Solve {l}{20.7=r/50E+6/E}{6r/E=144.9-20.7/0.5}

$$\left. \begin{array} { l } { 20.7 = \frac { r } { 50 E } + \frac { 6 } { E } } \\ { \frac { 6 r } { E } = \frac { 144.9 - 20.7 } { 0.5 } } \end{array} \right.$$

Solve for r, E

$$r=12.5$$

$$E=\frac{125}{414}\approx 0.301932367$$

Steps Using Substitution

Consider the first equation. Variable E cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 50E, the least common multiple of 50E,E.

$$1035E=r+50\times 6$$

Multiply 50 and 6 to get 300.

$$1035E=r+300$$

Divide both sides by 1035.

$$E=\frac{1}{1035}\left(r+300\right)$$

Multiply \frac{1}{1035} times r+300.

$$E=\frac{1}{1035}r+\frac{20}{69}$$

Substitute \frac{20}{69}+\frac{r}{1035} for E in the other equation, -\frac{1242}{5}E+6r=0.

$$-\frac{1242}{5}\left(\frac{1}{1035}r+\frac{20}{69}\right)+6r=0$$

Multiply -\frac{1242}{5} times \frac{20}{69}+\frac{r}{1035}.

$$-\frac{6}{25}r-72+6r=0$$

Add -\frac{6r}{25} to 6r.

$$\frac{144}{25}r-72=0$$

Add 72 to both sides of the equation.

$$\frac{144}{25}r=72$$

Divide both sides of the equation by \frac{144}{25}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$r=\frac{25}{2}$$

Substitute \frac{25}{2} for r in E=\frac{1}{1035}r+\frac{20}{69}. Because the resulting equation contains only one variable, you can solve for E directly.

$$E=\frac{1}{1035}\times \frac{25}{2}+\frac{20}{69}$$

Multiply \frac{1}{1035} times \frac{25}{2} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$E=\frac{5}{414}+\frac{20}{69}$$

Add \frac{20}{69} to \frac{5}{414} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$E=\frac{125}{414}$$

The system is now solved.

$$E=\frac{125}{414},r=\frac{25}{2}$$

Steps Using Matrices

Consider the first equation. Variable E cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 50E, the least common multiple of 50E,E.

$$1035E=r+50\times 6$$

Multiply 50 and 6 to get 300.

$$1035E=r+300$$

Subtract r from both sides.

$$1035E-r=300$$

Consider the second equation. Variable E cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by E.

$$6r=E\times \frac{144.9-20.7}{0.5}$$

Subtract 20.7 from 144.9 to get 124.2.

$$6r=E\times \frac{124.2}{0.5}$$

Expand \frac{124.2}{0.5} by multiplying both numerator and the denominator by 10.

$$6r=E\times \frac{1242}{5}$$

Subtract E\times \frac{1242}{5} from both sides.

$$6r-E\times \frac{1242}{5}=0$$

Multiply -1 and \frac{1242}{5} to get -\frac{1242}{5}.

$$6r-\frac{1242}{5}E=0$$

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

$$1035E-r=300,-\frac{1242}{5}E+6r=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}1035&-1\\-\frac{1242}{5}&6\end{matrix}\right)\left(\begin{matrix}E\\r\end{matrix}\right)=\left(\begin{matrix}300\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1035&-1\\-\frac{1242}{5}&6\end{matrix}\right).

$$inverse(\left(\begin{matrix}1035&-1\\-\frac{1242}{5}&6\end{matrix}\right))\left(\begin{matrix}1035&-1\\-\frac{1242}{5}&6\end{matrix}\right)\left(\begin{matrix}E\\r\end{matrix}\right)=inverse(\left(\begin{matrix}1035&-1\\-\frac{1242}{5}&6\end{matrix}\right))\left(\begin{matrix}300\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}E\\r\end{matrix}\right)=inverse(\left(\begin{matrix}1035&-1\\-\frac{1242}{5}&6\end{matrix}\right))\left(\begin{matrix}300\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}E\\r\end{matrix}\right)=inverse(\left(\begin{matrix}1035&-1\\-\frac{1242}{5}&6\end{matrix}\right))\left(\begin{matrix}300\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}E\\r\end{matrix}\right)=\left(\begin{matrix}\frac{6}{1035\times 6-\left(-\left(-\frac{1242}{5}\right)\right)}&-\frac{-1}{1035\times 6-\left(-\left(-\frac{1242}{5}\right)\right)}\\-\frac{-\frac{1242}{5}}{1035\times 6-\left(-\left(-\frac{1242}{5}\right)\right)}&\frac{1035}{1035\times 6-\left(-\left(-\frac{1242}{5}\right)\right)}\end{matrix}\right)\left(\begin{matrix}300\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}E\\r\end{matrix}\right)=\left(\begin{matrix}\frac{5}{4968}&\frac{5}{29808}\\\frac{1}{24}&\frac{25}{144}\end{matrix}\right)\left(\begin{matrix}300\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}E\\r\end{matrix}\right)=\left(\begin{matrix}\frac{5}{4968}\times 300\\\frac{1}{24}\times 300\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}E\\r\end{matrix}\right)=\left(\begin{matrix}\frac{125}{414}\\\frac{25}{2}\end{matrix}\right)$$

Extract the matrix elements E and r.

$$E=\frac{125}{414},r=\frac{25}{2}$$

Steps Using Elimination

Consider the first equation. Variable E cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 50E, the least common multiple of 50E,E.

$$1035E=r+50\times 6$$

Multiply 50 and 6 to get 300.

$$1035E=r+300$$

Subtract r from both sides.

$$1035E-r=300$$

Consider the second equation. Variable E cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by E.

$$6r=E\times \frac{144.9-20.7}{0.5}$$

Subtract 20.7 from 144.9 to get 124.2.

$$6r=E\times \frac{124.2}{0.5}$$

Expand \frac{124.2}{0.5} by multiplying both numerator and the denominator by 10.

$$6r=E\times \frac{1242}{5}$$

Subtract E\times \frac{1242}{5} from both sides.

$$6r-E\times \frac{1242}{5}=0$$

Multiply -1 and \frac{1242}{5} to get -\frac{1242}{5}.

$$6r-\frac{1242}{5}E=0$$

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.

$$1035E-r=300,-\frac{1242}{5}E+6r=0$$

To make 1035E and -\frac{1242E}{5} equal, multiply all terms on each side of the first equation by -\frac{1242}{5} and all terms on each side of the second by 1035.

$$-\frac{1242}{5}\times 1035E-\frac{1242}{5}\left(-1\right)r=-\frac{1242}{5}\times 300,1035\left(-\frac{1242}{5}\right)E+1035\times 6r=0$$

Simplify.

$$-257094E+\frac{1242}{5}r=-74520,-257094E+6210r=0$$

Subtract -257094E+6210r=0 from -257094E+\frac{1242}{5}r=-74520 by subtracting like terms on each side of the equal sign.

$$-257094E+257094E+\frac{1242}{5}r-6210r=-74520$$

Add -257094E to 257094E. Terms -257094E and 257094E cancel out, leaving an equation with only one variable that can be solved.

$$\frac{1242}{5}r-6210r=-74520$$

Add \frac{1242r}{5} to -6210r.

$$-\frac{29808}{5}r=-74520$$

Divide both sides of the equation by -\frac{29808}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$r=\frac{25}{2}$$

Substitute \frac{25}{2} for r in -\frac{1242}{5}E+6r=0. Because the resulting equation contains only one variable, you can solve for E directly.

$$-\frac{1242}{5}E+6\times \frac{25}{2}=0$$

Multiply 6 times \frac{25}{2}.

$$-\frac{1242}{5}E+75=0$$

Subtract 75 from both sides of the equation.

$$-\frac{1242}{5}E=-75$$

Divide both sides of the equation by -\frac{1242}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$E=\frac{125}{414}$$

The system is now solved.

$$E=\frac{125}{414},r=\frac{25}{2}$$