Question

y2+6 / 8y2 + 3=1 /5​

Answer 1

y²+ 6 /8y²+3=1/5

5(y²+6)=1(8y²+3)

5y²+30= 8y²+3

30-3=8y²-5y²

27=3y²

y²=27/3

y²=9

y= +3,-3

Answer 2

Given :

• $\displaystyle{\sf \frac{y^2+6}{8y^2+3}=\frac{1}{5}}$

Solution :

$\rightarrow \displaystyle{\sf \frac{y^2+6}{8y^2+3}=\frac{1}{5}}$

$\rightarrow \displaystyle{\sf 5(y^2+6)=1(8y^2+3)}$

$\rightarrow \displaystyle{\sf 5y^2+30=8y^2+3}$

$\rightarrow \displaystyle{\sf 5y^2-8y^2=3-30}$

$\rightarrow \displaystyle{\sf -3y^2=-27}$

$\rightarrow \displaystyle{\sf y^2=\frac{-27}{-3}}$

$\rightarrow \displaystyle{\sf y=\sqrt{9}}$

$\rightarrow \displaystyle{\sf y=\pm 3}$

Verification :

Substitute -3 in place of y

$\rightarrow \displaystyle{\sf \frac{y^2+6}{8y^2+3}=\frac{1}{5}}$

$\rightarrow \displaystyle{\sf \frac{(-3)^2+6}{8(-3)^2+3}=\frac{1}{5}}$

$\rightarrow \displaystyle{\sf \frac{9+6}{8(9)+3}=\frac{1}{5}}$

$\rightarrow \displaystyle{\sf \frac{15}{72+3}=\frac{1}{5}}$

$\rightarrow \displaystyle{\sf \frac{15}{75}=\frac{1}{5}}$

$\rightarrow \displaystyle{\sf \frac{1}{5}=\frac{1}{5}}$

Substitue 3 in place of y

$\rightarrow \displaystyle{\sf \frac{y^2+6}{8y^2+3}=\frac{1}{5}}$

$\rightarrow \displaystyle{\sf \frac{(3)^2+6}{8(3)^2+3}=\frac{1}{5}}$

$\rightarrow \displaystyle{\sf \frac{9+6}{8(9)+3}=\frac{1}{5}}$

$\rightarrow \displaystyle{\sf \frac{15}{72+3}=\frac{1}{5}}$

$\rightarrow \displaystyle{\sf \frac{15}{75}=\frac{1}{5}}$

$\rightarrow \displaystyle{\sf \frac{1}{5}=\frac{1}{5}}$

________________________________

Required Answer :

Solution of the given equation

$\displaystyle{\sf \frac{y^2+6}{8y^2+3}=\frac{1}{5}}$ is $\displaystyle{\sf y=\pm 3}$