Solve {l}{9A/25+16B=1}{18A/25+9B=1}

$$\left\{ \begin{array} { l } { \frac { 9 A } { 25 } + 16 B = 1 } \\ { \frac { 18 A } { 25 } + 9 B = 1 } \end{array} \right.$$

Solve for A, B

$$A=\frac{175}{207}\approx 0.845410628$$

$$B=\frac{1}{23}\approx 0.043478261$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 25.

$$9A+400B=25$$

Consider the second equation. Multiply both sides of the equation by 25.

$$18A+225B=25$$

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.

$$9A+400B=25,18A+225B=25$$

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

$$9A+400B=25$$

Subtract 400B from both sides of the equation.

$$9A=-400B+25$$

Divide both sides by 9.

$$A=\frac{1}{9}\left(-400B+25\right)$$

Multiply \frac{1}{9} times -400B+25.

$$A=-\frac{400}{9}B+\frac{25}{9}$$

Substitute \frac{-400B+25}{9} for A in the other equation, 18A+225B=25.

$$18\left(-\frac{400}{9}B+\frac{25}{9}\right)+225B=25$$

Multiply 18 times \frac{-400B+25}{9}.

$$-800B+50+225B=25$$

Add -800B to 225B.

$$-575B+50=25$$

Subtract 50 from both sides of the equation.

$$-575B=-25$$

Divide both sides by -575.

$$B=\frac{1}{23}$$

Substitute \frac{1}{23} for B in A=-\frac{400}{9}B+\frac{25}{9}. Because the resulting equation contains only one variable, you can solve for A directly.

$$A=-\frac{400}{9}\times \frac{1}{23}+\frac{25}{9}$$

Multiply -\frac{400}{9} times \frac{1}{23} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$A=-\frac{400}{207}+\frac{25}{9}$$

Add \frac{25}{9} to -\frac{400}{207} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$A=\frac{175}{207}$$

The system is now solved.

$$A=\frac{175}{207},B=\frac{1}{23}$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 25.

$$9A+400B=25$$

Consider the second equation. Multiply both sides of the equation by 25.

$$18A+225B=25$$

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

$$9A+400B=25,18A+225B=25$$

Write the equations in matrix form.

$$\left(\begin{matrix}9&400\\18&225\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}25\\25\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}9&400\\18&225\end{matrix}\right).

$$inverse(\left(\begin{matrix}9&400\\18&225\end{matrix}\right))\left(\begin{matrix}9&400\\18&225\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}9&400\\18&225\end{matrix}\right))\left(\begin{matrix}25\\25\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}9&400\\18&225\end{matrix}\right))\left(\begin{matrix}25\\25\end{matrix}\right)$$

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

$$\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}9&400\\18&225\end{matrix}\right))\left(\begin{matrix}25\\25\end{matrix}\right)$$

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.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}\frac{225}{9\times 225-400\times 18}&-\frac{400}{9\times 225-400\times 18}\\-\frac{18}{9\times 225-400\times 18}&\frac{9}{9\times 225-400\times 18}\end{matrix}\right)\left(\begin{matrix}25\\25\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{23}&\frac{16}{207}\\\frac{2}{575}&-\frac{1}{575}\end{matrix}\right)\left(\begin{matrix}25\\25\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{23}\times 25+\frac{16}{207}\times 25\\\frac{2}{575}\times 25-\frac{1}{575}\times 25\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}\frac{175}{207}\\\frac{1}{23}\end{matrix}\right)$$

Extract the matrix elements A and B.

$$A=\frac{175}{207},B=\frac{1}{23}$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 25.

$$9A+400B=25$$

Consider the second equation. Multiply both sides of the equation by 25.

$$18A+225B=25$$

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.

$$9A+400B=25,18A+225B=25$$

To make 9A and 18A equal, multiply all terms on each side of the first equation by 18 and all terms on each side of the second by 9.

$$18\times 9A+18\times 400B=18\times 25,9\times 18A+9\times 225B=9\times 25$$

Simplify.

$$162A+7200B=450,162A+2025B=225$$

Subtract 162A+2025B=225 from 162A+7200B=450 by subtracting like terms on each side of the equal sign.

$$162A-162A+7200B-2025B=450-225$$

Add 162A to -162A. Terms 162A and -162A cancel out, leaving an equation with only one variable that can be solved.

$$7200B-2025B=450-225$$

Add 7200B to -2025B.

$$5175B=450-225$$

Add 450 to -225.

$$5175B=225$$

Divide both sides by 5175.

$$B=\frac{1}{23}$$

Substitute \frac{1}{23} for B in 18A+225B=25. Because the resulting equation contains only one variable, you can solve for A directly.

$$18A+225\times \frac{1}{23}=25$$

Multiply 225 times \frac{1}{23}.

$$18A+\frac{225}{23}=25$$

Subtract \frac{225}{23} from both sides of the equation.

$$18A=\frac{350}{23}$$

Divide both sides by 18.

$$A=\frac{175}{207}$$

The system is now solved.

$$A=\frac{175}{207},B=\frac{1}{23}$$