by Simon
25. August 2010 11:48
If you haven't already, check out the post below this to see the problem I was having. Now, back to me.
My mistake was elementary. With the second term, when distributing the t into the other two, I distributed incorrectly. So we started with this:
(4)+t(t-1)(4)=(t-1)(4t-3) $)
Let's take a look at that second term more closely. First time around I distributed like this:
(4)\not\equiv t^2-t+4t $)
But what I should have done was this:
(4)\Rightarrow (t^2-t)(4)\Rightarrow 4t^2-4t $)
Let's look at it another way and commute
the multiplication in this term. I'll finish the problem after the break.
(4)\equiv 4t(t-1) $)
I can drop the parentheses around the (4) because when it multiplies with the t, it simply becomes 4t. And then, in one step,  $)
distributes to become 
.
So let's finish the problem.
(4)+t(t-1)(4)=(t-1)(4t-3) \\ 16t-12+4t^2-4t=4t^2-7t+3 \qquad\qquad\text{distribute the factors}\\ 4t^2+12t-12=4t^2-7t+3 \qquad\qquad\qquad\qquad\text{simplify each side} \\ 12t-12=-7t+3 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{subtract }4t^2\text{ from both sides} \\ 19t=15 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{add }7t+12\text{ to both sides} \\ t=\frac{15}{19} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{divide by 19 to get } t $)
And that's the answer. It's not a particularly nice answer. But it's an answer nonetheless. Thanks to my maths teachers for helping me understand where I made the mistake. It was easy to get confused with those extra terms and distributing the whole thing. Commuting the 4 and the t to be together simplified the process a great deal.
b3294c41-a826-4aaf-9be0-6cc33d7d18dc|1|5.0
Tags:
Fractions
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