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x+y=70,50x+70y=3800
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+y=70
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+70
Subtract y from both sides of the equation.
50\left(-y+70\right)+70y=3800
Substitute -y+70 for x in the other equation, 50x+70y=3800.
-50y+3500+70y=3800
Multiply 50 times -y+70.
20y+3500=3800
Add -50y to 70y.
20y=300
Subtract 3500 from both sides of the equation.
y=15
Divide both sides by 20.
x=-15+70
Substitute 15 for y in x=-y+70. Because the resulting equation contains only one variable, you can solve for x directly.
x=55
Add 70 to -15.
x=55,y=15
The system is now solved.
x+y=70,50x+70y=3800
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\50&70\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}70\\3800\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\50&70\end{matrix}\right))\left(\begin{matrix}1&1\\50&70\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\50&70\end{matrix}\right))\left(\begin{matrix}70\\3800\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\50&70\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&1\\50&70\end{matrix}\right))\left(\begin{matrix}70\\3800\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&1\\50&70\end{matrix}\right))\left(\begin{matrix}70\\3800\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{70}{70-50}&-\frac{1}{70-50}\\-\frac{50}{70-50}&\frac{1}{70-50}\end{matrix}\right)\left(\begin{matrix}70\\3800\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{7}{2}&-\frac{1}{20}\\-\frac{5}{2}&\frac{1}{20}\end{matrix}\right)\left(\begin{matrix}70\\3800\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{2}\times 70-\frac{1}{20}\times 3800\\-\frac{5}{2}\times 70+\frac{1}{20}\times 3800\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}55\\15\end{matrix}\right)
Do the arithmetic.
x=55,y=15
Extract the matrix elements x and y.
x+y=70,50x+70y=3800
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+50y=50\times 70,50x+70y=3800
To make x and 50x equal, multiply all terms on each side of the first equation by 50 and all terms on each side of the second by 1.
50x+50y=3500,50x+70y=3800
Simplify.
50x-50x+50y-70y=3500-3800
Subtract 50x+70y=3800 from 50x+50y=3500 by subtracting like terms on each side of the equal sign.
50y-70y=3500-3800
Add 50x to -50x. Terms 50x and -50x cancel out, leaving an equation with only one variable that can be solved.
-20y=3500-3800
Add 50y to -70y.
-20y=-300
Add 3500 to -3800.
y=15
Divide both sides by -20.
50x+70\times 15=3800
Substitute 15 for y in 50x+70y=3800. Because the resulting equation contains only one variable, you can solve for x directly.
50x+1050=3800
Multiply 70 times 15.
50x=2750
Subtract 1050 from both sides of the equation.
x=55
Divide both sides by 50.
x=55,y=15
The system is now solved.