Solve {l}{50a+40b=40050}{60a+52b=5000}

$$\left\{ \begin{array} { l } { 50 a + 40 b = 40050 } \\ { 60 a + 52 b = 5000 } \end{array} \right.$$

Solve for a, b

$$a=9413$$

$$b=-10765$$

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.

$$50a+40b=40050,60a+52b=5000$$

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

$$50a+40b=40050$$

Subtract 40b from both sides of the equation.

$$50a=-40b+40050$$

Divide both sides by 50.

$$a=\frac{1}{50}\left(-40b+40050\right)$$

Multiply \frac{1}{50} times -40b+40050.

$$a=-\frac{4}{5}b+801$$

Substitute -\frac{4b}{5}+801 for a in the other equation, 60a+52b=5000.

$$60\left(-\frac{4}{5}b+801\right)+52b=5000$$

Multiply 60 times -\frac{4b}{5}+801.

$$-48b+48060+52b=5000$$

Add -48b to 52b.

$$4b+48060=5000$$

Subtract 48060 from both sides of the equation.

$$4b=-43060$$

Divide both sides by 4.

$$b=-10765$$

Substitute -10765 for b in a=-\frac{4}{5}b+801. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-\frac{4}{5}\left(-10765\right)+801$$

Multiply -\frac{4}{5} times -10765.

$$a=8612+801$$

Add 801 to 8612.

$$a=9413$$

The system is now solved.

$$a=9413,b=-10765$$

Steps Using Matrices

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

$$50a+40b=40050,60a+52b=5000$$

Write the equations in matrix form.

$$\left(\begin{matrix}50&40\\60&52\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}40050\\5000\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}50&40\\60&52\end{matrix}\right).

$$inverse(\left(\begin{matrix}50&40\\60&52\end{matrix}\right))\left(\begin{matrix}50&40\\60&52\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}50&40\\60&52\end{matrix}\right))\left(\begin{matrix}40050\\5000\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}50&40\\60&52\end{matrix}\right))\left(\begin{matrix}40050\\5000\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}50&40\\60&52\end{matrix}\right))\left(\begin{matrix}40050\\5000\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{52}{50\times 52-40\times 60}&-\frac{40}{50\times 52-40\times 60}\\-\frac{60}{50\times 52-40\times 60}&\frac{50}{50\times 52-40\times 60}\end{matrix}\right)\left(\begin{matrix}40050\\5000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{13}{50}&-\frac{1}{5}\\-\frac{3}{10}&\frac{1}{4}\end{matrix}\right)\left(\begin{matrix}40050\\5000\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{13}{50}\times 40050-\frac{1}{5}\times 5000\\-\frac{3}{10}\times 40050+\frac{1}{4}\times 5000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}9413\\-10765\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=9413,b=-10765$$

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.

$$50a+40b=40050,60a+52b=5000$$

To make 50a and 60a equal, multiply all terms on each side of the first equation by 60 and all terms on each side of the second by 50.

$$60\times 50a+60\times 40b=60\times 40050,50\times 60a+50\times 52b=50\times 5000$$

Simplify.

$$3000a+2400b=2403000,3000a+2600b=250000$$

Subtract 3000a+2600b=250000 from 3000a+2400b=2403000 by subtracting like terms on each side of the equal sign.

$$3000a-3000a+2400b-2600b=2403000-250000$$

Add 3000a to -3000a. Terms 3000a and -3000a cancel out, leaving an equation with only one variable that can be solved.

$$2400b-2600b=2403000-250000$$

Add 2400b to -2600b.

$$-200b=2403000-250000$$

Add 2403000 to -250000.

$$-200b=2153000$$

Divide both sides by -200.

$$b=-10765$$

Substitute -10765 for b in 60a+52b=5000. Because the resulting equation contains only one variable, you can solve for a directly.

$$60a+52\left(-10765\right)=5000$$

Multiply 52 times -10765.

$$60a-559780=5000$$

Add 559780 to both sides of the equation.

$$60a=564780$$

Divide both sides by 60.

$$a=9413$$

The system is now solved.

$$a=9413,b=-10765$$