Question

if *If x/y = 2/3 then 2x² + 3y² / 2x² - 3y² = ?*

1️⃣ 35/19
2️⃣ 19/35
3️⃣ -19/35
4️⃣ -35/19​

Answer 1

Given data :

• $\frac{x}{y}=\frac{2}{3}$

To find :

The value of the term $\frac{2x^{2}+3y^{2}}{2x^{2}-3y^{2}}$

Step-by-step explanation :

Now, $\frac{2x^{2}+3y^{2}}{2x^{2}-3y^{2}}$

• Now divide both the numerator and the denominator by $y^{2}$ with the condition that $y$ is non-zero, i.e., $y\neq 0$

$\displaystyle=\frac{\frac{2x^{2}+3y^{2}}{y^{2}}}{\frac{2x^{2}-3y^{2}}{y^{2}}}$

$\displaystyle=\frac{2\frac{x^{2}}{y^{2}}+3}{2\frac{x^{2}}{y^{2}}-3}$

$\displaystyle=\frac{2(\frac{x}{y})^{2}+3}{2(\frac{x}{y})^{2}-3}$

$\displaystyle=\frac{2(\frac{2}{3})^{2}+3}{2(\frac{2}{3})^{2}-3}$

$\displaystyle=\frac{2*\frac{4}{9}+3}{2*\frac{4}{9}-3}$

$\displaystyle=\frac{\frac{8+27}{3}}{\frac{8-27}{3}}$

$\displaystyle=\frac{35}{-19}$

$\displaystyle=-\frac{35}{19}$

Answer :

Therefore Option 4 $-\frac{35}{19}$ is correct.