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x+y=150,x+0.13y=6300
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=150
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+150
Subtract y from both sides of the equation.
-y+150+0.13y=6300
Substitute -y+150 for x in the other equation, x+0.13y=6300.
-0.87y+150=6300
Add -y to \frac{13y}{100}.
-0.87y=6150
Subtract 150 from both sides of the equation.
y=-\frac{205000}{29}
Divide both sides of the equation by -0.87, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\left(-\frac{205000}{29}\right)+150
Substitute -\frac{205000}{29} for y in x=-y+150. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{205000}{29}+150
Multiply -1 times -\frac{205000}{29}.
x=\frac{209350}{29}
Add 150 to \frac{205000}{29}.
x=\frac{209350}{29},y=-\frac{205000}{29}
The system is now solved.
x+y=150,x+0.13y=6300
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\1&0.13\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}150\\6300\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\1&0.13\end{matrix}\right))\left(\begin{matrix}1&1\\1&0.13\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1&0.13\end{matrix}\right))\left(\begin{matrix}150\\6300\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\1&0.13\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&0.13\end{matrix}\right))\left(\begin{matrix}150\\6300\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&0.13\end{matrix}\right))\left(\begin{matrix}150\\6300\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{0.13}{0.13-1}&-\frac{1}{0.13-1}\\-\frac{1}{0.13-1}&\frac{1}{0.13-1}\end{matrix}\right)\left(\begin{matrix}150\\6300\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{13}{87}&\frac{100}{87}\\\frac{100}{87}&-\frac{100}{87}\end{matrix}\right)\left(\begin{matrix}150\\6300\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{13}{87}\times 150+\frac{100}{87}\times 6300\\\frac{100}{87}\times 150-\frac{100}{87}\times 6300\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{209350}{29}\\-\frac{205000}{29}\end{matrix}\right)
Do the arithmetic.
x=\frac{209350}{29},y=-\frac{205000}{29}
Extract the matrix elements x and y.
x+y=150,x+0.13y=6300
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-0.13y=150-6300
Subtract x+0.13y=6300 from x+y=150 by subtracting like terms on each side of the equal sign.
y-0.13y=150-6300
Add x to -x. Terms x and -x cancel out, leaving an equation with only one variable that can be solved.
0.87y=150-6300
Add y to -\frac{13y}{100}.
0.87y=-6150
Add 150 to -6300.
y=-\frac{205000}{29}
Divide both sides of the equation by 0.87, which is the same as multiplying both sides by the reciprocal of the fraction.
x+0.13\left(-\frac{205000}{29}\right)=6300
Substitute -\frac{205000}{29} for y in x+0.13y=6300. Because the resulting equation contains only one variable, you can solve for x directly.
x-\frac{26650}{29}=6300
Multiply 0.13 times -\frac{205000}{29} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=\frac{209350}{29}
Add \frac{26650}{29} to both sides of the equation.
x=\frac{209350}{29},y=-\frac{205000}{29}
The system is now solved.