Solve {l}{2018x+2017y=2016}{2016x+2015y=2014}

$$\left\{ \begin{array} { l } { 2018 x + 2017 y = 2016 } \\ { 2016 x + 2015 y = 2014 } \end{array} \right.$$

Solve for x, y

$$x=-1$$

$$y=2$$

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.

$$2018x+2017y=2016,2016x+2015y=2014$$

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

$$2018x+2017y=2016$$

Subtract 2017y from both sides of the equation.

$$2018x=-2017y+2016$$

Divide both sides by 2018.

$$x=\frac{1}{2018}\left(-2017y+2016\right)$$

Multiply \frac{1}{2018} times -2017y+2016.

$$x=-\frac{2017}{2018}y+\frac{1008}{1009}$$

Substitute -\frac{2017y}{2018}+\frac{1008}{1009} for x in the other equation, 2016x+2015y=2014.

$$2016\left(-\frac{2017}{2018}y+\frac{1008}{1009}\right)+2015y=2014$$

Multiply 2016 times -\frac{2017y}{2018}+\frac{1008}{1009}.

$$-\frac{2033136}{1009}y+\frac{2032128}{1009}+2015y=2014$$

Add -\frac{2033136y}{1009} to 2015y.

$$-\frac{1}{1009}y+\frac{2032128}{1009}=2014$$

Subtract \frac{2032128}{1009} from both sides of the equation.

$$-\frac{1}{1009}y=-\frac{2}{1009}$$

Multiply both sides by -1009.

$$y=2$$

Substitute 2 for y in x=-\frac{2017}{2018}y+\frac{1008}{1009}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{2017}{2018}\times 2+\frac{1008}{1009}$$

Multiply -\frac{2017}{2018} times 2.

$$x=\frac{-2017+1008}{1009}$$

Add \frac{1008}{1009} to -\frac{2017}{1009} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-1$$

The system is now solved.

$$x=-1,y=2$$

Steps Using Matrices

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

$$2018x+2017y=2016,2016x+2015y=2014$$

Write the equations in matrix form.

$$\left(\begin{matrix}2018&2017\\2016&2015\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2016\\2014\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2018&2017\\2016&2015\end{matrix}\right).

$$inverse(\left(\begin{matrix}2018&2017\\2016&2015\end{matrix}\right))\left(\begin{matrix}2018&2017\\2016&2015\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2018&2017\\2016&2015\end{matrix}\right))\left(\begin{matrix}2016\\2014\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2018&2017\\2016&2015\end{matrix}\right))\left(\begin{matrix}2016\\2014\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2018&2017\\2016&2015\end{matrix}\right))\left(\begin{matrix}2016\\2014\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2015}{2018\times 2015-2017\times 2016}&-\frac{2017}{2018\times 2015-2017\times 2016}\\-\frac{2016}{2018\times 2015-2017\times 2016}&\frac{2018}{2018\times 2015-2017\times 2016}\end{matrix}\right)\left(\begin{matrix}2016\\2014\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2015}{2}&\frac{2017}{2}\\1008&-1009\end{matrix}\right)\left(\begin{matrix}2016\\2014\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2015}{2}\times 2016+\frac{2017}{2}\times 2014\\1008\times 2016-1009\times 2014\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-1\\2\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-1,y=2$$

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.

$$2018x+2017y=2016,2016x+2015y=2014$$

To make 2018x and 2016x equal, multiply all terms on each side of the first equation by 2016 and all terms on each side of the second by 2018.

$$2016\times 2018x+2016\times 2017y=2016\times 2016,2018\times 2016x+2018\times 2015y=2018\times 2014$$

Simplify.

$$4068288x+4066272y=4064256,4068288x+4066270y=4064252$$

Subtract 4068288x+4066270y=4064252 from 4068288x+4066272y=4064256 by subtracting like terms on each side of the equal sign.

$$4068288x-4068288x+4066272y-4066270y=4064256-4064252$$

Add 4068288x to -4068288x. Terms 4068288x and -4068288x cancel out, leaving an equation with only one variable that can be solved.

$$4066272y-4066270y=4064256-4064252$$

Add 4066272y to -4066270y.

$$2y=4064256-4064252$$

Add 4064256 to -4064252.

$$2y=4$$

Divide both sides by 2.

$$y=2$$

Substitute 2 for y in 2016x+2015y=2014. Because the resulting equation contains only one variable, you can solve for x directly.

$$2016x+2015\times 2=2014$$

Multiply 2015 times 2.

$$2016x+4030=2014$$

Subtract 4030 from both sides of the equation.

$$2016x=-2016$$

Divide both sides by 2016.

$$x=-1$$

The system is now solved.

$$x=-1,y=2$$