Question

solve the pair of equations 7/3^x - 6/2^y = 15 and 8/3^x = 9/2^y.

Answer 1

Answer:

x = - 2    and   y = - 3

Step-by-step explanation:

$\large Given \ \dfrac{7}{3^x} -\dfrac{6}{2^y}=15 \ and \ \dfrac{8}{3^x} =\dfrac{9}{2^y} \\\\\\\large \text{we have to find value of x and y}\\\\\\\large Let \ 3^x=a \ and \ 2^y=b\\\\\\\\\large \text{Now equation become \dfrac{7}{a} -\dfrac{6}{b}=15 \ and \ \dfrac{8}{a} =\dfrac{9}{b} }\\\\\\ \large \text{Now simplify the equation}\\\\\\\large \dfrac{7b-6a}{ab}=15 \\\\\\\large 7b-6a=15ab \ ...(i)\\\\\\\large 8a=9b\\\\\\\large b=\dfrac{8a}{9} \ .(ii)$

$\large \text{now putting b value in (i) we get}\\\\\\\large 7b-6\times\dfrac{8b}{9}=15b\times\dfrac{8b}{9}\\\\\\\large \dfrac{63b-48b}{9}=\dfrac{15b\times 8b}{9}\\\\\\\large -15b=15\times8b^2\\\\\\\large b=\dfrac{-1}{8} \\\\\\\large \text{Now putting b value in (ii)}\\\\\\\large \text{ a=\dfrac{8\times\frac{-1}{8} }{9} }\\\\\\\large a=\dfrac{-1}{9} \\\\\\\large \text{We have value of a and b }$

$\large \text{putting values here}\\\\\\\large 3^x=\dfrac{-1}{9} \\\\\\\large 3^x=3^{-2}\\\\\\\large \text{comparing both side we get}\\\\\\\large x=-2\\\\\\\large \text{Now we have}\\\\\\\large 2^y=\dfrac{-1}{8} \\\\\\\large 2^y=2^{-3}\\\\\\\large \text{Again comparing both side we get}\\\\\\\large y=-3\\\\\\\large \text{Thus we get answer x=-2 \ and \ y=-3 }$