<p>You're right, you can multiply both sides by <span class="math-container">$q^2$</span> and you get <span class="math-container">$-100+q^2$</span> on the left-hand side and <span class="math-container">$0 \cdot q^2 = 0$</span> on the right.</p>
<p>Another way.</p>
<p><span class="math-container">$$-100/q^2 + 1 = 0 \iff 1 = 100/q^2 \iff q^2 = 100 \iff q^2 - 100 = 0$$</span></p>

CC BY-SA 4.0
2023-04-13T18:44:49.993Z
2023-04-13T18:44:49.993Z
Ifartforaliving
<p>What you actually do is multiply both sides of the equation by <span class="math-container">$q^2$</span>.</p>
<p>This works because <span class="math-container">$0 \cdot q^2 = 0$</span>.</p>

CC BY-SA 4.0
2023-04-13T18:38:49.273Z
2023-04-13T18:38:49.273Z
Nicolas Torres
<p>This may be a simple question, but how is <a href="https://i.sstatic.net/tLkjs.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/tLkjs.png" alt="enter image description here" /></a> simplified to <a href="https://i.sstatic.net/ozq3D.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ozq3D.png" alt="enter image description here" /></a></p>
<p>With the rules of simplification, do both the terms on the left hand side get multiplied by q^2?</p>

<mathematical-economics>
Basic simplifcation
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# Basic simplifcation

(Asked by: username_0 on 2023-04-13)

This may be a simple question, but how is [![enter image description here](https://i.sstatic.net/tLkjs.png)](https://i.sstatic.net/tLkjs.png) simplified to [![enter image description here](https://i.sstatic.net/ozq3D.png)](https://i.sstatic.net/ozq3D.png)

With the rules of simplification, do both the terms on the left hand side get multiplied by q^2?

**Tags:** mathematical-economics


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## 2 answers

### Answer by username_1 on 2023-04-13. Score: 1 | Accepted answer

You're right, you can multiply both sides by $q^2$ and you get $-100+q^2$ on the left-hand side and $0 \cdot q^2 = 0$ on the right.

Another way.

$$-100/q^2 + 1 = 0 \iff 1 = 100/q^2 \iff q^2 = 100 \iff q^2 - 100 = 0$$


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### Answer by username_2 on 2023-04-13. Score: 1

What you actually do is multiply both sides of the equation by $q^2$.

This works because $0 \cdot q^2 = 0$.


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