Question

please solve this equation​

Answer 1

→ y² + 4 = 1/2(3y² + 7)

→ y² + 4 = $\frac{3y^2 + 7}{2}$

→ 2y² + 8 = 3y² + 7

→ 3y² + 7 - 2y² = 8

→ y² + 7 = 8

→ y² = 8 - 7

→ y² = 1

∴ y = 1

Answer 2

$\huge{\text{\underline{\underline{\red{Answer:-}}}}}$

$\bf{\large{ {y}^{2} + 4 = \frac{1}{2} (3 {y}^{2} + 7)}}$

$\bf{\large{ {y}^{2} + 4 =\frac{ {3y + 7}^{2} }{2 } }}$

$\bf{\large{ {2y}^{2} + 8 = { 3y }^{2} + 7}}$

$\bf{\large{3 {y}^{2} + 7 - 2 {y}^{2} = 8}}$

$\bf{\large{{y}^{2} + 7 = 8}}$

$\bf{\large{{y}^{2} = 8 - 7}}$

$\bf{\large{{y}^{2} = 1}}$

$\bf{\large{y=1}}$

$\huge{\bold{\boxed{\boxed{\blue{Therefore\:y=1}}}}}$

#Bal