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3y+2x=75,y+x=50
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.
3y+2x=75
Choose one of the equations and solve it for y by isolating y on the left hand side of the equal sign.
3y=-2x+75
Subtract 2x from both sides of the equation.
y=\frac{1}{3}\left(-2x+75\right)
Divide both sides by 3.
y=-\frac{2}{3}x+25
Multiply \frac{1}{3} times -2x+75.
-\frac{2}{3}x+25+x=50
Substitute -\frac{2x}{3}+25 for y in the other equation, y+x=50.
\frac{1}{3}x+25=50
Add -\frac{2x}{3} to x.
\frac{1}{3}x=25
Subtract 25 from both sides of the equation.
x=75
Multiply both sides by 3.
y=-\frac{2}{3}\times 75+25
Substitute 75 for x in y=-\frac{2}{3}x+25. Because the resulting equation contains only one variable, you can solve for y directly.
y=-50+25
Multiply -\frac{2}{3} times 75.
y=-25
Add 25 to -50.
y=-25,x=75
The system is now solved.
3y+2x=75,y+x=50
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}3&2\\1&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}75\\50\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}3&2\\1&1\end{matrix}\right))\left(\begin{matrix}3&2\\1&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\1&1\end{matrix}\right))\left(\begin{matrix}75\\50\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&2\\1&1\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\1&1\end{matrix}\right))\left(\begin{matrix}75\\50\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\1&1\end{matrix}\right))\left(\begin{matrix}75\\50\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3-2}&-\frac{2}{3-2}\\-\frac{1}{3-2}&\frac{3}{3-2}\end{matrix}\right)\left(\begin{matrix}75\\50\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}y\\x\end{matrix}\right)=\left(\begin{matrix}1&-2\\-1&3\end{matrix}\right)\left(\begin{matrix}75\\50\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}75-2\times 50\\-75+3\times 50\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-25\\75\end{matrix}\right)
Do the arithmetic.
y=-25,x=75
Extract the matrix elements y and x.
3y+2x=75,y+x=50
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.
3y+2x=75,3y+3x=3\times 50
To make 3y and y equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 3.
3y+2x=75,3y+3x=150
Simplify.
3y-3y+2x-3x=75-150
Subtract 3y+3x=150 from 3y+2x=75 by subtracting like terms on each side of the equal sign.
2x-3x=75-150
Add 3y to -3y. Terms 3y and -3y cancel out, leaving an equation with only one variable that can be solved.
-x=75-150
Add 2x to -3x.
-x=-75
Add 75 to -150.
x=75
Divide both sides by -1.
y+75=50
Substitute 75 for x in y+x=50. Because the resulting equation contains only one variable, you can solve for y directly.
y=-25
Subtract 75 from both sides of the equation.
y=-25,x=75
The system is now solved.