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x+y=50,7x+3y=278
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=50
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+50
Subtract y from both sides of the equation.
7\left(-y+50\right)+3y=278
Substitute -y+50 for x in the other equation, 7x+3y=278.
-7y+350+3y=278
Multiply 7 times -y+50.
-4y+350=278
Add -7y to 3y.
-4y=-72
Subtract 350 from both sides of the equation.
y=18
Divide both sides by -4.
x=-18+50
Substitute 18 for y in x=-y+50. Because the resulting equation contains only one variable, you can solve for x directly.
x=32
Add 50 to -18.
x=32,y=18
The system is now solved.
x+y=50,7x+3y=278
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\7&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}50\\278\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\7&3\end{matrix}\right))\left(\begin{matrix}1&1\\7&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\7&3\end{matrix}\right))\left(\begin{matrix}50\\278\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\7&3\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\\7&3\end{matrix}\right))\left(\begin{matrix}50\\278\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\\7&3\end{matrix}\right))\left(\begin{matrix}50\\278\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{3}{3-7}&-\frac{1}{3-7}\\-\frac{7}{3-7}&\frac{1}{3-7}\end{matrix}\right)\left(\begin{matrix}50\\278\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{3}{4}&\frac{1}{4}\\\frac{7}{4}&-\frac{1}{4}\end{matrix}\right)\left(\begin{matrix}50\\278\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{4}\times 50+\frac{1}{4}\times 278\\\frac{7}{4}\times 50-\frac{1}{4}\times 278\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}32\\18\end{matrix}\right)
Do the arithmetic.
x=32,y=18
Extract the matrix elements x and y.
x+y=50,7x+3y=278
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.
7x+7y=7\times 50,7x+3y=278
To make x and 7x equal, multiply all terms on each side of the first equation by 7 and all terms on each side of the second by 1.
7x+7y=350,7x+3y=278
Simplify.
7x-7x+7y-3y=350-278
Subtract 7x+3y=278 from 7x+7y=350 by subtracting like terms on each side of the equal sign.
7y-3y=350-278
Add 7x to -7x. Terms 7x and -7x cancel out, leaving an equation with only one variable that can be solved.
4y=350-278
Add 7y to -3y.
4y=72
Add 350 to -278.
y=18
Divide both sides by 4.
7x+3\times 18=278
Substitute 18 for y in 7x+3y=278. Because the resulting equation contains only one variable, you can solve for x directly.
7x+54=278
Multiply 3 times 18.
7x=224
Subtract 54 from both sides of the equation.
x=32
Divide both sides by 7.
x=32,y=18
The system is now solved.