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2020x+2021y=2019,2021x+2020y=202
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.
2020x+2021y=2019
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
2020x=-2021y+2019
Subtract 2021y from both sides of the equation.
x=\frac{1}{2020}\left(-2021y+2019\right)
Divide both sides by 2020.
x=-\frac{2021}{2020}y+\frac{2019}{2020}
Multiply \frac{1}{2020} times -2021y+2019.
2021\left(-\frac{2021}{2020}y+\frac{2019}{2020}\right)+2020y=202
Substitute \frac{-2021y+2019}{2020} for x in the other equation, 2021x+2020y=202.
-\frac{4084441}{2020}y+\frac{4080399}{2020}+2020y=202
Multiply 2021 times \frac{-2021y+2019}{2020}.
-\frac{4041}{2020}y+\frac{4080399}{2020}=202
Add -\frac{4084441y}{2020} to 2020y.
-\frac{4041}{2020}y=-\frac{3672359}{2020}
Subtract \frac{4080399}{2020} from both sides of the equation.
y=\frac{3672359}{4041}
Divide both sides of the equation by -\frac{4041}{2020}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{2021}{2020}\times \frac{3672359}{4041}+\frac{2019}{2020}
Substitute \frac{3672359}{4041} for y in x=-\frac{2021}{2020}y+\frac{2019}{2020}. Because the resulting equation contains only one variable, you can solve for x directly.
x=-\frac{7421837539}{8162820}+\frac{2019}{2020}
Multiply -\frac{2021}{2020} times \frac{3672359}{4041} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=-\frac{3670138}{4041}
Add \frac{2019}{2020} to -\frac{7421837539}{8162820} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=-\frac{3670138}{4041},y=\frac{3672359}{4041}
The system is now solved.
2020x+2021y=2019,2021x+2020y=202
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}2020&2021\\2021&2020\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2019\\202\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}2020&2021\\2021&2020\end{matrix}\right))\left(\begin{matrix}2020&2021\\2021&2020\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2020&2021\\2021&2020\end{matrix}\right))\left(\begin{matrix}2019\\202\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}2020&2021\\2021&2020\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}2020&2021\\2021&2020\end{matrix}\right))\left(\begin{matrix}2019\\202\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}2020&2021\\2021&2020\end{matrix}\right))\left(\begin{matrix}2019\\202\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{2020}{2020\times 2020-2021\times 2021}&-\frac{2021}{2020\times 2020-2021\times 2021}\\-\frac{2021}{2020\times 2020-2021\times 2021}&\frac{2020}{2020\times 2020-2021\times 2021}\end{matrix}\right)\left(\begin{matrix}2019\\202\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{2020}{4041}&\frac{2021}{4041}\\\frac{2021}{4041}&-\frac{2020}{4041}\end{matrix}\right)\left(\begin{matrix}2019\\202\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2020}{4041}\times 2019+\frac{2021}{4041}\times 202\\\frac{2021}{4041}\times 2019-\frac{2020}{4041}\times 202\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3670138}{4041}\\\frac{3672359}{4041}\end{matrix}\right)
Do the arithmetic.
x=-\frac{3670138}{4041},y=\frac{3672359}{4041}
Extract the matrix elements x and y.
2020x+2021y=2019,2021x+2020y=202
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.
2021\times 2020x+2021\times 2021y=2021\times 2019,2020\times 2021x+2020\times 2020y=2020\times 202
To make 2020x and 2021x equal, multiply all terms on each side of the first equation by 2021 and all terms on each side of the second by 2020.
4082420x+4084441y=4080399,4082420x+4080400y=408040
Simplify.
4082420x-4082420x+4084441y-4080400y=4080399-408040
Subtract 4082420x+4080400y=408040 from 4082420x+4084441y=4080399 by subtracting like terms on each side of the equal sign.
4084441y-4080400y=4080399-408040
Add 4082420x to -4082420x. Terms 4082420x and -4082420x cancel out, leaving an equation with only one variable that can be solved.
4041y=4080399-408040
Add 4084441y to -4080400y.
4041y=3672359
Add 4080399 to -408040.
y=\frac{3672359}{4041}
Divide both sides by 4041.
2021x+2020\times \frac{3672359}{4041}=202
Substitute \frac{3672359}{4041} for y in 2021x+2020y=202. Because the resulting equation contains only one variable, you can solve for x directly.
2021x+\frac{7418165180}{4041}=202
Multiply 2020 times \frac{3672359}{4041}.
2021x=-\frac{7417348898}{4041}
Subtract \frac{7418165180}{4041} from both sides of the equation.
x=-\frac{3670138}{4041}
Divide both sides by 2021.
x=-\frac{3670138}{4041},y=\frac{3672359}{4041}
The system is now solved.