Q.(a) Show that the equation $2x^2+7xy+3y^2+5x+5y+2=0$ represent two straight lines and find their separate equations.

(a) The equation splits into the lines $2x+y+1=0$ and $x+3y+2=0$. (b) The valid solution is $x=\tfrac12$.

(a) Pair of lines

$2x^{2}+7xy+3y^{2}+5x+5y+2=0.$

Homogeneous part: $2x^{2}+7xy+3y^{2}=(2x+y)(x+3y).$

Assume $(2x+y+c_1)(x+3y+c_2)=0$; expanding, match: $c_1c_2=2,\ 2c_2+c_1=5,\ c_2+3c_1=5.$ Solving: $c_2=2,\ c_1=1$ (check $c_1c_2=2$ ✓). So

$(2x+y+1)(x+3y+2)=0\Rightarrow \boxed{2x+y+1=0},\ \boxed{x+3y+2=0}.$

(b) Solve $\tan^{-1}(x+1)+\tan^{-1}(x-1)=\tan^{-1}\tfrac47$

Combine LHS: $\tan^{-1}\dfrac{(x+1)+(x-1)}{1-(x+1)(x-1)}=\tan^{-1}\dfrac{2x}{2-x^{2}}.$

Equate: $\dfrac{2x}{2-x^{2}}=\dfrac47\Rightarrow14x=8-4x^{2}\Rightarrow2x^{2}+7x-4=0.$

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