Please help *Algebra*
- Thread starter Anifex
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You should express this as:
a = w/(kv)
This way we are sure both the factors \(k\) and \(v\) are in the denominator on the RHS...
or:
[MATH]a=\frac{w}{kv}[/MATH]
but if I am interpreting your result correctly, then you're right. Why do you think it's wrong?
a = w/(kv)
This way we are sure both the factors \(k\) and \(v\) are in the denominator on the RHS...
or:
[MATH]a=\frac{w}{kv}[/MATH]
but if I am interpreting your result correctly, then you're right. Why do you think it's wrong?
Dr.Peterson
Elite Member
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- Nov 12, 2017
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To solve for a, you need to undo the multiplication by k. You can do that, as has been said, by dividing both sides by k. But division by k is the same as multiplication by 1/k. So you get
ak * 1/k = w/v * 1/k
where "*" means multiplication. (That's how we type math when we don't want to bother with fancy stuff. ...)
And when you carry out those multiplications, you get
a = w/(vk)
It seems to me like the presence of so many variables is what's confusing you. Let's dispense with as many of them as we can. What if we replace \(k\) by 2, \(w\) by 3, and \(v\) by 4. Then you'd have:
\(\displaystyle 2a = \frac{3}{4}\)
How would you solve this equation for \(a\)? Well, you'd divide both sides of the equation by 2, so as to isolate \(a\) on one side:
\(\displaystyle a = \frac{3}{4} \div 2\)
Now, can we clean that up a bit? Dividing by 2 is the same as multiplying by \(\frac{1}{2}\) (why?):
\(\displaystyle a = \frac{3}{4} \times \frac{1}{2} = \frac{3 \times 1}{4 \times 2} = \frac{3}{8}\)
Let's do it again, still replacing \(w\) by 3, and \(v\) by 4, but this time replacing \(k\) by 7:
\(\displaystyle 7a = \frac{3}{4}\)
How would you solve this equation for \(a\)? Well, you'd divide both sides of the equation by 7 (i.e. multiply by \(\frac{1}{7}\)), so as to isolate \(a\) on one side:
\(\displaystyle a = \frac{3}{4} \times \frac{1}{7} = \frac{3 \times 1}{4 \times 7} = \frac{3}{28}\)
That seems like it should be sufficient to get the core concept down, so now let's re-introduce our variables, one at a time. First, re-introduce \(v\):
\(\displaystyle a = \frac{3}{v} \times \frac{1}{7} = \frac{3 \times 1}{v \times 7} = \frac{3}{7v}\)
Next, re-introduce \(w\):
\(\displaystyle a = \frac{w}{v} \times \frac{1}{7} = \frac{w \times 1}{v \times 7} = \frac{w}{7v}\)
And finally, re-introduce \(k\):
\(\displaystyle a = \frac{w}{v} \times \frac{1}{k} = \frac{w \times 1}{v \times k} = \frac{w}{kv}\)
In each of the two cases with "real" values, you simply divided by the value of \(k\) which undid the multiplication. Any reason to believe the exact same process wouldn't apply here? You may not know the exact value of \(k\), but you can still work with it all the same.
Thank you everyone who replied, it all just clicked for me. Very much appreciated!