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How to solve these equations? $$\begin{cases} 3-(y+1)^2 = \sqrt{x-y}\\ x+8y = \sqrt{x-y-9} \end{cases}$$

I've tried solving this using the substitution and elimination methods without any success. I also tried plotting these equations and I got $x = 8$, $y = -1$.

Can someone show me the steps required to solve this?

Thanks in advance.

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  • $\begingroup$ You can use the Rational Root Theorem to find the solutions $x = 8$ and $y = -1$. But is the question asking to solve over the Reals or Rationals? $\endgroup$ – Hourglass Aug 24 '20 at 12:45
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Guess and check is a valid method of solving equations. Notice that the domain of square root necessitates that

$$x-y \geq 9$$

so what happens if $x-y = 9$ ?

$$\begin{cases}3-(y+1)^2 = 3 \\ x+8y = 0\end{cases}$$

Can you take it from here?

Notice that the first equation also mandates that $-1-\sqrt{3} \leq y \leq -1+\sqrt{3}$ because the outcome of a square root always needs to be positive. Can you use this to prove there can't be any other solutions?

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Since $x-y\geq9,$ we obtain: $$3-(y+1)^2=\sqrt{x-y}\geq3,$$ which gives $$y=-1,$$ $$x-y=9$$ and $$x=8.$$ Now, it's enough to check that the second equation holds for these values of $x$ and $y$.

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Observing the right-hand sides, we get: $$(x+8y)^2 + 9 = (3 - (y+1)^2 )^2$$ $$(x+8y)^2 -(3 - (y+1)^2 )^2 = -9$$

and now using the difference of two squares: $$(x+8y+3-(y+1)^2)(x+8y-3 + (y+1)^2 )= -9$$

If there is a clean solution where $x, y$ are integers, then the two brackets must be integers themselves. There are only a few possibilities: which are: $$(\text{left}, \text{right}) = (-1, 9), (1, -9), (-3, 3), (3, -3), (-9, 1), (9, -1).$$

Some of these solutions are extraneous or have non-integer solutions (that can be expressed in radicals). With the pair $(3, -3)$, you get an integer solution $(x,y) = (1,8)$, and now substitute into the original equations to verify them.

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  • $\begingroup$ Are you using the Rational Root Theorem here? $\endgroup$ – ThermalRaindrops62 Aug 24 '20 at 13:46
  • $\begingroup$ No, the rational root theorem cannot be used here because we have polynomials in two variables. $\endgroup$ – Toby Mak Aug 24 '20 at 13:55

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