sides by 50.
x=-\frac{1}{50}y+4
Multiply \frac{1}{50} times -y+200.
60\left(-\frac{1}{50}y+4\right)+y=260
Substitute -\frac{y}{50}+4 for x in the other equation, 60x+y=260.
-\frac{6}{5}y+240+y=260
Multiply 60 times -\frac{y}{50}+4.
-\frac{1}{5}y+240=260
Add -\frac{6y}{5} to y.
-\frac{1}{5}y=20
Subtract 240 from both sides of the equation.
y=-100
Multiply both sides by -5.
x=-\frac{1}{50}\left(-100\right)+4
Substitute -100 for y in x=-\frac{1}{50}y+4. Because the resulting equation contains only one variable, you can solve for x directly.
x=2+4
Multiply -\frac{1}{50} times -100.
x=6
Add 4 to 2.
x=6,y=-100
The system is now solved.
50x+y=200,60x+y=260
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}50&1\\60&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}200\\260\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}50&1\\60&1\end{matrix}\right))\left(\begin{matrix}50&1\\60&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}50&1\\60&1\end{matrix}\right))\left(\begin{matrix}200\\260\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}50&1\\60&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}50&1\\60&1\end{matrix}\right))\left(\begin{matrix}200\\260\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}50&1\\60&1\end{matrix}\right))\left(\begin{matrix}200\\260\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}{50-60}&-\frac{1}{50-60}\\-\frac{60}{50-60}&\frac{50}{50-60}\end{matrix}\right)\left(\begin{matrix}200\\260\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}{10}&\frac{1}{10}\\6&-5\end{matrix}\right)\left(\begin{matrix}200\\260\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{10}\times 200+\frac{1}{10}\times 260\\6\times 200-5\times 260\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}6\\-100\end{matrix}\right)
Do the arithmetic.
x=6,y=-100
Extract the matrix elements x and y.
50x+y=200,60x+y=260
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.
50x-60x+y-y=200-260
Subtract 60x+y=260 from 50x+y=200 by subtracting like terms on each side of the equal sign.
50x-60x=200-260
Add y to -y. Terms y and -y cancel out, leaving an equation with only one variable that can be solved.
-10x=200-260
Add 50x to -60x.
-10x=-60
Add 200 to -260.
x=6
Divide both sides by -10.
60\times 6+y=260
Substitute 6 for x in 60x+y=260. Because the resulting equation contains only one variable, you can solve for y directly.
360+y=260
Multiply 60 times 6.
y=-100
Subtract 360 from both sides of the equation.
x=6,y=-100
The system is now solved.