Solve {l}{a+b=50}{1500a+2100b=90000}

$$\left\{ \begin{array} { l } { a + b = 50 } \\ { 1500 a + 2100 b = 90000 } \end{array} \right.$$

Solve for a, b

$$a=25$$

$$b=25$$

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.

$$a+b=50,1500a+2100b=90000$$

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

$$a+b=50$$

Subtract b from both sides of the equation.

$$a=-b+50$$

Substitute -b+50 for a in the other equation, 1500a+2100b=90000.

$$1500\left(-b+50\right)+2100b=90000$$

Multiply 1500 times -b+50.

$$-1500b+75000+2100b=90000$$

Add -1500b to 2100b.

$$600b+75000=90000$$

Subtract 75000 from both sides of the equation.

$$600b=15000$$

Divide both sides by 600.

$$b=25$$

Substitute 25 for b in a=-b+50. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-25+50$$

Add 50 to -25.

$$a=25$$

The system is now solved.

$$a=25,b=25$$

Steps Using Matrices

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

$$a+b=50,1500a+2100b=90000$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\1500&2100\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}50\\90000\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\1500&2100\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\1500&2100\end{matrix}\right))\left(\begin{matrix}1&1\\1500&2100\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1500&2100\end{matrix}\right))\left(\begin{matrix}50\\90000\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1500&2100\end{matrix}\right))\left(\begin{matrix}50\\90000\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1500&2100\end{matrix}\right))\left(\begin{matrix}50\\90000\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{2100}{2100-1500}&-\frac{1}{2100-1500}\\-\frac{1500}{2100-1500}&\frac{1}{2100-1500}\end{matrix}\right)\left(\begin{matrix}50\\90000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{7}{2}&-\frac{1}{600}\\-\frac{5}{2}&\frac{1}{600}\end{matrix}\right)\left(\begin{matrix}50\\90000\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{7}{2}\times 50-\frac{1}{600}\times 90000\\-\frac{5}{2}\times 50+\frac{1}{600}\times 90000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}25\\25\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=25,b=25$$

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.

$$a+b=50,1500a+2100b=90000$$

To make a and 1500a equal, multiply all terms on each side of the first equation by 1500 and all terms on each side of the second by 1.

$$1500a+1500b=1500\times 50,1500a+2100b=90000$$

Simplify.

$$1500a+1500b=75000,1500a+2100b=90000$$

Subtract 1500a+2100b=90000 from 1500a+1500b=75000 by subtracting like terms on each side of the equal sign.

$$1500a-1500a+1500b-2100b=75000-90000$$

Add 1500a to -1500a. Terms 1500a and -1500a cancel out, leaving an equation with only one variable that can be solved.

$$1500b-2100b=75000-90000$$

Add 1500b to -2100b.

$$-600b=75000-90000$$

Add 75000 to -90000.

$$-600b=-15000$$

Divide both sides by -600.

$$b=25$$

Substitute 25 for b in 1500a+2100b=90000. Because the resulting equation contains only one variable, you can solve for a directly.

$$1500a+2100\times 25=90000$$

Multiply 2100 times 25.

$$1500a+52500=90000$$

Subtract 52500 from both sides of the equation.

$$1500a=37500$$

Divide both sides by 1500.

$$a=25$$

The system is now solved.

$$a=25,b=25$$