Solve {l}{4L+2m=6400}{6L+2m=9000}

$$\left. \begin{array} { l } { 4 L + 2 m = 6400 } \\ { 6 L + 2 m = 9000 } \end{array} \right.$$

Solve for L, m

$$L=1300$$

$$m=600$$

Steps Using Substitution

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.

$$4L+2m=6400,6L+2m=9000$$

Choose one of the equations and solve it for L by isolating L on the left hand side of the equal sign.

$$4L+2m=6400$$

Subtract 2m from both sides of the equation.

$$4L=-2m+6400$$

Divide both sides by 4.

$$L=\frac{1}{4}\left(-2m+6400\right)$$

Multiply \frac{1}{4} times -2m+6400.

$$L=-\frac{1}{2}m+1600$$

Substitute -\frac{m}{2}+1600 for L in the other equation, 6L+2m=9000.

$$6\left(-\frac{1}{2}m+1600\right)+2m=9000$$

Multiply 6 times -\frac{m}{2}+1600.

$$-3m+9600+2m=9000$$

Add -3m to 2m.

$$-m+9600=9000$$

Subtract 9600 from both sides of the equation.

$$-m=-600$$

Divide both sides by -1.

$$m=600$$

Substitute 600 for m in L=-\frac{1}{2}m+1600. Because the resulting equation contains only one variable, you can solve for L directly.

$$L=-\frac{1}{2}\times 600+1600$$

Multiply -\frac{1}{2} times 600.

$$L=-300+1600$$

Add 1600 to -300.

$$L=1300$$

The system is now solved.

$$L=1300,m=600$$

Steps Using Matrices

Put the equations in standard form and then use matrices to solve the system of equations.

$$4L+2m=6400,6L+2m=9000$$

Write the equations in matrix form.

$$\left(\begin{matrix}4&2\\6&2\end{matrix}\right)\left(\begin{matrix}L\\m\end{matrix}\right)=\left(\begin{matrix}6400\\9000\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&2\\6&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}4&2\\6&2\end{matrix}\right))\left(\begin{matrix}4&2\\6&2\end{matrix}\right)\left(\begin{matrix}L\\m\end{matrix}\right)=inverse(\left(\begin{matrix}4&2\\6&2\end{matrix}\right))\left(\begin{matrix}6400\\9000\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}L\\m\end{matrix}\right)=inverse(\left(\begin{matrix}4&2\\6&2\end{matrix}\right))\left(\begin{matrix}6400\\9000\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}L\\m\end{matrix}\right)=inverse(\left(\begin{matrix}4&2\\6&2\end{matrix}\right))\left(\begin{matrix}6400\\9000\end{matrix}\right)$$

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.

$$\left(\begin{matrix}L\\m\end{matrix}\right)=\left(\begin{matrix}\frac{2}{4\times 2-2\times 6}&-\frac{2}{4\times 2-2\times 6}\\-\frac{6}{4\times 2-2\times 6}&\frac{4}{4\times 2-2\times 6}\end{matrix}\right)\left(\begin{matrix}6400\\9000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}L\\m\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{2}&\frac{1}{2}\\\frac{3}{2}&-1\end{matrix}\right)\left(\begin{matrix}6400\\9000\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}L\\m\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{2}\times 6400+\frac{1}{2}\times 9000\\\frac{3}{2}\times 6400-9000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}L\\m\end{matrix}\right)=\left(\begin{matrix}1300\\600\end{matrix}\right)$$

Extract the matrix elements L and m.

$$L=1300,m=600$$

Steps Using Elimination

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.

$$4L+2m=6400,6L+2m=9000$$

Subtract 6L+2m=9000 from 4L+2m=6400 by subtracting like terms on each side of the equal sign.

$$4L-6L+2m-2m=6400-9000$$

Add 2m to -2m. Terms 2m and -2m cancel out, leaving an equation with only one variable that can be solved.

$$4L-6L=6400-9000$$

Add 4L to -6L.

$$-2L=6400-9000$$

Add 6400 to -9000.

$$-2L=-2600$$

Divide both sides by -2.

$$L=1300$$

Substitute 1300 for L in 6L+2m=9000. Because the resulting equation contains only one variable, you can solve for m directly.

$$6\times 1300+2m=9000$$

Multiply 6 times 1300.

$$7800+2m=9000$$

Subtract 7800 from both sides of the equation.

$$2m=1200$$

Divide both sides by 2.

$$m=600$$

The system is now solved.

$$L=1300,m=600$$