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205x+2016y=2014,2016x+2015y=2017
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.
205x+2016y=2014
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
205x=-2016y+2014
Subtract 2016y from both sides of the equation.
x=\frac{1}{205}\left(-2016y+2014\right)
Divide both sides by 205.
x=-\frac{2016}{205}y+\frac{2014}{205}
Multiply \frac{1}{205} times -2016y+2014.
2016\left(-\frac{2016}{205}y+\frac{2014}{205}\right)+2015y=2017
Substitute \frac{-2016y+2014}{205} for x in the other equation, 2016x+2015y=2017.
-\frac{4064256}{205}y+\frac{4060224}{205}+2015y=2017
Multiply 2016 times \frac{-2016y+2014}{205}.
-\frac{3651181}{205}y+\frac{4060224}{205}=2017
Add -\frac{4064256y}{205} to 2015y.
-\frac{3651181}{205}y=-\frac{3646739}{205}
Subtract \frac{4060224}{205} from both sides of the equation.
y=\frac{3646739}{3651181}
Divide both sides of the equation by -\frac{3651181}{205}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{2016}{205}\times \frac{3646739}{3651181}+\frac{2014}{205}
Substitute \frac{3646739}{3651181} for y in x=-\frac{2016}{205}y+\frac{2014}{205}. Because the resulting equation contains only one variable, you can solve for x directly.
x=-\frac{7351825824}{748492105}+\frac{2014}{205}
Multiply -\frac{2016}{205} times \frac{3646739}{3651181} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=\frac{8062}{3651181}
Add \frac{2014}{205} to -\frac{7351825824}{748492105} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=\frac{8062}{3651181},y=\frac{3646739}{3651181}
The system is now solved.
205x+2016y=2014,2016x+2015y=2017
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}205&2016\\2016&2015\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2014\\2017\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}205&2016\\2016&2015\end{matrix}\right))\left(\begin{matrix}205&2016\\2016&2015\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}205&2016\\2016&2015\end{matrix}\right))\left(\begin{matrix}2014\\2017\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}205&2016\\2016&2015\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}205&2016\\2016&2015\end{matrix}\right))\left(\begin{matrix}2014\\2017\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}205&2016\\2016&2015\end{matrix}\right))\left(\begin{matrix}2014\\2017\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{2015}{205\times 2015-2016\times 2016}&-\frac{2016}{205\times 2015-2016\times 2016}\\-\frac{2016}{205\times 2015-2016\times 2016}&\frac{205}{205\times 2015-2016\times 2016}\end{matrix}\right)\left(\begin{matrix}2014\\2017\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}{3651181}&\frac{2016}{3651181}\\\frac{2016}{3651181}&-\frac{205}{3651181}\end{matrix}\right)\left(\begin{matrix}2014\\2017\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2015}{3651181}\times 2014+\frac{2016}{3651181}\times 2017\\\frac{2016}{3651181}\times 2014-\frac{205}{3651181}\times 2017\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{8062}{3651181}\\\frac{3646739}{3651181}\end{matrix}\right)
Do the arithmetic.
x=\frac{8062}{3651181},y=\frac{3646739}{3651181}
Extract the matrix elements x and y.
205x+2016y=2014,2016x+2015y=2017
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.
2016\times 205x+2016\times 2016y=2016\times 2014,205\times 2016x+205\times 2015y=205\times 2017
To make 205x 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 205.
413280x+4064256y=4060224,413280x+413075y=413485
Simplify.
413280x-413280x+4064256y-413075y=4060224-413485
Subtract 413280x+413075y=413485 from 413280x+4064256y=4060224 by subtracting like terms on each side of the equal sign.
4064256y-413075y=4060224-413485
Add 413280x to -413280x. Terms 413280x and -413280x cancel out, leaving an equation with only one variable that can be solved.
3651181y=4060224-413485
Add 4064256y to -413075y.
3651181y=3646739
Add 4060224 to -413485.
y=\frac{3646739}{3651181}
Divide both sides by 3651181.
2016x+2015\times \frac{3646739}{3651181}=2017
Substitute \frac{3646739}{3651181} for y in 2016x+2015y=2017. Because the resulting equation contains only one variable, you can solve for x directly.
2016x+\frac{7348179085}{3651181}=2017
Multiply 2015 times \frac{3646739}{3651181}.
2016x=\frac{16252992}{3651181}
Subtract \frac{7348179085}{3651181} from both sides of the equation.
x=\frac{8062}{3651181}
Divide both sides by 2016.
x=\frac{8062}{3651181},y=\frac{3646739}{3651181}
The system is now solved.