Question

(i) (2p-3q) = 10 and (8p3 - 27q) = 100 , pq =?​

Answer 1

(2p - 3q) = 10

2p - 3q = 10

2p - q = 10/3

p - q = $\frac{10}{3} / 2$

p - q = 10/3 * 1/2

p- q = $\frac{10}{3} *\frac{1}{2}$

p-q = 5/3

p = 5/3 + q

value of p = 5/3+ q

8p³ - 27q = 100

Put the value of 'p' to find out 'q' value.

8(5/3 + q)³  - 27q = 100

8 * (125/27 + q³) - 27q = 100

125/27 + q³ - 27q = 100* 8

125/27 + q³ - q = 800/27

125 / 27 + q² =800 27

$\frac{125+27q^{2} }{27} = \frac{800}{27}$

125 + 27q² = $\frac{800}{27} * 27$

125 + 27q² = 800

27q² = 800 - 125

27*q² = 675

q² = 675/27

q² = 25

q = $\sqrt{25} = \sqrt{5*5}$

q = 5

Value of 'q' = 5

Value of 'p' is -

2p - 3q = 10

2p - 3*5 = 10

2p - 15 = 10

2p = 10+15

2p = 25

p= 25/2

p = 12.5

p*q = 12.5 * 5

      = 62.5

Answer - 62.5

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Answer 2

$\sf {\underline \red{ Given: -}}$

$\tt2p+3q=102p+3q=10 \: \: \: \: \: \: \: \: \: ...[Equation 1]$

$\tt {8p}^{3} +27q^3={1008p}^{3}+ {27q}^{3} =100 \: \: \: \: ...[Equation \: 2]$

Factorization of Equation 2

$\tt\dashrightarrow (2p+3q)(4p^2-6pq+9p^2)=100$

$\tt\dashrightarrow 10(4p^2-6pq+9q^2)=100$

$\tt\dashrightarrow 4p^2-6pq+9q^2=10 \: \: \: \: \: \: ...[Equation 3]$

Since this question requires identities, for convenience, let the sum and product of 2p,3q2p,3q be m,nm,n .

Then Equation 1, 3 are

$\sf\begin{gathered} \sf\dashrightarrow \begin {cases} & m=10 \\ & m^2-3n=10 \end {cases}\end{gathered}$

$\tt\dashrightarrow m=10,n=30$

$\tt \therefore 6pq=30$

$\therefore \tt \:pq=5$