✅ \(x = \frac{9}{13}, y = -\frac{5}{13}\)

From \(9x = 2y+7\) \(⇒ x = \frac{2y+7}{9}\). Substitute in \(3x-5y=4\):
\(3\left(\frac{2y+7}{9}\right)-5y=4\) \(⇒\frac{2y+7}{3}-5y=4\) ⇒ \(\frac{2y+7-15y}{3}=4\) \(⇒7-13y=12\) \(⇒ y=-\frac{5}{13}\).
Then \(x = \frac{2(-\frac{5}{13})+7}{9}\) \(= \frac{-\frac{10}{13}+7}{9} \)\(= \frac{\frac{91-10}{13}}{9}\)\( = \frac{81}{13\times9} \)\(= \frac{9}{13}\).

✅ Roots: \(-2, \frac{3}{2}\)

\(2x^2+4x-3x-6=0\) \(⇒2x(x+2)-3(x+2)=0\) \(⇒(x+2)(2x-3)=0\) \(⇒x=-2\) or \(x=\frac{3}{2}\).

\(\sec\theta = \frac{13}{12}\) ⇒ Hypotenuse = 13, Base = 12.
Perpendicular = \(\sqrt{13^2-12^2}\)\( = \sqrt{169-144}\)\( = \sqrt{25} = 5\).
\(\sin\theta \)\(= \frac{5}{13},\; \cos\theta \)\(= \frac{12}{13},\; \tan\theta \)\(= \frac{5}{12},\; \csc\theta \)\(= \frac{13}{5},\; \cot\theta \)\(= \frac{12}{5}\).

✅ Sphere Volume: \(113.14\) cm³
\(V = \frac{4}{3}\pi r^3 \)\(= \frac{4}{3} \times \frac{22}{7} \times 27\)\( =\frac{4\times22\times27}{21}\)\( = 113.14\) cm³.

OR Cylinder CSA: \(2\pi rh \)\(= 2\times\frac{22}{7}\times2\times7 \)\(= 2\times22\times2 \)\(= 88 cm².\)

✅ y = 3 or -9

\(\sqrt{(10-2)^2 + (y+3)^2} = 10\) \(⇒64 + (y+3)^2 = 100\) \(⇒(y+3)^2 = 36\) \(⇒y+3 = \pm 6\) \(⇒ y=3\) or \(y=-9\).

OR Isosceles check: AB = \(\sqrt{(5-6)^2+(-2-4)^2}\)\( = \sqrt{1+36}\)\( = \sqrt{37}\)
BC = \(\sqrt{(6-7)^2+(4+2)^2}\)\( = \sqrt{1+36} \)\(= \sqrt{37}\) \(⇒ AB = BC\)\( ⇒ isosceles.\)

✅ \(3x^2 – 3\sqrt{2}x + 1\)

General form: \(k[x^2 – (\text{sum})x + \text{product}]\)
\(= k[x^2 – \sqrt{2}x + \frac{1}{3}]\). Taking \(k=3\): \(3x^2 – 3\sqrt{2}x + 1\).