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6y+4x=27,y+x=50
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.
6y+4x=27
Choose one of the equations and solve it for y by isolating y on the left hand side of the equal sign.
6y=-4x+27
Subtract 4x from both sides of the equation.
y=\frac{1}{6}\left(-4x+27\right)
Divide both sides by 6.
y=-\frac{2}{3}x+\frac{9}{2}
Multiply \frac{1}{6} times -4x+27.
-\frac{2}{3}x+\frac{9}{2}+x=50
Substitute -\frac{2x}{3}+\frac{9}{2} for y in the other equation, y+x=50.
\frac{1}{3}x+\frac{9}{2}=50
Add -\frac{2x}{3} to x.
\frac{1}{3}x=\frac{91}{2}
Subtract \frac{9}{2} from both sides of the equation.
x=\frac{273}{2}
Multiply both sides by 3.
y=-\frac{2}{3}\times \frac{273}{2}+\frac{9}{2}
Substitute \frac{273}{2} for x in y=-\frac{2}{3}x+\frac{9}{2}. Because the resulting equation contains only one variable, you can solve for y directly.
y=-91+\frac{9}{2}
Multiply -\frac{2}{3} times \frac{273}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
y=-\frac{173}{2}
Add \frac{9}{2} to -91.
y=-\frac{173}{2},x=\frac{273}{2}
The system is now solved.
6y+4x=27,y+x=50
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}6&4\\1&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}27\\50\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}6&4\\1&1\end{matrix}\right))\left(\begin{matrix}6&4\\1&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}6&4\\1&1\end{matrix}\right))\left(\begin{matrix}27\\50\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&4\\1&1\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}6&4\\1&1\end{matrix}\right))\left(\begin{matrix}27\\50\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}6&4\\1&1\end{matrix}\right))\left(\begin{matrix}27\\50\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{6-4}&-\frac{4}{6-4}\\-\frac{1}{6-4}&\frac{6}{6-4}\end{matrix}\right)\left(\begin{matrix}27\\50\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}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}&-2\\-\frac{1}{2}&3\end{matrix}\right)\left(\begin{matrix}27\\50\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}\times 27-2\times 50\\-\frac{1}{2}\times 27+3\times 50\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{173}{2}\\\frac{273}{2}\end{matrix}\right)
Do the arithmetic.
y=-\frac{173}{2},x=\frac{273}{2}
Extract the matrix elements y and x.
6y+4x=27,y+x=50
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.
6y+4x=27,6y+6x=6\times 50
To make 6y and y equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 6.
6y+4x=27,6y+6x=300
Simplify.
6y-6y+4x-6x=27-300
Subtract 6y+6x=300 from 6y+4x=27 by subtracting like terms on each side of the equal sign.
4x-6x=27-300
Add 6y to -6y. Terms 6y and -6y cancel out, leaving an equation with only one variable that can be solved.
-2x=27-300
Add 4x to -6x.
-2x=-273
Add 27 to -300.
x=\frac{273}{2}
Divide both sides by -2.
y+\frac{273}{2}=50
Substitute \frac{273}{2} for x in y+x=50. Because the resulting equation contains only one variable, you can solve for y directly.
y=-\frac{173}{2}
Subtract \frac{273}{2} from both sides of the equation.
y=-\frac{173}{2},x=\frac{273}{2}
The system is now solved.