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6x+2y=300,3x+5y=600
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.
6x+2y=300
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
6x=-2y+300
Subtract 2y from both sides of the equation.
x=\frac{1}{6}\left(-2y+300\right)
Divide both sides by 6.
x=-\frac{1}{3}y+50
Multiply \frac{1}{6} times -2y+300.
3\left(-\frac{1}{3}y+50\right)+5y=600
Substitute -\frac{y}{3}+50 for x in the other equation, 3x+5y=600.
-y+150+5y=600
Multiply 3 times -\frac{y}{3}+50.
4y+150=600
Add -y to 5y.
4y=450
Subtract 150 from both sides of the equation.
y=\frac{225}{2}
Divide both sides by 4.
x=-\frac{1}{3}\times \frac{225}{2}+50
Substitute \frac{225}{2} for y in x=-\frac{1}{3}y+50. Because the resulting equation contains only one variable, you can solve for x directly.
x=-\frac{75}{2}+50
Multiply -\frac{1}{3} times \frac{225}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=\frac{25}{2}
Add 50 to -\frac{75}{2}.
x=\frac{25}{2},y=\frac{225}{2}
The system is now solved.
6x+2y=300,3x+5y=600
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}6&2\\3&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}300\\600\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}6&2\\3&5\end{matrix}\right))\left(\begin{matrix}6&2\\3&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&2\\3&5\end{matrix}\right))\left(\begin{matrix}300\\600\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&2\\3&5\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}6&2\\3&5\end{matrix}\right))\left(\begin{matrix}300\\600\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}6&2\\3&5\end{matrix}\right))\left(\begin{matrix}300\\600\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{5}{6\times 5-2\times 3}&-\frac{2}{6\times 5-2\times 3}\\-\frac{3}{6\times 5-2\times 3}&\frac{6}{6\times 5-2\times 3}\end{matrix}\right)\left(\begin{matrix}300\\600\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{5}{24}&-\frac{1}{12}\\-\frac{1}{8}&\frac{1}{4}\end{matrix}\right)\left(\begin{matrix}300\\600\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{24}\times 300-\frac{1}{12}\times 600\\-\frac{1}{8}\times 300+\frac{1}{4}\times 600\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{25}{2}\\\frac{225}{2}\end{matrix}\right)
Do the arithmetic.
x=\frac{25}{2},y=\frac{225}{2}
Extract the matrix elements x and y.
6x+2y=300,3x+5y=600
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.
3\times 6x+3\times 2y=3\times 300,6\times 3x+6\times 5y=6\times 600
To make 6x and 3x equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 6.
18x+6y=900,18x+30y=3600
Simplify.
18x-18x+6y-30y=900-3600
Subtract 18x+30y=3600 from 18x+6y=900 by subtracting like terms on each side of the equal sign.
6y-30y=900-3600
Add 18x to -18x. Terms 18x and -18x cancel out, leaving an equation with only one variable that can be solved.
-24y=900-3600
Add 6y to -30y.
-24y=-2700
Add 900 to -3600.
y=\frac{225}{2}
Divide both sides by -24.
3x+5\times \frac{225}{2}=600
Substitute \frac{225}{2} for y in 3x+5y=600. Because the resulting equation contains only one variable, you can solve for x directly.
3x+\frac{1125}{2}=600
Multiply 5 times \frac{225}{2}.
3x=\frac{75}{2}
Subtract \frac{1125}{2} from both sides of the equation.
x=\frac{25}{2}
Divide both sides by 3.
x=\frac{25}{2},y=\frac{225}{2}
The system is now solved.