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x+y=84,2x+y=154
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=84
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+84
Subtract y from both sides of the equation.
2\left(-y+84\right)+y=154
Substitute -y+84 for x in the other equation, 2x+y=154.
-2y+168+y=154
Multiply 2 times -y+84.
-y+168=154
Add -2y to y.
-y=-14
Subtract 168 from both sides of the equation.
y=14
Divide both sides by -1.
x=-14+84
Substitute 14 for y in x=-y+84. Because the resulting equation contains only one variable, you can solve for x directly.
x=70
Add 84 to -14.
x=70,y=14
The system is now solved.
x+y=84,2x+y=154
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}84\\154\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\2&1\end{matrix}\right))\left(\begin{matrix}1&1\\2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\2&1\end{matrix}\right))\left(\begin{matrix}84\\154\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\2&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}1&1\\2&1\end{matrix}\right))\left(\begin{matrix}84\\154\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\\2&1\end{matrix}\right))\left(\begin{matrix}84\\154\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}{1-2}&-\frac{1}{1-2}\\-\frac{2}{1-2}&\frac{1}{1-2}\end{matrix}\right)\left(\begin{matrix}84\\154\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&1\\2&-1\end{matrix}\right)\left(\begin{matrix}84\\154\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-84+154\\2\times 84-154\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}70\\14\end{matrix}\right)
Do the arithmetic.
x=70,y=14
Extract the matrix elements x and y.
x+y=84,2x+y=154
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.
x-2x+y-y=84-154
Subtract 2x+y=154 from x+y=84 by subtracting like terms on each side of the equal sign.
x-2x=84-154
Add y to -y. Terms y and -y cancel out, leaving an equation with only one variable that can be solved.
-x=84-154
Add x to -2x.
-x=-70
Add 84 to -154.
x=70
Divide both sides by -1.
2\times 70+y=154
Substitute 70 for x in 2x+y=154. Because the resulting equation contains only one variable, you can solve for y directly.
140+y=154
Multiply 2 times 70.
y=14
Subtract 140 from both sides of the equation.
x=70,y=14
The system is now solved.