Chapter 46, Problem 69A

To determine

To solve for the unknown value in the given equation.

Answer to Problem 69A

  $C=\pm 4.434$.

Explanation of Solution

Given information:

The given equation is $\frac{12.8}{5-3C^2}=\frac{7.43}{-31.29}$.

Calculation:

We have been given the equation as $\frac{12.8}{5-3C^2}=\frac{7.43}{-31.29}$.

We will perform some basic mathematical operation to solve the given equation as below.

  $\frac{12.8}{5-3C^2}=\frac{7.43}{-31.29}$

  $\Rightarrow 7.43(5-3C^2)=-31.29\times 12.8\left\{\text{Cross multiply}\right\}$

  $\Rightarrow 37.15-22.29C^2=-400.512$

  $\Rightarrow -22.29C^2+37.15-37.15=-400.512-37.15\{\text{Subtracting 37}\text{.15 on both sides}}$

  $\Rightarrow -22.29C^2=-437.662\left\{\text{Simplify}\right\}$

  $\Rightarrow \frac{-22.29}{-22.29}{C}^2=\frac{-437.662}{-22.29}\{\text{Dividing}\text{by}-22.29\text{on}\text{both}\text{sides}\}$

  $\Rightarrow C^2=19.661$

  $\Rightarrow C=\pm \sqrt{19.661}$

  $\Rightarrow C=\pm 4.434$

  $\text{Check:}$

  $\frac{12.8}{5-3C^2}=\frac{7.43}{-31.29}$

  $\frac{12.8}{5-3{(\pm 4.434)}^2}=\frac{7.43}{-31.29}$

  $\frac{12.8}{5-3\times 4.434}=\frac{7.43}{-31.29}$

  $\frac{12.8}{-8.302}=\frac{7.43}{-31.29}$

  $\frac{7.43}{-31.29}=\frac{7.43}{-31.29}$

Hence, solution of the equation will be given by-

  $C=\pm 4.434$.