Solve {l}{57*190=9A+45B}{289.168=45A+285B}

$$\left. \begin{array} { l } { 57 \cdot 190 = 9 A + 45 B } \\ { 289.168 = 45 A + 285 B } \end{array} \right.$$

Solve for A, B

$$A=5691.736$$

$$B = -\frac{1683151}{1875} = -897\frac{1276}{1875} \approx -897.680533333$$

Steps Using Substitution

Consider the first equation. Multiply 57 and 190 to get 10830.

$$10830=9A+45B$$

Swap sides so that all variable terms are on the left hand side.

$$9A+45B=10830$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$45A+285B=289.168$$

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.

$$9A+45B=10830,45A+285B=289.168$$

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

$$9A+45B=10830$$

Subtract 45B from both sides of the equation.

$$9A=-45B+10830$$

Divide both sides by 9.

$$A=\frac{1}{9}\left(-45B+10830\right)$$

Multiply \frac{1}{9} times -45B+10830.

$$A=-5B+\frac{3610}{3}$$

Substitute -5B+\frac{3610}{3} for A in the other equation, 45A+285B=289.168.

$$45\left(-5B+\frac{3610}{3}\right)+285B=289.168$$

Multiply 45 times -5B+\frac{3610}{3}.

$$-225B+54150+285B=289.168$$

Add -225B to 285B.

$$60B+54150=289.168$$

Subtract 54150 from both sides of the equation.

$$60B=-53860.832$$

Divide both sides by 60.

$$B=-\frac{1683151}{1875}$$

Substitute -\frac{1683151}{1875} for B in A=-5B+\frac{3610}{3}. Because the resulting equation contains only one variable, you can solve for A directly.

$$A=-5\left(-\frac{1683151}{1875}\right)+\frac{3610}{3}$$

Multiply -5 times -\frac{1683151}{1875}.

$$A=\frac{1683151}{375}+\frac{3610}{3}$$

Add \frac{3610}{3} to \frac{1683151}{375} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$A=\frac{711467}{125}$$

The system is now solved.

$$A=\frac{711467}{125},B=-\frac{1683151}{1875}$$

Steps Using Matrices

Consider the first equation. Multiply 57 and 190 to get 10830.

$$10830=9A+45B$$

Swap sides so that all variable terms are on the left hand side.

$$9A+45B=10830$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$45A+285B=289.168$$

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

$$9A+45B=10830,45A+285B=289.168$$

Write the equations in matrix form.

$$\left(\begin{matrix}9&45\\45&285\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}10830\\289.168\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}9&45\\45&285\end{matrix}\right).

$$inverse(\left(\begin{matrix}9&45\\45&285\end{matrix}\right))\left(\begin{matrix}9&45\\45&285\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}9&45\\45&285\end{matrix}\right))\left(\begin{matrix}10830\\289.168\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}9&45\\45&285\end{matrix}\right))\left(\begin{matrix}10830\\289.168\end{matrix}\right)$$

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

$$\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}9&45\\45&285\end{matrix}\right))\left(\begin{matrix}10830\\289.168\end{matrix}\right)$$

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.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}\frac{285}{9\times 285-45\times 45}&-\frac{45}{9\times 285-45\times 45}\\-\frac{45}{9\times 285-45\times 45}&\frac{9}{9\times 285-45\times 45}\end{matrix}\right)\left(\begin{matrix}10830\\289.168\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}\frac{19}{36}&-\frac{1}{12}\\-\frac{1}{12}&\frac{1}{60}\end{matrix}\right)\left(\begin{matrix}10830\\289.168\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}\frac{19}{36}\times 10830-\frac{1}{12}\times 289.168\\-\frac{1}{12}\times 10830+\frac{1}{60}\times 289.168\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}\frac{711467}{125}\\-\frac{1683151}{1875}\end{matrix}\right)$$

Extract the matrix elements A and B.

$$A=\frac{711467}{125},B=-\frac{1683151}{1875}$$

Steps Using Elimination

Consider the first equation. Multiply 57 and 190 to get 10830.

$$10830=9A+45B$$

Swap sides so that all variable terms are on the left hand side.

$$9A+45B=10830$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$45A+285B=289.168$$

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.

$$9A+45B=10830,45A+285B=289.168$$

To make 9A and 45A equal, multiply all terms on each side of the first equation by 45 and all terms on each side of the second by 9.

$$45\times 9A+45\times 45B=45\times 10830,9\times 45A+9\times 285B=9\times 289.168$$

Simplify.

$$405A+2025B=487350,405A+2565B=2602.512$$

Subtract 405A+2565B=2602.512 from 405A+2025B=487350 by subtracting like terms on each side of the equal sign.

$$405A-405A+2025B-2565B=487350-2602.512$$

Add 405A to -405A. Terms 405A and -405A cancel out, leaving an equation with only one variable that can be solved.

$$2025B-2565B=487350-2602.512$$

Add 2025B to -2565B.

$$-540B=487350-2602.512$$

Add 487350 to -2602.512.

$$-540B=484747.488$$

Divide both sides by -540.

$$B=-\frac{1683151}{1875}$$

Substitute -\frac{1683151}{1875} for B in 45A+285B=289.168. Because the resulting equation contains only one variable, you can solve for A directly.

$$45A+285\left(-\frac{1683151}{1875}\right)=289.168$$

Multiply 285 times -\frac{1683151}{1875}.

$$45A-\frac{31979869}{125}=289.168$$

Add \frac{31979869}{125} to both sides of the equation.

$$45A=\frac{6403203}{25}$$

Divide both sides by 45.

$$A=\frac{711467}{125}$$

The system is now solved.

$$A=\frac{711467}{125},B=-\frac{1683151}{1875}$$