Question

Find the solution set of each equation. If the equation has no solution write 0/ --- If the equation is an identity state so.

w^2-7w=0

Answer 1

In order to solve: \(w^2-7w=0\)
we will make use of a proposition called "common factor" which can be called a reverse step from the distributive axiom.
Now, common factor is pretty much dividing and multiplying the whole thing by the variable in question, o by associative:
\[w^2-7w=0 \iff (w^2-7w)=0\]
Notice that there is an axiom called the "Neutral of multiplication" it states that there exists a number which multiplied by any quantity, gives as a result, the quantity it multiplies more mathematically written: \(\exists "1" \in \mathbb{R} / 1.a=a\)
so, applying it:
\[(w^2-7w)=0 \iff 1.(w^2-7w)=0\]
\[\frac{ w }{ w } =1\]
So, combining these two statements, we get:
\[\frac{ w }{ w }(w^2-7w)=0\]
Now it's just a matter of operations:
\[w(\frac{ w^2 }{ w }-\frac{ 7w }{ w })=0\]
\[\iff w(w-7)=0\]
Now, from this last step I'll let you continue, try using hankel's property which is as follows: \[a.b=0 \iff a=0; b=0\]

Answer 2

Tbh, I don't even know what Hankel's Property is. This question just has me completely turned upside down

Answer 3

So, beginning from
\[w(w-7)=0\]
Hankel's property is also called the "zero product rule" by some teachers, which I do not really support.
Anywho, if we apply hankel's peroperty, we will have two cases in order for the equality \(w(w-7)=0\) to be satisfied, this being either \(w=0\) is equal to zero or \(w-7=0\)
So, we have now the following separate equalities:
\[w=0\]
\[w-7=0 \]
Which I believe you can solve.

Answer 4

w=7?

Answer 5

Good, that is a solution indeed.

Answer 6

When I put that in my answer system it says it's wrong

Answer 7

Well, it must say it's rong because there is not only one solution, one of them is as you stated w=7.
Which is the second solution?