Solving D S.'s system of equation

Note that $24x^3-10x^2y-3xy^2+y^3=(2x-y)(4x-y)(3x+y)$ .

Hence there are three cases to consider:

Case 1 : $y=2x$

Substituting into $x^2+5x=y^2$ , we get $x^2+5x=4x^2 \Rightarrow x(3x-5)=0$ Hence $x=0$ or $\frac {5}{3}$ . The corresponding $y$ values are $0$ and $\frac {10}{3}$ respectively.

Case 2 : $y=4x$

Substituting into $x^2+5x=y^2$ , we get $x^2+5x=16x^2 \Rightarrow x(15x-5)=0$ Hence $x=0$ or $\frac {1}{3}$ . The corresponding $y$ values are $0$ and $\frac {4}{3}$ respectively.

Case 3 : $y=-3x$

Substituting into $x^2+5x=y^2$ , we get $x^2+5x=9x^2 \Rightarrow x(8x-5)=0$ Hence $x=0$ or $\frac {5}{8}$ . The corresponding $y$ values are $0$ and $-\frac {15}{8}$ respectively.

Hence, the solutions are $(0,0)$ , $(\frac {5}{3},\frac {10}{3})$ , $(\frac {1}{3},\frac {4}{3})$ and $(\frac {5}{8},-\frac {15}{8})$ .

The solution with the largest value of $x$ is $(\frac {5}{3},\frac {10}{3})$ , so $a+b=\frac {5}{3}+\frac {10}{3}=5$ .

Let $y=ax$ for some non zero real number $a$ . Our equations transform to: $x^2+5x=a^2x^2 \Rightarrow x+5=a^2x$ and $24x^3-10x^3a-3x^3a^2+a^3x^3=0$ $a^3-3a^2-10a+24=(a-2)(a^2-a-12)=(a-2)(a-4)(a+3)=0$

We know that : $x=\frac{5}{a^2-1}$ Clearly, $x$ is maximised when $a=2$ $\Rightarrow x=\frac{5}{3}$ .

Thus, $x+y=3x=\boxed{5}$ .

Note: We can divide by $x$ in all the equations in which I have divided with $x$ , since $x \neq 0$ .

The first equation can be factorised as $(2x-y)(4x-y)(3x+y)=0$ . This gives us 3 cases:

Case 1: $2x=y$ Using this substitution in the second equation gives us $x^2+5x=4x^2$ . This can be rearranged to give us $3x^2-5x=0$ , which gives us $x=0, \frac{5}{3}$

Case 2: $4x=y$ Using this substitution in the second equation gives us $x^2+5x=16x^2$ . This can be rearranged to give us $15x^2-5x=0$ , which gives us $x=0, \frac{5}{15}$ or simplified to $\frac{1}{3}$ .

Case 3: $3x=-y$ Using this substitution in the second equation gives us $x^2+5x=9x^2$ . This can be rearranged to give us $8x^2-5x=0$ , which gives us $x=0, \frac{5}{8}$ .

From these 3 cases, we have 4 possible values of $x: 0, \frac{5}{3}, \frac{1}{3}, \frac{5}{8}$ . The first case gives us the largest value of $x$ , namely $x=\frac{5}{3}$ . From here, we have $y=2x=2*\frac{5}{3}=\frac{10}{3}$ and thus the solution $(x,y)=(\frac{5}{3},\frac{10}{3})$

from 24x^3-10x^2y-3xy^2+y^3 = 0 we get (2x-y)(4x-y)(3x+y)=0 it's easily to see that x is max when 2x=y so x=5/3 and y=10/3 so a+b=5

Divide equation 1 by y^3 it becomes a cubic equation with roots 1/2, 1/4, -1/3 now for these ratios of x/y solve equation 2