Solve {l}{I_{p}=2.1*10^18*1.6*10^-19/1}{I_{c}=1.6*10^-19*4.15*10^18/1}

$$\left. \begin{array} { l } { I _ { p } = \frac { 2.1 \times 10 ^ { 18 } \times 1.6 \times 10 ^ { - 19 } } { 1 } } \\ { I _ { c } = \frac { 1.6 \times 10 ^ { - 19 } \times 4.15 \times 10 ^ { 18 } } { 1 } } \end{array} \right.$$

Solve for I_p, I_c

$$I_{p}=0.336$$

$$I_{c}=0.664$$

Solution Steps

Consider the first equation. To multiply powers of the same base, add their exponents. Add 18 and -19 to get -1.

$$I_{p}=\frac{2.1\times 10^{-1}\times 1.6}{1}$$

Calculate 10 to the power of -1 and get \frac{1}{10}.

$$I_{p}=\frac{2.1\times \frac{1}{10}\times 1.6}{1}$$

Multiply 2.1 and \frac{1}{10} to get \frac{21}{100}.

$$I_{p}=\frac{\frac{21}{100}\times 1.6}{1}$$

Multiply \frac{21}{100} and 1.6 to get \frac{42}{125}.

$$I_{p}=\frac{\frac{42}{125}}{1}$$

Anything divided by one gives itself.

$$I_{p}=\frac{42}{125}$$

Consider the second equation. To multiply powers of the same base, add their exponents. Add -19 and 18 to get -1.

$$I_{c}=\frac{1.6\times 10^{-1}\times 4.15}{1}$$

Calculate 10 to the power of -1 and get \frac{1}{10}.

$$I_{c}=\frac{1.6\times \frac{1}{10}\times 4.15}{1}$$

Multiply 1.6 and \frac{1}{10} to get \frac{4}{25}.

$$I_{c}=\frac{\frac{4}{25}\times 4.15}{1}$$

Multiply \frac{4}{25} and 4.15 to get \frac{83}{125}.

$$I_{c}=\frac{\frac{83}{125}}{1}$$

Anything divided by one gives itself.

$$I_{c}=\frac{83}{125}$$

The system is now solved.

$$I_{p}=\frac{42}{125} I_{c}=\frac{83}{125}$$