$|ax - 11| = 4x - 10$ has a positive integral solution, what is $x, a$?

Question

I was given $$|ax - 11| = 4x - 10$$ has a positive integral solution and $a$ is a positive integer.

I was asked what was $x, a$

$$ax > 11 $$ then we have $x=\frac{1}{a-4}$ $$ax < 11$$ then we have $x=\frac{22}{a+4}$

What should I do now? I don't understand what does the "one solution" mean because all I know there are some possibilities for $x,a$ to be the solutions. The answer keys are $x, a =3$

Answer 1

safer to split into cases and confirm with the original item.

We can have the thing in the absolute value either positive or negative,

(I) positive $$ ax-11 = 4x-10 $$ $$ (a-4)x = 1 $$ Since we demand positive integers, we get $a=5, x=1.$ then check: $$ |ax-11| = |5 - 11| = |-6| = 6, $$ while $$ 4x-10 = -6 $$ So this type failed we can't have $ax-11$ positive

(II) negative $$ 11 - ax = 4x-10 $$ $$ 21 = (a+4)x $$ we can't have $a+4$ come out either $1$ or $3,$ the remaining choices are: $a+4 = 7,$ $x=3,$ so $a = 3, x = 3.$ OR, $a+4 = 21,$ $x=1,$ so $a=17, x=1.$ Checking: $a=3,x=3,$ $$ |ax-11| = |9 - 11| = |-2| = 2, $$ while $$ 4x-10 = 12 - 10 = 2. $$ This one works.

Finally $a=17,x=1,$ $$ |ax-11| = |17 - 11| = 6, $$ while $$ 4x-10 = 8 - 10 = -2. $$ This one fails.

Answer 2

I think you made a little error because you had to have $x=\frac{21}{a+4}$ when $ax<11$. Hence $(x,a)=(3,3)$.