Question

Solve the following equations:
$3x^2-5y^2=7$,
$3xy-4y^2=2$.

• A. $x=\pm 2;y=\pm 1$
• B. $x=\pm 3;y=\pm 2$
• C. $x=\pm 4;y=\pm 3$
• D. $x=\pm 1;y=\pm 2$

Answer 1

Answer: (A), (B)

$x=\pm 2;y=\pm 1$
$x=\pm 3;y=\pm 2$

Given equations are $3x^2-{5y}^2=7$ .....(i)

and $3xy-4y^2=2$

Put $y=vx$

$3x^2-5v^2{x}^2=7$ ......(ii)

$3v{x}^2-{4v}^2{x}^2=2$ .....(iii)

Dividing (ii) by (iii), we get

$\frac{3x^2-5v^2{x}^2}{3v{x}^2-{4v}^2{x}^2}=\frac{7}{2}\Rightarrow \frac{3-5v^2}{3v-{4v}^2}=\frac{7}{2}\Rightarrow 6-10v^2=21v-28v^2\Rightarrow 18v^2-21v+6=0\Rightarrow 6v^2-7v+2=0\Rightarrow 6v^2-3v-4v+2=0\Rightarrow 3v(2v-1)-2(2v-1)=0\Rightarrow (3v-2)(2v-1)=0\Rightarrow v=\frac{2}{3},\frac{1}{2}\Rightarrow y=\frac{2x}{3},\frac{x}{2}$

Substituting $y$ in (i), we get

$3x^2-{5y}^2=7$

(i) Put $y=\frac{2x}{3}$

Therefore, $3x^2-5{\left(\frac{2x}{3}\right)}^2=7$

$\Rightarrow \frac{7x^2}{9}=7\Rightarrow x=\pm 3$

Thus $y=\frac{2(\pm 3)}{3}=\pm 2$

(ii) Put $y=\frac{x}{2}$

Therefore, $3x^2-5{\left(\frac{x}{2}\right)}^2=7$

$\Rightarrow \frac{7x^2}{4}=7\Rightarrow x=\pm 2$

Thus $y=\frac{\pm 2}{2}=\pm 1$