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x-y=200,4x-3y=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.
x-y=200
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=y+200
Add y to both sides of the equation.
4\left(y+200\right)-3y=260
Substitute y+200 for x in the other equation, 4x-3y=260.
4y+800-3y=260
Multiply 4 times y+200.
y+800=260
Add 4y to -3y.
y=-540
Subtract 800 from both sides of the equation.
x=-540+200
Substitute -540 for y in x=y+200. Because the resulting equation contains only one variable, you can solve for x directly.
x=-340
Add 200 to -540.
x=-340,y=-540
The system is now solved.
x-y=200,4x-3y=260
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&-1\\4&-3\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}1&-1\\4&-3\end{matrix}\right))\left(\begin{matrix}1&-1\\4&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\4&-3\end{matrix}\right))\left(\begin{matrix}200\\260\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-1\\4&-3\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\\4&-3\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}1&-1\\4&-3\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{3}{-3-\left(-4\right)}&-\frac{-1}{-3-\left(-4\right)}\\-\frac{4}{-3-\left(-4\right)}&\frac{1}{-3-\left(-4\right)}\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}-3&1\\-4&1\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}-3\times 200+260\\-4\times 200+260\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-340\\-540\end{matrix}\right)
Do the arithmetic.
x=-340,y=-540
Extract the matrix elements x and y.
x-y=200,4x-3y=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.
4x+4\left(-1\right)y=4\times 200,4x-3y=260
To make x and 4x equal, multiply all terms on each side of the first equation by 4 and all terms on each side of the second by 1.
4x-4y=800,4x-3y=260
Simplify.
4x-4x-4y+3y=800-260
Subtract 4x-3y=260 from 4x-4y=800 by subtracting like terms on each side of the equal sign.
-4y+3y=800-260
Add 4x to -4x. Terms 4x and -4x cancel out, leaving an equation with only one variable that can be solved.
-y=800-260
Add -4y to 3y.
-y=540
Add 800 to -260.
y=-540
Divide both sides by -1.
4x-3\left(-540\right)=260
Substitute -540 for y in 4x-3y=260. Because the resulting equation contains only one variable, you can solve for x directly.
4x+1620=260
Multiply -3 times -540.
4x=-1360
Subtract 1620 from both sides of the equation.
x=-340
Divide both sides by 4.
x=-340,y=-540
The system is now solved.