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6x+3y=24,8x+y=23
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.
6x+3y=24
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
6x=-3y+24
Subtract 3y from both sides of the equation.
x=\frac{1}{6}\left(-3y+24\right)
Divide both sides by 6.
x=-\frac{1}{2}y+4
Multiply \frac{1}{6} times -3y+24.
8\left(-\frac{1}{2}y+4\right)+y=23
Substitute -\frac{y}{2}+4 for x in the other equation, 8x+y=23.
-4y+32+y=23
Multiply 8 times -\frac{y}{2}+4.
-3y+32=23
Add -4y to y.
-3y=-9
Subtract 32 from both sides of the equation.
y=3
Divide both sides by -3.
x=-\frac{1}{2}\times 3+4
Substitute 3 for y in x=-\frac{1}{2}y+4. Because the resulting equation contains only one variable, you can solve for x directly.
x=-\frac{3}{2}+4
Multiply -\frac{1}{2} times 3.
x=\frac{5}{2}
Add 4 to -\frac{3}{2}.
x=\frac{5}{2},y=3
The system is now solved.
6x+3y=24,8x+y=23
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}6&3\\8&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}24\\23\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}6&3\\8&1\end{matrix}\right))\left(\begin{matrix}6&3\\8&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&3\\8&1\end{matrix}\right))\left(\begin{matrix}24\\23\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&3\\8&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}6&3\\8&1\end{matrix}\right))\left(\begin{matrix}24\\23\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}6&3\\8&1\end{matrix}\right))\left(\begin{matrix}24\\23\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}{6-3\times 8}&-\frac{3}{6-3\times 8}\\-\frac{8}{6-3\times 8}&\frac{6}{6-3\times 8}\end{matrix}\right)\left(\begin{matrix}24\\23\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}{18}&\frac{1}{6}\\\frac{4}{9}&-\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}24\\23\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{18}\times 24+\frac{1}{6}\times 23\\\frac{4}{9}\times 24-\frac{1}{3}\times 23\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{2}\\3\end{matrix}\right)
Do the arithmetic.
x=\frac{5}{2},y=3
Extract the matrix elements x and y.
6x+3y=24,8x+y=23
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.
8\times 6x+8\times 3y=8\times 24,6\times 8x+6y=6\times 23
To make 6x and 8x equal, multiply all terms on each side of the first equation by 8 and all terms on each side of the second by 6.
48x+24y=192,48x+6y=138
Simplify.
48x-48x+24y-6y=192-138
Subtract 48x+6y=138 from 48x+24y=192 by subtracting like terms on each side of the equal sign.
24y-6y=192-138
Add 48x to -48x. Terms 48x and -48x cancel out, leaving an equation with only one variable that can be solved.
18y=192-138
Add 24y to -6y.
18y=54
Add 192 to -138.
y=3
Divide both sides by 18.
8x+3=23
Substitute 3 for y in 8x+y=23. Because the resulting equation contains only one variable, you can solve for x directly.
8x=20
Subtract 3 from both sides of the equation.
x=\frac{5}{2}
Divide both sides by 8.
x=\frac{5}{2},y=3
The system is now solved.