Solve {l}{258=0*5A*10^2+A*10}{a=6-2*6^2}{b=a}{text{Solvefor}ctext{where}}{c=b}

$$\left. \begin{array} { l } { 258 = 0 \cdot 5 A \cdot 10 ^ {2} + A \cdot 10 }\\ { a = 6 - 2 \cdot 6 ^ {2} }\\ { b = a }\\ { \text{Solve for } c \text{ where} } \\ { c = b } \end{array} \right.$$

Solve for A, a, b, c

$$c=-66$$

Solution Steps

Consider the first equation. Multiply 0 and 5 to get 0.

$$258=0A\times 10^{2}+A\times 10$$

Calculate 10 to the power of 2 and get 100.

$$258=0A\times 100+A\times 10$$

Multiply 0 and 100 to get 0.

$$258=0A+A\times 10$$

Anything times zero gives zero.

$$258=0+A\times 10$$

Anything plus zero gives itself.

$$258=A\times 10$$

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

$$A\times 10=258$$

Divide both sides by 10.

$$A=\frac{258}{10}$$

Reduce the fraction \frac{258}{10} to lowest terms by extracting and canceling out 2.

$$A=\frac{129}{5}$$

Consider the second equation. Calculate 6 to the power of 2 and get 36.

$$a=6-2\times 36$$

Multiply 2 and 36 to get 72.

$$a=6-72$$

Subtract 72 from 6 to get -66.

$$a=-66$$

Consider the third equation. Insert the known values of variables into the equation.

$$b=-66$$

Consider the fourth equation. Insert the known values of variables into the equation.

$$c=-66$$

The system is now solved.

$$A=\frac{129}{5} a=-66 b=-66 c=-66$$