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2\left(x+y\right)+3\left(y-x\right)=54
Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.
2x+2y+3\left(y-x\right)=54
Use the distributive property to multiply 2 by x+y.
2x+2y+3y-3x=54
Use the distributive property to multiply 3 by y-x.
2x+5y-3x=54
Combine 2y and 3y to get 5y.
-x+5y=54
Combine 2x and -3x to get -x.
2x+x+y=60
Consider the second equation. Multiply both sides of the equation by 4, the least common multiple of 2,4.
3x+y=60
Combine 2x and x to get 3x.
-x+5y=54,3x+y=60
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.
-x+5y=54
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
-x=-5y+54
Subtract 5y from both sides of the equation.
x=-\left(-5y+54\right)
Divide both sides by -1.
x=5y-54
Multiply -1 times -5y+54.
3\left(5y-54\right)+y=60
Substitute 5y-54 for x in the other equation, 3x+y=60.
15y-162+y=60
Multiply 3 times 5y-54.
16y-162=60
Add 15y to y.
16y=222
Add 162 to both sides of the equation.
y=\frac{111}{8}
Divide both sides by 16.
x=5\times \frac{111}{8}-54
Substitute \frac{111}{8} for y in x=5y-54. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{555}{8}-54
Multiply 5 times \frac{111}{8}.
x=\frac{123}{8}
Add -54 to \frac{555}{8}.
x=\frac{123}{8},y=\frac{111}{8}
The system is now solved.
2\left(x+y\right)+3\left(y-x\right)=54
Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.
2x+2y+3\left(y-x\right)=54
Use the distributive property to multiply 2 by x+y.
2x+2y+3y-3x=54
Use the distributive property to multiply 3 by y-x.
2x+5y-3x=54
Combine 2y and 3y to get 5y.
-x+5y=54
Combine 2x and -3x to get -x.
2x+x+y=60
Consider the second equation. Multiply both sides of the equation by 4, the least common multiple of 2,4.
3x+y=60
Combine 2x and x to get 3x.
-x+5y=54,3x+y=60
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}-1&5\\3&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}54\\60\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}-1&5\\3&1\end{matrix}\right))\left(\begin{matrix}-1&5\\3&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-1&5\\3&1\end{matrix}\right))\left(\begin{matrix}54\\60\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}-1&5\\3&1\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}-1&5\\3&1\end{matrix}\right))\left(\begin{matrix}54\\60\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}-1&5\\3&1\end{matrix}\right))\left(\begin{matrix}54\\60\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{1}{-1-5\times 3}&-\frac{5}{-1-5\times 3}\\-\frac{3}{-1-5\times 3}&-\frac{1}{-1-5\times 3}\end{matrix}\right)\left(\begin{matrix}54\\60\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{1}{16}&\frac{5}{16}\\\frac{3}{16}&\frac{1}{16}\end{matrix}\right)\left(\begin{matrix}54\\60\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{16}\times 54+\frac{5}{16}\times 60\\\frac{3}{16}\times 54+\frac{1}{16}\times 60\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{123}{8}\\\frac{111}{8}\end{matrix}\right)
Do the arithmetic.
x=\frac{123}{8},y=\frac{111}{8}
Extract the matrix elements x and y.
2\left(x+y\right)+3\left(y-x\right)=54
Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.
2x+2y+3\left(y-x\right)=54
Use the distributive property to multiply 2 by x+y.
2x+2y+3y-3x=54
Use the distributive property to multiply 3 by y-x.
2x+5y-3x=54
Combine 2y and 3y to get 5y.
-x+5y=54
Combine 2x and -3x to get -x.
2x+x+y=60
Consider the second equation. Multiply both sides of the equation by 4, the least common multiple of 2,4.
3x+y=60
Combine 2x and x to get 3x.
-x+5y=54,3x+y=60
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\left(-1\right)x+3\times 5y=3\times 54,-3x-y=-60
To make -x 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 -1.
-3x+15y=162,-3x-y=-60
Simplify.
-3x+3x+15y+y=162+60
Subtract -3x-y=-60 from -3x+15y=162 by subtracting like terms on each side of the equal sign.
15y+y=162+60
Add -3x to 3x. Terms -3x and 3x cancel out, leaving an equation with only one variable that can be solved.
16y=162+60
Add 15y to y.
16y=222
Add 162 to 60.
y=\frac{111}{8}
Divide both sides by 16.
3x+\frac{111}{8}=60
Substitute \frac{111}{8} for y in 3x+y=60. Because the resulting equation contains only one variable, you can solve for x directly.
3x=\frac{369}{8}
Subtract \frac{111}{8} from both sides of the equation.
x=\frac{123}{8}
Divide both sides by 3.
x=\frac{123}{8},y=\frac{111}{8}
The system is now solved.