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6x+5y=70,3x+7y=710
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+5y=70
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
6x=-5y+70
Subtract 5y from both sides of the equation.
x=\frac{1}{6}\left(-5y+70\right)
Divide both sides by 6.
x=-\frac{5}{6}y+\frac{35}{3}
Multiply \frac{1}{6} times -5y+70.
3\left(-\frac{5}{6}y+\frac{35}{3}\right)+7y=710
Substitute -\frac{5y}{6}+\frac{35}{3} for x in the other equation, 3x+7y=710.
-\frac{5}{2}y+35+7y=710
Multiply 3 times -\frac{5y}{6}+\frac{35}{3}.
\frac{9}{2}y+35=710
Add -\frac{5y}{2} to 7y.
\frac{9}{2}y=675
Subtract 35 from both sides of the equation.
y=150
Divide both sides of the equation by \frac{9}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{5}{6}\times 150+\frac{35}{3}
Substitute 150 for y in x=-\frac{5}{6}y+\frac{35}{3}. Because the resulting equation contains only one variable, you can solve for x directly.
x=-125+\frac{35}{3}
Multiply -\frac{5}{6} times 150.
x=-\frac{340}{3}
Add \frac{35}{3} to -125.
x=-\frac{340}{3},y=150
The system is now solved.
6x+5y=70,3x+7y=710
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}6&5\\3&7\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}70\\710\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}6&5\\3&7\end{matrix}\right))\left(\begin{matrix}6&5\\3&7\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&5\\3&7\end{matrix}\right))\left(\begin{matrix}70\\710\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&5\\3&7\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&5\\3&7\end{matrix}\right))\left(\begin{matrix}70\\710\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&5\\3&7\end{matrix}\right))\left(\begin{matrix}70\\710\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{7}{6\times 7-5\times 3}&-\frac{5}{6\times 7-5\times 3}\\-\frac{3}{6\times 7-5\times 3}&\frac{6}{6\times 7-5\times 3}\end{matrix}\right)\left(\begin{matrix}70\\710\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{7}{27}&-\frac{5}{27}\\-\frac{1}{9}&\frac{2}{9}\end{matrix}\right)\left(\begin{matrix}70\\710\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{27}\times 70-\frac{5}{27}\times 710\\-\frac{1}{9}\times 70+\frac{2}{9}\times 710\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{340}{3}\\150\end{matrix}\right)
Do the arithmetic.
x=-\frac{340}{3},y=150
Extract the matrix elements x and y.
6x+5y=70,3x+7y=710
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 5y=3\times 70,6\times 3x+6\times 7y=6\times 710
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+15y=210,18x+42y=4260
Simplify.
18x-18x+15y-42y=210-4260
Subtract 18x+42y=4260 from 18x+15y=210 by subtracting like terms on each side of the equal sign.
15y-42y=210-4260
Add 18x to -18x. Terms 18x and -18x cancel out, leaving an equation with only one variable that can be solved.
-27y=210-4260
Add 15y to -42y.
-27y=-4050
Add 210 to -4260.
y=150
Divide both sides by -27.
3x+7\times 150=710
Substitute 150 for y in 3x+7y=710. Because the resulting equation contains only one variable, you can solve for x directly.
3x+1050=710
Multiply 7 times 150.
3x=-340
Subtract 1050 from both sides of the equation.
x=-\frac{340}{3}
Divide both sides by 3.
x=-\frac{340}{3},y=150
The system is now solved.