CAT 2024 Slot 2 · Quant · Question 21

One equation in two unknowns normally has infinitely many solutions, so the fact that a unique value is asked for signals a sum of perfect squares.

The terms 4{x}^{2}-4xy suggest completing {\left( 2x-y \right)}^{2}=4{x}^{2}-4xy+{y}^{2}.

Pulling that out uses up one of the four {y}^{2} terms, leaving {\left( 2x-y \right)}^{2}+3{y}^{2}-6y+3=0.

The remainder factors as 3\left({y}^{2}-2y+1 \right)=3{\left( y-1 \right)}^{2}.

So the equation is {\left( 2x-y \right)}^{2}+3{\left( y-1 \right)}^{2}=0.

A sum of two non-negative quantities is zero only when each is zero.

From {\left( y-1 \right)}^{2}=0, y=1.

From {\left( 2x-y \right)}^{2}=0, 2x=y=1, so x=\frac{1}{2}.

Therefore 4x+5y=4\times \frac{1}{2}+5\times 1=2+5=7.

Shortcut. Treat the equation as a quadratic in x: 4{x}^{2}-4xy+\left( 4{y}^{2}-6y+3 \right)=0. For a real x the discriminant must be non-negative, giving 16{y}^{2}-16\left( 4{y}^{2}-6y+3 \right)\ge 0, which simplifies to {\left( y-1 \right)}^{2}\le 0 and forces y=1 immediately.