Question

5.3 adding & subtracting polynomials find the value of each expression when ( a = -2, b = -3, c = 4, d = 5 ) ( ad + bc ) ( \frac{1}{4}a - \frac{1}{6}b + 5(c - d) ) ( \frac{3d - b}{a + 2c} )

Explanation:

Step1: Substitute values for $ad+bc$

Substitute $a=-2, b=-3, c=4, d=5$:
$(-2)(5) + (-3)(4)$

Step2: Calculate $ad+bc$

$$\begin{align*} (-2)(5) + (-3)(4) &= -10 + (-12) \\ &= -22 \end{align*}$$

Step3: Substitute values for $\frac{1}{4}a-\frac{1}{6}b+5(c-d)$

Substitute $a=-2, b=-3, c=4, d=5$:
$\frac{1}{4}(-2) - \frac{1}{6}(-3) + 5(4-5)$

Step4: Calculate the expression

$$\begin{align*} \frac{1}{4}(-2) - \frac{1}{6}(-3) + 5(4-5) &= -\frac{1}{2} + \frac{1}{2} + 5(-1) \\ &= 0 -5 \\ &= -5 \end{align*}$$

Step5: Substitute values for $\frac{3d-b}{a+2c}$

Substitute $a=-2, b=-3, c=4, d=5$:
$\frac{3(5)-(-3)}{-2+2(4)}$

Step6: Calculate the rational expression

$$\begin{align*} \frac{3(5)-(-3)}{-2+2(4)} &= \frac{15+3}{-2+8} \\ &= \frac{18}{6} \\ &= 3 \end{align*}$$

Answer:

1. For $ad+bc$: $-22$
2. For $\frac{1}{4}a-\frac{1}{6}b+5(c-d)$: $-5$
3. For $\frac{3d-b}{a+2c}$: $3$