Looking at the first two equations, my thought was that set needs to be symmetric in order for it to be satisfied. Then that gives us 5 ordered pairs. But the answer is 6. How do I incorporate the third equation into this?
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1Do you mean $a, b \in {0,1,2,3,4}$, i.e. $a$ and $b$ are both in that set, or $a$ and $b$ are real numbers whose product is in that set? – Robert Israel Apr 17 '20 at 16:52
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@RobertIsrael I think it means that both a and b belong to the set – Aditya Apr 17 '20 at 18:16
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Subtracting the first two equations gives you $(a-b)x = b-a$, so either $x=-1$ (and then $y = a+b$) or $a=b$ (and then $y = a(1-x)$).
In the first case, the third equation becomes $b^2 = a$.
In the second case, the third equation becomes $a (1-a) x = 0$, so $x=0$ always works.
Thus, assuming you meant $a, b \in \{0,1,2,3,4\}$, the cases corresponding to the first case are $(a,b) = \{(0,0), (1,1), (4,2)\}$; the cases corresponding to the second case are $\{(0,0),(1,1),(2,2),(3,3),(4,4)\}$. That's a total of $6$ ordered pairs.
Robert Israel
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$ax+y = b$, then $x = \frac{b-y}{a}$
$bx+y = a$, $b\frac{b-y}{a}+y = a$
$b^2-by+ay = a^2$, $b^2+y(a-b) = a^2$
$ax+by = ab$, $a\frac{b-y}{a}+by = ab$ $b-y+by = ab$, then $y = \frac{ab-b}{b-1}$
$b^2+(a-b)\cdot{b}\cdot\frac{a-1}{b-1} = a^2$, simply further to get $(b-a)\cdot(b^2-a) = 0$
Meaning $a=b$, and $a=b^2$ but $a,b \in {0,1,2,3,4}$
$(a,b) = ${$(0,0),(1,1),(2,2),(2,4),(3,3),(4,4)$} Which are 6 ordered pairs
