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$Q1$: $\frac{2x-7}{x^2-7x+12} = \frac{2x-7}{x^2-7x+10}$

I was taught that when the values of the fractions of two sides are equal and when numerator of the two sides are equal, but denominators are unequal, then the value of the numerator is zero.

So, $2x-7 = 0$

Or, $x=\frac{7}{2}$

$Q2$: $x-4=\frac{x-4}{x}$

If we apply the above-mentioned knowledge in this equation, we get

$x=4$

and at the same time, we miss the solution for $x=1$

Again, I was also taught that if $\frac{a}{b}=\frac{a}{c}$, then $b=c$.

According to this, we get the solution

$x=1$, but we miss $x=4$

Could someone please tell me what am I missing here?

Russell
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1 Answers1

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To expand on my comment, we're trying to understand when $$\frac ab = \frac ac,$$ assuming the fractions make sense. You are saying that your only solution is $a=0$. But this is wrong. Remember that (by "cross-multiplying") $$\frac ab = \frac AB \quad\text{if and only if} \quad aB = Ab.$$ In our case, $$\frac ab = \frac ac \quad\text{if and only if} \quad ac = ab \quad\text{if and only if} \quad a(b-c) = 0.$$ This means that the fractions are equal when EITHER $a=0$ OR $b=c$. In your first example, $b$ cannot be equal to $c$; in your second example, it certainly can ($x=1$).

Ted Shifrin
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