Consider the system of equations $ax+y=b$, $bx+y=a$, $ax+by=ab$, where $ab\in \{0,1,2,3,4\}$, then find no. of pairs for which system is consistent

Question

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?

Answer 1

$ax+y = b$, 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

Answer 2

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.