Question 1

Hey can you help me with this one ?

Question 2

Answer:

a = 8 or
a = -(14)/(3)

Explanation:

13(a+4)+24(a+5) = 3(a^(2) +9a+20)

3a^(2) -10a-112=0

(3a+14)(a-8)=0

a = 8 or
a = -(14)/(3)

Question 3

${\qquad\qquad\huge\underline{{\sf Answer}}}$

$\qquad \sf  \dashrightarrow \: \cfrac{13}{a + 5} + \cfrac{24}{a + 4} = 3$

$\qquad \sf  \dashrightarrow \: \cfrac{13(a + 4) + 24(a + 5)}{(a + 5)(a + 4)} = 3$

$\qquad \sf  \dashrightarrow \: \cfrac{13a + 52+ 24a + 120}{a {}^(2) + 5a + 4a + 20} = 3$

$\qquad \sf  \dashrightarrow \: \cfrac{ 37a + 172}{a {}^(2) +9a + 20} = 3$

$\qquad \sf  \dashrightarrow \: 3( {a }^(2) + 9a + 20) = 37a + 172$

$\qquad \sf  \dashrightarrow \: 3 {a}^(2) + 27a + 60 = 37a + 172$

$\qquad \sf  \dashrightarrow \: 3 {a}^(2) + 27a - 37a + 60 - 172 = 0$

$\qquad \sf  \dashrightarrow \: 3 {a}^(2) - 10a - 112= 0$

$\qquad \sf  \dashrightarrow \: 3 {a}^(2) - 24a + 14a - 112 = 0$

$\qquad \sf  \dashrightarrow \: 3a(a - 8) + 14(a - 8) = 0$

$\qquad \sf  \dashrightarrow \: (a - 8) + (3a + 14) = 0$

So, required values of " a " are :

$\qquad \sf  \dashrightarrow \: a = 8 \: \: \: or \: \: \: a = - \cfrac{ 14}{3}$

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