Solve {l}{2012x-2013y=2015}{2011x-2012y=2014}

$$\left\{ \begin{array} { l } { 2012 x - 2013 y = 2015 } \\ { 2011 x - 2012 y = 2014 } \end{array} \right.$$

Solve for x, y

$$x=-2$$

$$y=-3$$

Steps Using Substitution

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.

$$2012x-2013y=2015,2011x-2012y=2014$$

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

$$2012x-2013y=2015$$

Add 2013y to both sides of the equation.

$$2012x=2013y+2015$$

Divide both sides by 2012.

$$x=\frac{1}{2012}\left(2013y+2015\right)$$

Multiply \frac{1}{2012} times 2013y+2015.

$$x=\frac{2013}{2012}y+\frac{2015}{2012}$$

Substitute \frac{2013y+2015}{2012} for x in the other equation, 2011x-2012y=2014.

$$2011\left(\frac{2013}{2012}y+\frac{2015}{2012}\right)-2012y=2014$$

Multiply 2011 times \frac{2013y+2015}{2012}.

$$\frac{4048143}{2012}y+\frac{4052165}{2012}-2012y=2014$$

Add \frac{4048143y}{2012} to -2012y.

$$-\frac{1}{2012}y+\frac{4052165}{2012}=2014$$

Subtract \frac{4052165}{2012} from both sides of the equation.

$$-\frac{1}{2012}y=\frac{3}{2012}$$

Multiply both sides by -2012.

$$y=-3$$

Substitute -3 for y in x=\frac{2013}{2012}y+\frac{2015}{2012}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{2013}{2012}\left(-3\right)+\frac{2015}{2012}$$

Multiply \frac{2013}{2012} times -3.

$$x=\frac{-6039+2015}{2012}$$

Add \frac{2015}{2012} to -\frac{6039}{2012} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-2$$

The system is now solved.

$$x=-2,y=-3$$

Steps Using Matrices

Put the equations in standard form and then use matrices to solve the system of equations.

$$2012x-2013y=2015,2011x-2012y=2014$$

Write the equations in matrix form.

$$\left(\begin{matrix}2012&-2013\\2011&-2012\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2015\\2014\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2012&-2013\\2011&-2012\end{matrix}\right).

$$inverse(\left(\begin{matrix}2012&-2013\\2011&-2012\end{matrix}\right))\left(\begin{matrix}2012&-2013\\2011&-2012\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2012&-2013\\2011&-2012\end{matrix}\right))\left(\begin{matrix}2015\\2014\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2012&-2013\\2011&-2012\end{matrix}\right))\left(\begin{matrix}2015\\2014\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2012&-2013\\2011&-2012\end{matrix}\right))\left(\begin{matrix}2015\\2014\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2012}{2012\left(-2012\right)-\left(-2013\times 2011\right)}&-\frac{-2013}{2012\left(-2012\right)-\left(-2013\times 2011\right)}\\-\frac{2011}{2012\left(-2012\right)-\left(-2013\times 2011\right)}&\frac{2012}{2012\left(-2012\right)-\left(-2013\times 2011\right)}\end{matrix}\right)\left(\begin{matrix}2015\\2014\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2012&-2013\\2011&-2012\end{matrix}\right)\left(\begin{matrix}2015\\2014\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2012\times 2015-2013\times 2014\\2011\times 2015-2012\times 2014\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-2\\-3\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-2,y=-3$$

Steps Using Elimination

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.

$$2012x-2013y=2015,2011x-2012y=2014$$

To make 2012x and 2011x equal, multiply all terms on each side of the first equation by 2011 and all terms on each side of the second by 2012.

$$2011\times 2012x+2011\left(-2013\right)y=2011\times 2015,2012\times 2011x+2012\left(-2012\right)y=2012\times 2014$$

Simplify.

$$4046132x-4048143y=4052165,4046132x-4048144y=4052168$$

Subtract 4046132x-4048144y=4052168 from 4046132x-4048143y=4052165 by subtracting like terms on each side of the equal sign.

$$4046132x-4046132x-4048143y+4048144y=4052165-4052168$$

Add 4046132x to -4046132x. Terms 4046132x and -4046132x cancel out, leaving an equation with only one variable that can be solved.

$$-4048143y+4048144y=4052165-4052168$$

Add -4048143y to 4048144y.

$$y=4052165-4052168$$

Add 4052165 to -4052168.

$$y=-3$$

Substitute -3 for y in 2011x-2012y=2014. Because the resulting equation contains only one variable, you can solve for x directly.

$$2011x-2012\left(-3\right)=2014$$

Multiply -2012 times -3.

$$2011x+6036=2014$$

Subtract 6036 from both sides of the equation.

$$2011x=-4022$$

Divide both sides by 2011.

$$x=-2$$

The system is now solved.

$$x=-2,y=-3$$