Solve {l}{0.001*100=0.002*(a+b)}{(b+100)0.00=0.009a}

$$\left. \begin{array} { l } { 0.001 \times 100 = 0.002 \times ( a + b ) } \\ { ( b + 100 ) 0.00 = 0.009 a } \end{array} \right.$$

Solve for a, b

$$a=0$$

$$b=50$$

Solution Steps

Consider the first equation. Multiply 0.001 and 100 to get 0.1.

$$0.1=0.002\left(a+b\right)$$

Use the distributive property to multiply 0.002 by a+b.

$$0.1=0.002a+0.002b$$

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

$$0.002a+0.002b=0.1$$

Consider the second equation. Anything times zero gives zero.

$$0=0.009a$$

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

$$0.009a=0$$

Divide both sides by 0.009. Zero divided by any non-zero number gives zero.

$$a=0$$

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

$$0.002\times 0+0.002b=0.1$$

Multiply 0.002 and 0 to get 0.

$$0+0.002b=0.1$$

Anything plus zero gives itself.

$$0.002b=0.1$$

Divide both sides by 0.002.

$$b=\frac{0.1}{0.002}$$

Expand \frac{0.1}{0.002} by multiplying both numerator and the denominator by 1000.

$$b=\frac{100}{2}$$

Divide 100 by 2 to get 50.

$$b=50$$

The system is now solved.

$$a=0 b=50$$