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2x+2y=-104
Consider the first equation. Add 2y to both sides.
3x+2y=-3
Consider the second equation. Add 2y to both sides.
2x+2y=-104,3x+2y=-3
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.
2x+2y=-104
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
2x=-2y-104
Subtract 2y from both sides of the equation.
x=\frac{1}{2}\left(-2y-104\right)
Divide both sides by 2.
x=-y-52
Multiply \frac{1}{2} times -2y-104.
3\left(-y-52\right)+2y=-3
Substitute -y-52 for x in the other equation, 3x+2y=-3.
-3y-156+2y=-3
Multiply 3 times -y-52.
-y-156=-3
Add -3y to 2y.
-y=153
Add 156 to both sides of the equation.
y=-153
Divide both sides by -1.
x=-\left(-153\right)-52
Substitute -153 for y in x=-y-52. Because the resulting equation contains only one variable, you can solve for x directly.
x=153-52
Multiply -1 times -153.
x=101
Add -52 to 153.
x=101,y=-153
The system is now solved.
2x+2y=-104
Consider the first equation. Add 2y to both sides.
3x+2y=-3
Consider the second equation. Add 2y to both sides.
2x+2y=-104,3x+2y=-3
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}2&2\\3&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-104\\-3\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}2&2\\3&2\end{matrix}\right))\left(\begin{matrix}2&2\\3&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&2\\3&2\end{matrix}\right))\left(\begin{matrix}-104\\-3\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&2\\3&2\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}2&2\\3&2\end{matrix}\right))\left(\begin{matrix}-104\\-3\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}2&2\\3&2\end{matrix}\right))\left(\begin{matrix}-104\\-3\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{2}{2\times 2-2\times 3}&-\frac{2}{2\times 2-2\times 3}\\-\frac{3}{2\times 2-2\times 3}&\frac{2}{2\times 2-2\times 3}\end{matrix}\right)\left(\begin{matrix}-104\\-3\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}-1&1\\\frac{3}{2}&-1\end{matrix}\right)\left(\begin{matrix}-104\\-3\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\left(-104\right)-3\\\frac{3}{2}\left(-104\right)-\left(-3\right)\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}101\\-153\end{matrix}\right)
Do the arithmetic.
x=101,y=-153
Extract the matrix elements x and y.
2x+2y=-104
Consider the first equation. Add 2y to both sides.
3x+2y=-3
Consider the second equation. Add 2y to both sides.
2x+2y=-104,3x+2y=-3
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.
2x-3x+2y-2y=-104+3
Subtract 3x+2y=-3 from 2x+2y=-104 by subtracting like terms on each side of the equal sign.
2x-3x=-104+3
Add 2y to -2y. Terms 2y and -2y cancel out, leaving an equation with only one variable that can be solved.
-x=-104+3
Add 2x to -3x.
-x=-101
Add -104 to 3.
x=101
Divide both sides by -1.
3\times 101+2y=-3
Substitute 101 for x in 3x+2y=-3. Because the resulting equation contains only one variable, you can solve for y directly.
303+2y=-3
Multiply 3 times 101.
2y=-306
Subtract 303 from both sides of the equation.
y=-153
Divide both sides by 2.
x=101,y=-153
The system is now solved.