\left\{ \begin{array} { l } { 2 ( x + y ) - 3 ( x - y ) = 4 } \\ { 5 ( x + y ) - 7 ( x - y ) = 2 } \end{array} \right.
Solve for x, y
x=-19
y=-3
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2x+2y-3\left(x-y\right)=4
Consider the first equation. Use the distributive property to multiply 2 by x+y.
2x+2y-3x+3y=4
Use the distributive property to multiply -3 by x-y.
-x+2y+3y=4
Combine 2x and -3x to get -x.
-x+5y=4
Combine 2y and 3y to get 5y.
5x+5y-7\left(x-y\right)=2
Consider the second equation. Use the distributive property to multiply 5 by x+y.
5x+5y-7x+7y=2
Use the distributive property to multiply -7 by x-y.
-2x+5y+7y=2
Combine 5x and -7x to get -2x.
-2x+12y=2
Combine 5y and 7y to get 12y.
-x+5y=4,-2x+12y=2
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=4
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
-x=-5y+4
Subtract 5y from both sides of the equation.
x=-\left(-5y+4\right)
Divide both sides by -1.
x=5y-4
Multiply -1 times -5y+4.
-2\left(5y-4\right)+12y=2
Substitute 5y-4 for x in the other equation, -2x+12y=2.
-10y+8+12y=2
Multiply -2 times 5y-4.
2y+8=2
Add -10y to 12y.
2y=-6
Subtract 8 from both sides of the equation.
y=-3
Divide both sides by 2.
x=5\left(-3\right)-4
Substitute -3 for y in x=5y-4. Because the resulting equation contains only one variable, you can solve for x directly.
x=-15-4
Multiply 5 times -3.
x=-19
Add -4 to -15.
x=-19,y=-3
The system is now solved.
2x+2y-3\left(x-y\right)=4
Consider the first equation. Use the distributive property to multiply 2 by x+y.
2x+2y-3x+3y=4
Use the distributive property to multiply -3 by x-y.
-x+2y+3y=4
Combine 2x and -3x to get -x.
-x+5y=4
Combine 2y and 3y to get 5y.
5x+5y-7\left(x-y\right)=2
Consider the second equation. Use the distributive property to multiply 5 by x+y.
5x+5y-7x+7y=2
Use the distributive property to multiply -7 by x-y.
-2x+5y+7y=2
Combine 5x and -7x to get -2x.
-2x+12y=2
Combine 5y and 7y to get 12y.
-x+5y=4,-2x+12y=2
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}-1&5\\-2&12\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4\\2\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}-1&5\\-2&12\end{matrix}\right))\left(\begin{matrix}-1&5\\-2&12\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-1&5\\-2&12\end{matrix}\right))\left(\begin{matrix}4\\2\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}-1&5\\-2&12\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\\-2&12\end{matrix}\right))\left(\begin{matrix}4\\2\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\\-2&12\end{matrix}\right))\left(\begin{matrix}4\\2\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{12}{-12-5\left(-2\right)}&-\frac{5}{-12-5\left(-2\right)}\\-\frac{-2}{-12-5\left(-2\right)}&-\frac{1}{-12-5\left(-2\right)}\end{matrix}\right)\left(\begin{matrix}4\\2\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}-6&\frac{5}{2}\\-1&\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}4\\2\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-6\times 4+\frac{5}{2}\times 2\\-4+\frac{1}{2}\times 2\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-19\\-3\end{matrix}\right)
Do the arithmetic.
x=-19,y=-3
Extract the matrix elements x and y.
2x+2y-3\left(x-y\right)=4
Consider the first equation. Use the distributive property to multiply 2 by x+y.
2x+2y-3x+3y=4
Use the distributive property to multiply -3 by x-y.
-x+2y+3y=4
Combine 2x and -3x to get -x.
-x+5y=4
Combine 2y and 3y to get 5y.
5x+5y-7\left(x-y\right)=2
Consider the second equation. Use the distributive property to multiply 5 by x+y.
5x+5y-7x+7y=2
Use the distributive property to multiply -7 by x-y.
-2x+5y+7y=2
Combine 5x and -7x to get -2x.
-2x+12y=2
Combine 5y and 7y to get 12y.
-x+5y=4,-2x+12y=2
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.
-2\left(-1\right)x-2\times 5y=-2\times 4,-\left(-2\right)x-12y=-2
To make -x and -2x equal, multiply all terms on each side of the first equation by -2 and all terms on each side of the second by -1.
2x-10y=-8,2x-12y=-2
Simplify.
2x-2x-10y+12y=-8+2
Subtract 2x-12y=-2 from 2x-10y=-8 by subtracting like terms on each side of the equal sign.
-10y+12y=-8+2
Add 2x to -2x. Terms 2x and -2x cancel out, leaving an equation with only one variable that can be solved.
2y=-8+2
Add -10y to 12y.
2y=-6
Add -8 to 2.
y=-3
Divide both sides by 2.
-2x+12\left(-3\right)=2
Substitute -3 for y in -2x+12y=2. Because the resulting equation contains only one variable, you can solve for x directly.
-2x-36=2
Multiply 12 times -3.
-2x=38
Add 36 to both sides of the equation.
x=-19
Divide both sides by -2.
x=-19,y=-3
The system is now solved.
Examples
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Simultaneous equation
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Limits
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