Solve {l}{20p+30q=160000}{p+q=2287000}

$$\left. \begin{array} { l } { 20 p + 30 q = 160000 } \\ { p + q = 2287000 } \end{array} \right.$$

Solve for p, q

$$p=6845000$$

$$q=-4558000$$

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.

$$20p+30q=160000,p+q=2287000$$

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

$$20p+30q=160000$$

Subtract 30q from both sides of the equation.

$$20p=-30q+160000$$

Divide both sides by 20.

$$p=\frac{1}{20}\left(-30q+160000\right)$$

Multiply \frac{1}{20} times -30q+160000.

$$p=-\frac{3}{2}q+8000$$

Substitute -\frac{3q}{2}+8000 for p in the other equation, p+q=2287000.

$$-\frac{3}{2}q+8000+q=2287000$$

Add -\frac{3q}{2} to q.

$$-\frac{1}{2}q+8000=2287000$$

Subtract 8000 from both sides of the equation.

$$-\frac{1}{2}q=2279000$$

Multiply both sides by -2.

$$q=-4558000$$

Substitute -4558000 for q in p=-\frac{3}{2}q+8000. Because the resulting equation contains only one variable, you can solve for p directly.

$$p=-\frac{3}{2}\left(-4558000\right)+8000$$

Multiply -\frac{3}{2} times -4558000.

$$p=6837000+8000$$

Add 8000 to 6837000.

$$p=6845000$$

The system is now solved.

$$p=6845000,q=-4558000$$

Steps Using Matrices

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

$$20p+30q=160000,p+q=2287000$$

Write the equations in matrix form.

$$\left(\begin{matrix}20&30\\1&1\end{matrix}\right)\left(\begin{matrix}p\\q\end{matrix}\right)=\left(\begin{matrix}160000\\2287000\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}20&30\\1&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}20&30\\1&1\end{matrix}\right))\left(\begin{matrix}20&30\\1&1\end{matrix}\right)\left(\begin{matrix}p\\q\end{matrix}\right)=inverse(\left(\begin{matrix}20&30\\1&1\end{matrix}\right))\left(\begin{matrix}160000\\2287000\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}p\\q\end{matrix}\right)=inverse(\left(\begin{matrix}20&30\\1&1\end{matrix}\right))\left(\begin{matrix}160000\\2287000\end{matrix}\right)$$

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

$$\left(\begin{matrix}p\\q\end{matrix}\right)=inverse(\left(\begin{matrix}20&30\\1&1\end{matrix}\right))\left(\begin{matrix}160000\\2287000\end{matrix}\right)$$

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.

$$\left(\begin{matrix}p\\q\end{matrix}\right)=\left(\begin{matrix}\frac{1}{20-30}&-\frac{30}{20-30}\\-\frac{1}{20-30}&\frac{20}{20-30}\end{matrix}\right)\left(\begin{matrix}160000\\2287000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}p\\q\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{10}&3\\\frac{1}{10}&-2\end{matrix}\right)\left(\begin{matrix}160000\\2287000\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}p\\q\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{10}\times 160000+3\times 2287000\\\frac{1}{10}\times 160000-2\times 2287000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}p\\q\end{matrix}\right)=\left(\begin{matrix}6845000\\-4558000\end{matrix}\right)$$

Extract the matrix elements p and q.

$$p=6845000,q=-4558000$$

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.

$$20p+30q=160000,p+q=2287000$$

To make 20p and p equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 20.

$$20p+30q=160000,20p+20q=20\times 2287000$$

Simplify.

$$20p+30q=160000,20p+20q=45740000$$

Subtract 20p+20q=45740000 from 20p+30q=160000 by subtracting like terms on each side of the equal sign.

$$20p-20p+30q-20q=160000-45740000$$

Add 20p to -20p. Terms 20p and -20p cancel out, leaving an equation with only one variable that can be solved.

$$30q-20q=160000-45740000$$

Add 30q to -20q.

$$10q=160000-45740000$$

Add 160000 to -45740000.

$$10q=-45580000$$

Divide both sides by 10.

$$q=-4558000$$

Substitute -4558000 for q in p+q=2287000. Because the resulting equation contains only one variable, you can solve for p directly.

$$p-4558000=2287000$$

Add 4558000 to both sides of the equation.

$$p=6845000$$

The system is now solved.

$$p=6845000,q=-4558000$$