Chapter 39, Problem 20A

To determine

(a)

The value of $\frac{b}{d}+e-3$.

Answer to Problem 20A

The value of $\frac{b}{d}+e-3$ is $1$.

Explanation of Solution

Given:

The value of $b$ is $8$, $d$ is $4$ and $e$ is $2$.

Calculation:

The algebraic expression is given below:

  $\frac{b}{d}+e-3$  ... (1)

Substitute $8$ for $b$, $4$ for $d$ and $2$ for $e$ in equation (1) as follows:

  $\begin{array}{c}\frac{b}{d}+e-3=\frac{8}{4}+2-3\\ =2+2-3\\ =4-3\\ =1\end{array}$

Thus, the value of $\frac{b}{d}+e-3$ is $1$.

Conclusion:

The value of $\frac{b}{d}+e-3$ is $1$.

To determine

(b)

The value of $bd\left(3+4d-b\right)$.

Answer to Problem 20A

The value of $bd\left(3+4d-b\right)$ is $352$.

Explanation of Solution

Given:

The value of $b$ is $8$, $d$ is $4$ and $e$ is $2$.

Calculation:

The algebraic expression is given below:

  $bd\left(3+4d-b\right)$  ... (1)

Substitute $8$ for $b$ and $4$ for $d$ in equation (1) as follows:

  $\begin{array}{c}bd\left(3+4d-b\right)=8\times 4\left(3+4\times 4-8\right)\\ =32\left(3+16-8\right)\\ =32\times 11\\ =352\end{array}$

Thus, the value of $bd\left(3+4d-b\right)$ is $352$.

Conclusion:

The value of $bd\left(3+4d-b\right)$ is $352$.

To determine

(c)

The value of $5b-\left(bd+3\right)$.

Answer to Problem 20A

The value of $5b-\left(bd+3\right)$ is $5$.

Explanation of Solution

Given:

The value of $b$ is $8$, $d$ is $4$ and $e$ is $2$.

Calculation:

The algebraic expression is given below:

  $5b-\left(bd+3\right)$  ... (1)

Substitute $8$ for $b$ and $4$ for $d$ in equation (1) as follows:

  $\begin{array}{c}5b-\left(bd+3\right)=5\times 8-\left(8\times 4+3\right)\\ =40-\left(32+3\right)\\ =40-35\\ =5\end{array}$

Thus, the value of $5b-\left(bd+3\right)$ is $5$.

Conclusion:

The value of $5b-\left(bd+3\right)$ is $5$.

To determine

(d)

The value of $3e\left(b-e\right)-d\left(\frac{b}{2}\right)$.

Answer to Problem 20A

The value of $3e\left(b-e\right)-d\left(\frac{b}{2}\right)$ is $20$.

Explanation of Solution

Given:

The value of $b$ is $8$, $d$ is $4$ and $e$ is $2$.

Calculation:

The algebraic expression is given below:

  $3e\left(b-e\right)-d\left(\frac{b}{2}\right)$  ... (1)

Substitute $8$ for $b$, $4$ for $d$ and $2$ for $e$ in equation (1) as follows:

  $\begin{array}{c}3e\left(b-e\right)-d\left(\frac{b}{2}\right)=3\times 2\left(8-2\right)-4\times \left(\frac{8}{2}\right)\\ =\left(6\times 6\right)-\left(4\times 4\right)\\ =36-16\\ =20\end{array}$

Thus, the value of $3e\left(b-e\right)-d\left(\frac{b}{2}\right)$ is $20$.

Conclusion:

The value of $3e\left(b-e\right)-d\left(\frac{b}{2}\right)$ is $20$.

To determine

(e)

The value of $\frac{12d}{e}-\left[3b-\left(d+e\right)+4\right]$.

Answer to Problem 20A

The value of $\frac{12d}{e}-\left[3b-\left(d+e\right)+4\right]$ is $2$.

Explanation of Solution

Given:

The value of $b$ is $8$, $d$ is $4$ and $e$ is $2$.

Calculation:

The algebraic expression is given below:

  $\frac{12d}{e}-\left[3b-\left(d+e\right)+4\right]$  ... (1)

Substitute $8$ for $b$, $4$ for $d$ and $2$ for $e$ in equation (1) as follows:

  $\begin{array}{c}\frac{12d}{e}-\left[3b-\left(d+e\right)+4\right]=\frac{12\times 4}{2}-\left[3\times 8-\left(4+2\right)+4\right]\\ =24-\left[24-6+4\right]\\ =24-24+2\\ =2\end{array}$

Thus, the value of $\frac{12d}{e}-\left[3b-\left(d+e\right)+4\right]$ is $2$.

Conclusion:

The value of $\frac{12d}{e}-\left[3b-\left(d+e\right)+4\right]$ is $2$.

To determine

(f)

The value of $\frac{b+d+e}{d-e}+b{d}^2-d{e}^3$.

Answer to Problem 20A

The value of $\frac{b+d+e}{d-e}+b{d}^2-d{e}^3$ is $103$.

Explanation of Solution

Given:

The value of $b$ is $8$, $d$ is $4$ and $e$ is $2$.

Calculation:

The algebraic expression is given below:

  $\frac{b+d+e}{d-e}+b{d}^2-d{e}^3$  ... (1)

Substitute $8$ for $b$, $4$ for $d$ and $2$ for $e$ in equation (1) as follows:

  $\begin{array}{c}\frac{b+d+e}{d-e}+b{d}^2-d{e}^3=\frac{8+4+2}{4-2}+8{\left(4\right)}^2-4{\left(2\right)}^3\\ =\frac{14}{2}+128-32\\ =7+96\\ =103\end{array}$

Thus, the value of $\frac{b+d+e}{d-e}+b{d}^2-d{e}^3$ is $103$.

Conclusion:

The value of $\frac{b+d+e}{d-e}+b{d}^2-d{e}^3$ is $103$.