\left\{ \begin{array} { l } { 2 x + 3 y = 120 } \\ { x + y = 1 } \end{array} \right.
Solve for x, y
x=-117
y=118
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2x+3y=120,x+y=1
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+3y=120
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
2x=-3y+120
Subtract 3y from both sides of the equation.
x=\frac{1}{2}\left(-3y+120\right)
Divide both sides by 2.
x=-\frac{3}{2}y+60
Multiply \frac{1}{2} times -3y+120.
-\frac{3}{2}y+60+y=1
Substitute -\frac{3y}{2}+60 for x in the other equation, x+y=1.
-\frac{1}{2}y+60=1
Add -\frac{3y}{2} to y.
-\frac{1}{2}y=-59
Subtract 60 from both sides of the equation.
y=118
Multiply both sides by -2.
x=-\frac{3}{2}\times 118+60
Substitute 118 for y in x=-\frac{3}{2}y+60. Because the resulting equation contains only one variable, you can solve for x directly.
x=-177+60
Multiply -\frac{3}{2} times 118.
x=-117
Add 60 to -177.
x=-117,y=118
The system is now solved.
2x+3y=120,x+y=1
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}2&3\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}120\\1\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}2&3\\1&1\end{matrix}\right))\left(\begin{matrix}2&3\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&3\\1&1\end{matrix}\right))\left(\begin{matrix}120\\1\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&3\\1&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}2&3\\1&1\end{matrix}\right))\left(\begin{matrix}120\\1\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&3\\1&1\end{matrix}\right))\left(\begin{matrix}120\\1\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}{2-3}&-\frac{3}{2-3}\\-\frac{1}{2-3}&\frac{2}{2-3}\end{matrix}\right)\left(\begin{matrix}120\\1\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&3\\1&-2\end{matrix}\right)\left(\begin{matrix}120\\1\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-120+3\\120-2\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-117\\118\end{matrix}\right)
Do the arithmetic.
x=-117,y=118
Extract the matrix elements x and y.
2x+3y=120,x+y=1
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+3y=120,2x+2y=2
To make 2x and x equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 2.
2x-2x+3y-2y=120-2
Subtract 2x+2y=2 from 2x+3y=120 by subtracting like terms on each side of the equal sign.
3y-2y=120-2
Add 2x to -2x. Terms 2x and -2x cancel out, leaving an equation with only one variable that can be solved.
y=120-2
Add 3y to -2y.
y=118
Add 120 to -2.
x+118=1
Substitute 118 for y in x+y=1. Because the resulting equation contains only one variable, you can solve for x directly.
x=-117
Subtract 118 from both sides of the equation.
x=-117,y=118
The system is now solved.
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Simultaneous equation
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Limits
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