For which values of P and Q does the equation have infinitely many solutions?

I'm doing some online exercises and I got really confused here.

Can someone explain to me how to make sense of this? Thank you.

For which values of $P$ and $Q$ does the equation have infinitely many solutions? $$73x+P=Qx+71$$

• I'm not sure this really needed 3 identical solutions, but ok – qbert Oct 22 '17 at 0:34

If we take $Q=73$ and $P=71$ your equation reduces to the tautology $$(73-Q)x=71-P\implies 0=0$$ valid for any $x$ (we eliminated any $x$ dependence with our choice of $Q$).

• Hey qbert, I don't quite understand it yet, maybe I missed some part of Algebra, still working on it. Why can we just replace Q with 73 and P with 71? If its better to explain by video I would really appreciate a video link or a topic name I could google that would explain it to me. – Eli S. Oct 22 '17 at 6:03
• @EliS. the question is asking to find values $P,Q$ for which the equation has infinitely many solutions. $Q=73$ and $P=71$ are these values. – qbert Oct 22 '17 at 6:12
• Thank you qbert, English is not my first language and I skip through the lines more often than I should. – Eli S. Oct 22 '17 at 6:13
• @EliS. no big deal. no problem at all. – qbert Oct 22 '17 at 6:15

An algebraic equation has infinitely many solutions if the equation simplifies to a tautology, or something that looks trivially true. An easy way to tackle this problem would be to simply ask yourself, what values $P$ and $Q$ would lead to such a trivially true statement that doesn't yield an answer for $x$?

An easy answer would be to let $P = 71$ and $Q = 73$. Simplifying it yeilds $0 = 0$ which is obviously true and cannot be solved any further.

Rewrite the equation as $(Q-73)x=71-P$, so now you have two cases. If $Q-73 \neq 0$ then $x=\frac{71-P}{Q-73}$ so you have a unique solution.

In the second case you have $Q-73=0$, so the equation becomes $0*x=71-P$. If $P \neq 71$ you have no solution; if $P=71$ you get an infinite number of solutions, as in $x$ satisfy the equation.