Solve {l}{85a+90b=20800}{75(a+b)=18000}

$$\left\{ \begin{array} { l } { 85 a + 90 b = 20800 } \\ { 75 ( a + b ) = 18000 } \end{array} \right.$$

Solve for a, b

$$a=160$$

$$b=80$$

Steps Using Substitution

Consider the second equation. Divide both sides by 75.

$$a+b=\frac{18000}{75}$$

Divide 18000 by 75 to get 240.

$$a+b=240$$

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.

$$85a+90b=20800,a+b=240$$

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

$$85a+90b=20800$$

Subtract 90b from both sides of the equation.

$$85a=-90b+20800$$

Divide both sides by 85.

$$a=\frac{1}{85}\left(-90b+20800\right)$$

Multiply \frac{1}{85} times -90b+20800.

$$a=-\frac{18}{17}b+\frac{4160}{17}$$

Substitute \frac{-18b+4160}{17} for a in the other equation, a+b=240.

$$-\frac{18}{17}b+\frac{4160}{17}+b=240$$

Add -\frac{18b}{17} to b.

$$-\frac{1}{17}b+\frac{4160}{17}=240$$

Subtract \frac{4160}{17} from both sides of the equation.

$$-\frac{1}{17}b=-\frac{80}{17}$$

Multiply both sides by -17.

$$b=80$$

Substitute 80 for b in a=-\frac{18}{17}b+\frac{4160}{17}. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-\frac{18}{17}\times 80+\frac{4160}{17}$$

Multiply -\frac{18}{17} times 80.

$$a=\frac{-1440+4160}{17}$$

Add \frac{4160}{17} to -\frac{1440}{17} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a=160$$

The system is now solved.

$$a=160,b=80$$

Steps Using Matrices

Consider the second equation. Divide both sides by 75.

$$a+b=\frac{18000}{75}$$

Divide 18000 by 75 to get 240.

$$a+b=240$$

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

$$85a+90b=20800,a+b=240$$

Write the equations in matrix form.

$$\left(\begin{matrix}85&90\\1&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}20800\\240\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}85&90\\1&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}85&90\\1&1\end{matrix}\right))\left(\begin{matrix}85&90\\1&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}85&90\\1&1\end{matrix}\right))\left(\begin{matrix}20800\\240\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}85&90\\1&1\end{matrix}\right))\left(\begin{matrix}20800\\240\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}85&90\\1&1\end{matrix}\right))\left(\begin{matrix}20800\\240\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{1}{85-90}&-\frac{90}{85-90}\\-\frac{1}{85-90}&\frac{85}{85-90}\end{matrix}\right)\left(\begin{matrix}20800\\240\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{5}&18\\\frac{1}{5}&-17\end{matrix}\right)\left(\begin{matrix}20800\\240\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{5}\times 20800+18\times 240\\\frac{1}{5}\times 20800-17\times 240\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}160\\80\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=160,b=80$$

Steps Using Elimination

Consider the second equation. Divide both sides by 75.

$$a+b=\frac{18000}{75}$$

Divide 18000 by 75 to get 240.

$$a+b=240$$

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.

$$85a+90b=20800,a+b=240$$

To make 85a and a equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 85.

$$85a+90b=20800,85a+85b=85\times 240$$

Simplify.

$$85a+90b=20800,85a+85b=20400$$

Subtract 85a+85b=20400 from 85a+90b=20800 by subtracting like terms on each side of the equal sign.

$$85a-85a+90b-85b=20800-20400$$

Add 85a to -85a. Terms 85a and -85a cancel out, leaving an equation with only one variable that can be solved.

$$90b-85b=20800-20400$$

Add 90b to -85b.

$$5b=20800-20400$$

Add 20800 to -20400.

$$5b=400$$

Divide both sides by 5.

$$b=80$$

Substitute 80 for b in a+b=240. Because the resulting equation contains only one variable, you can solve for a directly.

$$a+80=240$$

Subtract 80 from both sides of the equation.

$$a=160$$

The system is now solved.

$$a=160,b=80$$