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x-y=4
Consider the first equation. Subtract y from both sides.
x-y=4,x+y=150
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=4
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=y+4
Add y to both sides of the equation.
y+4+y=150
Substitute y+4 for x in the other equation, x+y=150.
2y+4=150
Add y to y.
2y=146
Subtract 4 from both sides of the equation.
y=73
Divide both sides by 2.
x=73+4
Substitute 73 for y in x=y+4. Because the resulting equation contains only one variable, you can solve for x directly.
x=77
Add 4 to 73.
x=77,y=73
The system is now solved.
x-y=4
Consider the first equation. Subtract y from both sides.
x-y=4,x+y=150
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&-1\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4\\150\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&-1\\1&1\end{matrix}\right))\left(\begin{matrix}1&-1\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\1&1\end{matrix}\right))\left(\begin{matrix}4\\150\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-1\\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}1&-1\\1&1\end{matrix}\right))\left(\begin{matrix}4\\150\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\\1&1\end{matrix}\right))\left(\begin{matrix}4\\150\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-\left(-1\right)}&-\frac{-1}{1-\left(-1\right)}\\-\frac{1}{1-\left(-1\right)}&\frac{1}{1-\left(-1\right)}\end{matrix}\right)\left(\begin{matrix}4\\150\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}{2}&\frac{1}{2}\\-\frac{1}{2}&\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}4\\150\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}\times 4+\frac{1}{2}\times 150\\-\frac{1}{2}\times 4+\frac{1}{2}\times 150\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}77\\73\end{matrix}\right)
Do the arithmetic.
x=77,y=73
Extract the matrix elements x and y.
x-y=4
Consider the first equation. Subtract y from both sides.
x-y=4,x+y=150
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-x-y-y=4-150
Subtract x+y=150 from x-y=4 by subtracting like terms on each side of the equal sign.
-y-y=4-150
Add x to -x. Terms x and -x cancel out, leaving an equation with only one variable that can be solved.
-2y=4-150
Add -y to -y.
-2y=-146
Add 4 to -150.
y=73
Divide both sides by -2.
x+73=150
Substitute 73 for y in x+y=150. Because the resulting equation contains only one variable, you can solve for x directly.
x=77
Subtract 73 from both sides of the equation.
x=77,y=73
The system is now solved.