Solve {l}{503.4=5a+295b}{44880.3=295a+27315b}

$$\left. \begin{array}{l}{ 503.4 = 5 a + 295 b }\\{ 44880.3 = 295 a + 27315 b }\end{array} \right.$$

Solve for a, b

$$a = \frac{204273}{19820} = 10\frac{6073}{19820} \approx 10.306407669$$

$$b = \frac{151797}{99100} = 1\frac{52697}{99100} \approx 1.531755802$$

Steps Using Substitution

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

$$5a+295b=503.4$$

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

$$295a+27315b=44880.3$$

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.

$$5a+295b=503.4,295a+27315b=44880.3$$

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

$$5a+295b=503.4$$

Subtract 295b from both sides of the equation.

$$5a=-295b+503.4$$

Divide both sides by 5.

$$a=\frac{1}{5}\left(-295b+503.4\right)$$

Multiply \frac{1}{5} times -295b+503.4.

$$a=-59b+\frac{2517}{25}$$

Substitute -59b+\frac{2517}{25} for a in the other equation, 295a+27315b=44880.3.

$$295\left(-59b+\frac{2517}{25}\right)+27315b=44880.3$$

Multiply 295 times -59b+\frac{2517}{25}.

$$-17405b+\frac{148503}{5}+27315b=44880.3$$

Add -17405b to 27315b.

$$9910b+\frac{148503}{5}=44880.3$$

Subtract \frac{148503}{5} from both sides of the equation.

$$9910b=\frac{151797}{10}$$

Divide both sides by 9910.

$$b=\frac{151797}{99100}$$

Substitute \frac{151797}{99100} for b in a=-59b+\frac{2517}{25}. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-59\times \frac{151797}{99100}+\frac{2517}{25}$$

Multiply -59 times \frac{151797}{99100}.

$$a=-\frac{8956023}{99100}+\frac{2517}{25}$$

Add \frac{2517}{25} to -\frac{8956023}{99100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a=\frac{204273}{19820}$$

The system is now solved.

$$a=\frac{204273}{19820},b=\frac{151797}{99100}$$

Steps Using Matrices

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

$$5a+295b=503.4$$

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

$$295a+27315b=44880.3$$

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

$$5a+295b=503.4,295a+27315b=44880.3$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&295\\295&27315\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}503.4\\44880.3\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&295\\295&27315\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&295\\295&27315\end{matrix}\right))\left(\begin{matrix}5&295\\295&27315\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}5&295\\295&27315\end{matrix}\right))\left(\begin{matrix}503.4\\44880.3\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}5&295\\295&27315\end{matrix}\right))\left(\begin{matrix}503.4\\44880.3\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}5&295\\295&27315\end{matrix}\right))\left(\begin{matrix}503.4\\44880.3\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{27315}{5\times 27315-295\times 295}&-\frac{295}{5\times 27315-295\times 295}\\-\frac{295}{5\times 27315-295\times 295}&\frac{5}{5\times 27315-295\times 295}\end{matrix}\right)\left(\begin{matrix}503.4\\44880.3\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{5463}{9910}&-\frac{59}{9910}\\-\frac{59}{9910}&\frac{1}{9910}\end{matrix}\right)\left(\begin{matrix}503.4\\44880.3\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{5463}{9910}\times 503.4-\frac{59}{9910}\times 44880.3\\-\frac{59}{9910}\times 503.4+\frac{1}{9910}\times 44880.3\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{204273}{19820}\\\frac{151797}{99100}\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=\frac{204273}{19820},b=\frac{151797}{99100}$$

Steps Using Elimination

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

$$5a+295b=503.4$$

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

$$295a+27315b=44880.3$$

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.

$$5a+295b=503.4,295a+27315b=44880.3$$

To make 5a and 295a equal, multiply all terms on each side of the first equation by 295 and all terms on each side of the second by 5.

$$295\times 5a+295\times 295b=295\times 503.4,5\times 295a+5\times 27315b=5\times 44880.3$$

Simplify.

$$1475a+87025b=148503,1475a+136575b=224401.5$$

Subtract 1475a+136575b=224401.5 from 1475a+87025b=148503 by subtracting like terms on each side of the equal sign.

$$1475a-1475a+87025b-136575b=148503-224401.5$$

Add 1475a to -1475a. Terms 1475a and -1475a cancel out, leaving an equation with only one variable that can be solved.

$$87025b-136575b=148503-224401.5$$

Add 87025b to -136575b.

$$-49550b=148503-224401.5$$

Add 148503 to -224401.5.

$$-49550b=-75898.5$$

Divide both sides by -49550.

$$b=\frac{151797}{99100}$$

Substitute \frac{151797}{99100} for b in 295a+27315b=44880.3. Because the resulting equation contains only one variable, you can solve for a directly.

$$295a+27315\times \frac{151797}{99100}=44880.3$$

Multiply 27315 times \frac{151797}{99100}.

$$295a+\frac{829267011}{19820}=44880.3$$

Subtract \frac{829267011}{19820} from both sides of the equation.

$$295a=\frac{12052107}{3964}$$

Divide both sides by 295.

$$a=\frac{204273}{19820}$$

The system is now solved.

$$a=\frac{204273}{19820},b=\frac{151797}{99100}$$