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50x+25y=4500,-x+y=30
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.
50x+25y=4500
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
50x=-25y+4500
Subtract 25y from both sides of the equation.
x=\frac{1}{50}\left(-25y+4500\right)
Divide both sides by 50.
x=-\frac{1}{2}y+90
Multiply \frac{1}{50} times -25y+4500.
-\left(-\frac{1}{2}y+90\right)+y=30
Substitute -\frac{y}{2}+90 for x in the other equation, -x+y=30.
\frac{1}{2}y-90+y=30
Multiply -1 times -\frac{y}{2}+90.
\frac{3}{2}y-90=30
Add \frac{y}{2} to y.
\frac{3}{2}y=120
Add 90 to both sides of the equation.
y=80
Divide both sides of the equation by \frac{3}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{1}{2}\times 80+90
Substitute 80 for y in x=-\frac{1}{2}y+90. Because the resulting equation contains only one variable, you can solve for x directly.
x=-40+90
Multiply -\frac{1}{2} times 80.
x=50
Add 90 to -40.
x=50,y=80
The system is now solved.
50x+25y=4500,-x+y=30
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}50&25\\-1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4500\\30\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}50&25\\-1&1\end{matrix}\right))\left(\begin{matrix}50&25\\-1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}50&25\\-1&1\end{matrix}\right))\left(\begin{matrix}4500\\30\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}50&25\\-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}50&25\\-1&1\end{matrix}\right))\left(\begin{matrix}4500\\30\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}50&25\\-1&1\end{matrix}\right))\left(\begin{matrix}4500\\30\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}{50-25\left(-1\right)}&-\frac{25}{50-25\left(-1\right)}\\-\frac{-1}{50-25\left(-1\right)}&\frac{50}{50-25\left(-1\right)}\end{matrix}\right)\left(\begin{matrix}4500\\30\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}{75}&-\frac{1}{3}\\\frac{1}{75}&\frac{2}{3}\end{matrix}\right)\left(\begin{matrix}4500\\30\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{75}\times 4500-\frac{1}{3}\times 30\\\frac{1}{75}\times 4500+\frac{2}{3}\times 30\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}50\\80\end{matrix}\right)
Do the arithmetic.
x=50,y=80
Extract the matrix elements x and y.
50x+25y=4500,-x+y=30
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.
-50x-25y=-4500,50\left(-1\right)x+50y=50\times 30
To make 50x 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 50.
-50x-25y=-4500,-50x+50y=1500
Simplify.
-50x+50x-25y-50y=-4500-1500
Subtract -50x+50y=1500 from -50x-25y=-4500 by subtracting like terms on each side of the equal sign.
-25y-50y=-4500-1500
Add -50x to 50x. Terms -50x and 50x cancel out, leaving an equation with only one variable that can be solved.
-75y=-4500-1500
Add -25y to -50y.
-75y=-6000
Add -4500 to -1500.
y=80
Divide both sides by -75.
-x+80=30
Substitute 80 for y in -x+y=30. Because the resulting equation contains only one variable, you can solve for x directly.
-x=-50
Subtract 80 from both sides of the equation.
x=50
Divide both sides by -1.
x=50,y=80
The system is now solved.