Solve {l}{8B+6P=14400}{6B+15P=19200}

$$\left. \begin{array} { l } { 8 B + 6 P = 14400 } \\ { 6 B + 15 P = 19200 } \end{array} \right.$$

Solve for B, P

$$B=1200$$

$$P=800$$

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.

$$8B+6P=14400,6B+15P=19200$$

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

$$8B+6P=14400$$

Subtract 6P from both sides of the equation.

$$8B=-6P+14400$$

Divide both sides by 8.

$$B=\frac{1}{8}\left(-6P+14400\right)$$

Multiply \frac{1}{8} times -6P+14400.

$$B=-\frac{3}{4}P+1800$$

Substitute -\frac{3P}{4}+1800 for B in the other equation, 6B+15P=19200.

$$6\left(-\frac{3}{4}P+1800\right)+15P=19200$$

Multiply 6 times -\frac{3P}{4}+1800.

$$-\frac{9}{2}P+10800+15P=19200$$

Add -\frac{9P}{2} to 15P.

$$\frac{21}{2}P+10800=19200$$

Subtract 10800 from both sides of the equation.

$$\frac{21}{2}P=8400$$

Divide both sides of the equation by \frac{21}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$P=800$$

Substitute 800 for P in B=-\frac{3}{4}P+1800. Because the resulting equation contains only one variable, you can solve for B directly.

$$B=-\frac{3}{4}\times 800+1800$$

Multiply -\frac{3}{4} times 800.

$$B=-600+1800$$

Add 1800 to -600.

$$B=1200$$

The system is now solved.

$$B=1200,P=800$$

Steps Using Matrices

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

$$8B+6P=14400,6B+15P=19200$$

Write the equations in matrix form.

$$\left(\begin{matrix}8&6\\6&15\end{matrix}\right)\left(\begin{matrix}B\\P\end{matrix}\right)=\left(\begin{matrix}14400\\19200\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}8&6\\6&15\end{matrix}\right).

$$inverse(\left(\begin{matrix}8&6\\6&15\end{matrix}\right))\left(\begin{matrix}8&6\\6&15\end{matrix}\right)\left(\begin{matrix}B\\P\end{matrix}\right)=inverse(\left(\begin{matrix}8&6\\6&15\end{matrix}\right))\left(\begin{matrix}14400\\19200\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}B\\P\end{matrix}\right)=inverse(\left(\begin{matrix}8&6\\6&15\end{matrix}\right))\left(\begin{matrix}14400\\19200\end{matrix}\right)$$

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

$$\left(\begin{matrix}B\\P\end{matrix}\right)=inverse(\left(\begin{matrix}8&6\\6&15\end{matrix}\right))\left(\begin{matrix}14400\\19200\end{matrix}\right)$$

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.

$$\left(\begin{matrix}B\\P\end{matrix}\right)=\left(\begin{matrix}\frac{15}{8\times 15-6\times 6}&-\frac{6}{8\times 15-6\times 6}\\-\frac{6}{8\times 15-6\times 6}&\frac{8}{8\times 15-6\times 6}\end{matrix}\right)\left(\begin{matrix}14400\\19200\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}B\\P\end{matrix}\right)=\left(\begin{matrix}\frac{5}{28}&-\frac{1}{14}\\-\frac{1}{14}&\frac{2}{21}\end{matrix}\right)\left(\begin{matrix}14400\\19200\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}B\\P\end{matrix}\right)=\left(\begin{matrix}\frac{5}{28}\times 14400-\frac{1}{14}\times 19200\\-\frac{1}{14}\times 14400+\frac{2}{21}\times 19200\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}B\\P\end{matrix}\right)=\left(\begin{matrix}1200\\800\end{matrix}\right)$$

Extract the matrix elements B and P.

$$B=1200,P=800$$

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.

$$8B+6P=14400,6B+15P=19200$$

To make 8B and 6B equal, multiply all terms on each side of the first equation by 6 and all terms on each side of the second by 8.

$$6\times 8B+6\times 6P=6\times 14400,8\times 6B+8\times 15P=8\times 19200$$

Simplify.

$$48B+36P=86400,48B+120P=153600$$

Subtract 48B+120P=153600 from 48B+36P=86400 by subtracting like terms on each side of the equal sign.

$$48B-48B+36P-120P=86400-153600$$

Add 48B to -48B. Terms 48B and -48B cancel out, leaving an equation with only one variable that can be solved.

$$36P-120P=86400-153600$$

Add 36P to -120P.

$$-84P=86400-153600$$

Add 86400 to -153600.

$$-84P=-67200$$

Divide both sides by -84.

$$P=800$$

Substitute 800 for P in 6B+15P=19200. Because the resulting equation contains only one variable, you can solve for B directly.

$$6B+15\times 800=19200$$

Multiply 15 times 800.

$$6B+12000=19200$$

Subtract 12000 from both sides of the equation.

$$6B=7200$$

Divide both sides by 6.

$$B=1200$$

The system is now solved.

$$B=1200,P=800$$