Solve {l}{-a-4b+229a=113}{-b+4a+229b=0=}

$$\left. \begin{array} { l } { - a - 4 b + 229 a = 113 } \\ { - b + 4 a + 229 b = 0 = } \end{array} \right.$$

Solve for a, b

$$a=\frac{6441}{13000}\approx 0.495461538$$

$$b=-\frac{113}{13000}\approx -0.008692308$$

Steps Using Substitution

Consider the first equation. Combine -a and 229a to get 228a.

$$228a-4b=113$$

Consider the second equation. Combine -b and 229b to get 228b.

$$228b+4a=0$$

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.

$$228a-4b=113,4a+228b=0$$

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

$$228a-4b=113$$

Add 4b to both sides of the equation.

$$228a=4b+113$$

Divide both sides by 228.

$$a=\frac{1}{228}\left(4b+113\right)$$

Multiply \frac{1}{228} times 4b+113.

$$a=\frac{1}{57}b+\frac{113}{228}$$

Substitute \frac{b}{57}+\frac{113}{228} for a in the other equation, 4a+228b=0.

$$4\left(\frac{1}{57}b+\frac{113}{228}\right)+228b=0$$

Multiply 4 times \frac{b}{57}+\frac{113}{228}.

$$\frac{4}{57}b+\frac{113}{57}+228b=0$$

Add \frac{4b}{57} to 228b.

$$\frac{13000}{57}b+\frac{113}{57}=0$$

Subtract \frac{113}{57} from both sides of the equation.

$$\frac{13000}{57}b=-\frac{113}{57}$$

Divide both sides of the equation by \frac{13000}{57}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$b=-\frac{113}{13000}$$

Substitute -\frac{113}{13000} for b in a=\frac{1}{57}b+\frac{113}{228}. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=\frac{1}{57}\left(-\frac{113}{13000}\right)+\frac{113}{228}$$

Multiply \frac{1}{57} times -\frac{113}{13000} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$a=-\frac{113}{741000}+\frac{113}{228}$$

Add \frac{113}{228} to -\frac{113}{741000} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a=\frac{6441}{13000}$$

The system is now solved.

$$a=\frac{6441}{13000},b=-\frac{113}{13000}$$

Steps Using Matrices

Consider the first equation. Combine -a and 229a to get 228a.

$$228a-4b=113$$

Consider the second equation. Combine -b and 229b to get 228b.

$$228b+4a=0$$

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

$$228a-4b=113,4a+228b=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}228&-4\\4&228\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}113\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}228&-4\\4&228\end{matrix}\right).

$$inverse(\left(\begin{matrix}228&-4\\4&228\end{matrix}\right))\left(\begin{matrix}228&-4\\4&228\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}228&-4\\4&228\end{matrix}\right))\left(\begin{matrix}113\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}228&-4\\4&228\end{matrix}\right))\left(\begin{matrix}113\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}228&-4\\4&228\end{matrix}\right))\left(\begin{matrix}113\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{228}{228\times 228-\left(-4\times 4\right)}&-\frac{-4}{228\times 228-\left(-4\times 4\right)}\\-\frac{4}{228\times 228-\left(-4\times 4\right)}&\frac{228}{228\times 228-\left(-4\times 4\right)}\end{matrix}\right)\left(\begin{matrix}113\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{57}{13000}&\frac{1}{13000}\\-\frac{1}{13000}&\frac{57}{13000}\end{matrix}\right)\left(\begin{matrix}113\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{57}{13000}\times 113\\-\frac{1}{13000}\times 113\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{6441}{13000}\\-\frac{113}{13000}\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=\frac{6441}{13000},b=-\frac{113}{13000}$$

Steps Using Elimination

Consider the first equation. Combine -a and 229a to get 228a.

$$228a-4b=113$$

Consider the second equation. Combine -b and 229b to get 228b.

$$228b+4a=0$$

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.

$$228a-4b=113,4a+228b=0$$

To make 228a and 4a equal, multiply all terms on each side of the first equation by 4 and all terms on each side of the second by 228.

$$4\times 228a+4\left(-4\right)b=4\times 113,228\times 4a+228\times 228b=0$$

Simplify.

$$912a-16b=452,912a+51984b=0$$

Subtract 912a+51984b=0 from 912a-16b=452 by subtracting like terms on each side of the equal sign.

$$912a-912a-16b-51984b=452$$

Add 912a to -912a. Terms 912a and -912a cancel out, leaving an equation with only one variable that can be solved.

$$-16b-51984b=452$$

Add -16b to -51984b.

$$-52000b=452$$

Divide both sides by -52000.

$$b=-\frac{113}{13000}$$

Substitute -\frac{113}{13000} for b in 4a+228b=0. Because the resulting equation contains only one variable, you can solve for a directly.

$$4a+228\left(-\frac{113}{13000}\right)=0$$

Multiply 228 times -\frac{113}{13000}.

$$4a-\frac{6441}{3250}=0$$

Add \frac{6441}{3250} to both sides of the equation.

$$4a=\frac{6441}{3250}$$

Divide both sides by 4.

$$a=\frac{6441}{13000}$$

The system is now solved.

$$a=\frac{6441}{13000},b=-\frac{113}{13000}$$