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2013x+2949y=2015,2014x+2015y=2016
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.
2013x+2949y=2015
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
2013x=-2949y+2015
Subtract 2949y from both sides of the equation.
x=\frac{1}{2013}\left(-2949y+2015\right)
Divide both sides by 2013.
x=-\frac{983}{671}y+\frac{2015}{2013}
Multiply \frac{1}{2013} times -2949y+2015.
2014\left(-\frac{983}{671}y+\frac{2015}{2013}\right)+2015y=2016
Substitute -\frac{983y}{671}+\frac{2015}{2013} for x in the other equation, 2014x+2015y=2016.
-\frac{1979762}{671}y+\frac{4058210}{2013}+2015y=2016
Multiply 2014 times -\frac{983y}{671}+\frac{2015}{2013}.
-\frac{627697}{671}y+\frac{4058210}{2013}=2016
Add -\frac{1979762y}{671} to 2015y.
-\frac{627697}{671}y=-\frac{2}{2013}
Subtract \frac{4058210}{2013} from both sides of the equation.
y=\frac{2}{1883091}
Divide both sides of the equation by -\frac{627697}{671}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{983}{671}\times \frac{2}{1883091}+\frac{2015}{2013}
Substitute \frac{2}{1883091} for y in x=-\frac{983}{671}y+\frac{2015}{2013}. Because the resulting equation contains only one variable, you can solve for x directly.
x=-\frac{1966}{1263554061}+\frac{2015}{2013}
Multiply -\frac{983}{671} times \frac{2}{1883091} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=\frac{1884959}{1883091}
Add \frac{2015}{2013} to -\frac{1966}{1263554061} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=\frac{1884959}{1883091},y=\frac{2}{1883091}
The system is now solved.
2013x+2949y=2015,2014x+2015y=2016
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}2013&2949\\2014&2015\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2015\\2016\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}2013&2949\\2014&2015\end{matrix}\right))\left(\begin{matrix}2013&2949\\2014&2015\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2013&2949\\2014&2015\end{matrix}\right))\left(\begin{matrix}2015\\2016\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}2013&2949\\2014&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}2013&2949\\2014&2015\end{matrix}\right))\left(\begin{matrix}2015\\2016\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}2013&2949\\2014&2015\end{matrix}\right))\left(\begin{matrix}2015\\2016\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}{2013\times 2015-2949\times 2014}&-\frac{2949}{2013\times 2015-2949\times 2014}\\-\frac{2014}{2013\times 2015-2949\times 2014}&\frac{2013}{2013\times 2015-2949\times 2014}\end{matrix}\right)\left(\begin{matrix}2015\\2016\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}{1883091}&\frac{983}{627697}\\\frac{2014}{1883091}&-\frac{671}{627697}\end{matrix}\right)\left(\begin{matrix}2015\\2016\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2015}{1883091}\times 2015+\frac{983}{627697}\times 2016\\\frac{2014}{1883091}\times 2015-\frac{671}{627697}\times 2016\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1884959}{1883091}\\\frac{2}{1883091}\end{matrix}\right)
Do the arithmetic.
x=\frac{1884959}{1883091},y=\frac{2}{1883091}
Extract the matrix elements x and y.
2013x+2949y=2015,2014x+2015y=2016
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.
2014\times 2013x+2014\times 2949y=2014\times 2015,2013\times 2014x+2013\times 2015y=2013\times 2016
To make 2013x and 2014x equal, multiply all terms on each side of the first equation by 2014 and all terms on each side of the second by 2013.
4054182x+5939286y=4058210,4054182x+4056195y=4058208
Simplify.
4054182x-4054182x+5939286y-4056195y=4058210-4058208
Subtract 4054182x+4056195y=4058208 from 4054182x+5939286y=4058210 by subtracting like terms on each side of the equal sign.
5939286y-4056195y=4058210-4058208
Add 4054182x to -4054182x. Terms 4054182x and -4054182x cancel out, leaving an equation with only one variable that can be solved.
1883091y=4058210-4058208
Add 5939286y to -4056195y.
1883091y=2
Add 4058210 to -4058208.
y=\frac{2}{1883091}
Divide both sides by 1883091.
2014x+2015\times \frac{2}{1883091}=2016
Substitute \frac{2}{1883091} for y in 2014x+2015y=2016. Because the resulting equation contains only one variable, you can solve for x directly.
2014x+\frac{4030}{1883091}=2016
Multiply 2015 times \frac{2}{1883091}.
2014x=\frac{3796307426}{1883091}
Subtract \frac{4030}{1883091} from both sides of the equation.
x=\frac{1884959}{1883091}
Divide both sides by 2014.
x=\frac{1884959}{1883091},y=\frac{2}{1883091}
The system is now solved.