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6x+5y=60
Consider the first equation. Multiply both sides of the equation by 30, the least common multiple of 5,6.
6\times 2x-5\times 7y=-150
Consider the second equation. Multiply both sides of the equation by 30, the least common multiple of 5,6.
12x-5\times 7y=-150
Multiply 6 and 2 to get 12.
12x-35y=-150
Multiply -5 and 7 to get -35.
6x+5y=60,12x-35y=-150
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=60
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
6x=-5y+60
Subtract 5y from both sides of the equation.
x=\frac{1}{6}\left(-5y+60\right)
Divide both sides by 6.
x=-\frac{5}{6}y+10
Multiply \frac{1}{6} times -5y+60.
12\left(-\frac{5}{6}y+10\right)-35y=-150
Substitute -\frac{5y}{6}+10 for x in the other equation, 12x-35y=-150.
-10y+120-35y=-150
Multiply 12 times -\frac{5y}{6}+10.
-45y+120=-150
Add -10y to -35y.
-45y=-270
Subtract 120 from both sides of the equation.
y=6
Divide both sides by -45.
x=-\frac{5}{6}\times 6+10
Substitute 6 for y in x=-\frac{5}{6}y+10. Because the resulting equation contains only one variable, you can solve for x directly.
x=-5+10
Multiply -\frac{5}{6} times 6.
x=5
Add 10 to -5.
x=5,y=6
The system is now solved.
6x+5y=60
Consider the first equation. Multiply both sides of the equation by 30, the least common multiple of 5,6.
6\times 2x-5\times 7y=-150
Consider the second equation. Multiply both sides of the equation by 30, the least common multiple of 5,6.
12x-5\times 7y=-150
Multiply 6 and 2 to get 12.
12x-35y=-150
Multiply -5 and 7 to get -35.
6x+5y=60,12x-35y=-150
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}6&5\\12&-35\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}60\\-150\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}6&5\\12&-35\end{matrix}\right))\left(\begin{matrix}6&5\\12&-35\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&5\\12&-35\end{matrix}\right))\left(\begin{matrix}60\\-150\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&5\\12&-35\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\\12&-35\end{matrix}\right))\left(\begin{matrix}60\\-150\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\\12&-35\end{matrix}\right))\left(\begin{matrix}60\\-150\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{35}{6\left(-35\right)-5\times 12}&-\frac{5}{6\left(-35\right)-5\times 12}\\-\frac{12}{6\left(-35\right)-5\times 12}&\frac{6}{6\left(-35\right)-5\times 12}\end{matrix}\right)\left(\begin{matrix}60\\-150\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}{54}&\frac{1}{54}\\\frac{2}{45}&-\frac{1}{45}\end{matrix}\right)\left(\begin{matrix}60\\-150\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{54}\times 60+\frac{1}{54}\left(-150\right)\\\frac{2}{45}\times 60-\frac{1}{45}\left(-150\right)\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}5\\6\end{matrix}\right)
Do the arithmetic.
x=5,y=6
Extract the matrix elements x and y.
6x+5y=60
Consider the first equation. Multiply both sides of the equation by 30, the least common multiple of 5,6.
6\times 2x-5\times 7y=-150
Consider the second equation. Multiply both sides of the equation by 30, the least common multiple of 5,6.
12x-5\times 7y=-150
Multiply 6 and 2 to get 12.
12x-35y=-150
Multiply -5 and 7 to get -35.
6x+5y=60,12x-35y=-150
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.
12\times 6x+12\times 5y=12\times 60,6\times 12x+6\left(-35\right)y=6\left(-150\right)
To make 6x and 12x equal, multiply all terms on each side of the first equation by 12 and all terms on each side of the second by 6.
72x+60y=720,72x-210y=-900
Simplify.
72x-72x+60y+210y=720+900
Subtract 72x-210y=-900 from 72x+60y=720 by subtracting like terms on each side of the equal sign.
60y+210y=720+900
Add 72x to -72x. Terms 72x and -72x cancel out, leaving an equation with only one variable that can be solved.
270y=720+900
Add 60y to 210y.
270y=1620
Add 720 to 900.
y=6
Divide both sides by 270.
12x-35\times 6=-150
Substitute 6 for y in 12x-35y=-150. Because the resulting equation contains only one variable, you can solve for x directly.
12x-210=-150
Multiply -35 times 6.
12x=60
Add 210 to both sides of the equation.
x=5
Divide both sides by 12.
x=5,y=6
The system is now solved.