Multiplication and Division Algebra

Jason76

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No adding or subtracting on this one.

\(\displaystyle 5x = 10x^{2}\)

\(\displaystyle 5 = \dfrac{10x^{2}}{x}\)

\(\displaystyle 5 = 10x\)

\(\displaystyle x = \dfrac{1}{2}\) :confused: Right strategy?
 
No adding or subtracting on this one.

\(\displaystyle 5x = 10x^{2}\)

\(\displaystyle 5 = \dfrac{10x^{2}}{x}\)
What if x= 0? You can't divide by 0!

\(\displaystyle 5 = 10x\)

\(\displaystyle x = \dfrac{1}{2}\) :confused: Right strategy?
Yes, IF you can answer my question!

Or you could write it as \(\displaystyle 10x^2- 5x= 5x(2x- 1)= 0\).
That's a quadratic equation. Do you see that there are two values of x that satisfy it?
 
What if x= 0? You can't divide by 0!


Yes, IF you can answer my question!

Or you could write it as \(\displaystyle 10x^2- 5x= 5x(2x- 1)= 0\).
That's a quadratic equation. Do you see that there are two values of x that satisfy it?

How algebraically did you get from \(\displaystyle 5x = 10x^{2}\) to \(\displaystyle 10x^{2} - 5x = 0\) ???
 
Last edited:
How algebraically did you get from \(\displaystyle 5x = 10x^{2}\) to \(\displaystyle 10x^{2} - 5x = 0\) ???

\(\displaystyle 5x = 10x^2\)

\(\displaystyle 10x^2 = 5x\)

\(\displaystyle 10x^2 - 5x = 5x - 5x \)

\(\displaystyle 10x^2 - 5x = 0 \)
 
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