Question

1. $\frac{p^2 + p + 8}{p^2 - 7p + 6} + \frac{2}{p - 1} = \frac{2}{3}$
2. $\frac{1}{3} - \frac{1}{4v + 3} = \frac{v^2 - 3v - 18}{12v + 9}$

Explanation:

Problem 24

Step1: Factor the denominator

Factor $p^2-7p+6$:
$p^2-7p+6=(p-1)(p-6)$

Step2: Find common denominator

The common denominator is $3(p-1)(p-6)$. Multiply all terms by it:
$$3(p^2+p+8) + 6(p-6) = 2(p-1)(p-6)$$

Step3: Expand all expressions

Expand each part:
Left side: $3p^2+3p+24 + 6p-36 = 3p^2+9p-12$
Right side: $2(p^2-7p+6)=2p^2-14p+12$

Step4: Rearrange to standard quadratic form

Bring all terms to left side:
$3p^2+9p-12 -2p^2+14p-12=0$
$p^2+23p-24=0$

Step5: Solve quadratic equation

Factor the quadratic:
$(p+24)(p-1)=0$
Exclude $p=1$ (makes original denominator 0), so $p=-24$.

Step1: Factor all denominators

Factor $v^2-3v-18=(v-6)(v+3)$, $12v+9=3(4v+3)$.
Original equation: $\frac{1}{3}-\frac{1}{4v+3}=\frac{(v-6)(v+3)}{3(4v+3)}$

Step2: Find common denominator

Common denominator is $3(4v+3)$. Multiply all terms by it:
$$(4v+3) - 3 = (v-6)(v+3)$$

Step3: Simplify left side

$4v+3-3=4v$

Step4: Expand right side

$(v-6)(v+3)=v^2-3v-18$

Step5: Rearrange to standard quadratic form

$0=v^2-3v-18-4v$
$v^2-7v-18=0$

Step6: Solve quadratic equation

Factor the quadratic:
$(v-9)(v+2)=0$
Check: $v=9$ and $v=-2$ do not make any original denominator 0.

Answer:

$p=-24$

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Problem 25