Hii!

I have this other piece of practice work that I don't understand.

The method I do understand is for calculating the final amount:

The interest is 8.5 %
The years are 7

I know that you must do:

$P+ (i . P . n )$
= $21 860+ (0.085 . 21 860 . 7)$

That's easy.

What I'm stuck on is:

Final amount is 8 775
Interest is 7%
Years are 5
CALCULATE THE AMOUNT YOU STARTED WITH.

What would I do?

Thanks!

2. ## Re: Grade 8 Algebra

Go backwards. $P+i\cdot{P}\cdot{n}=F$, where F is the final amount.

Substitute the known values into the equation and try to solve for the unknown.

3. ## Re: Grade 8 Algebra

Originally Posted by Quacky
Go backwards. $P+i\cdot{P}\cdot{n}=F$, where F is the final amount.

Substitute the known values into the equation and try to solve for the unknown.
I don't quite understand.

If the final amount is 8775 ( $A$)
Interest is 0.07 ( $i$)
And years 5 ( $n$)

How would I substitute?

Would I go:
$8775- (0.07/ 8775 . 5)$
= (amount we started with which is $P$)

Like that?

4. ## Re: Grade 8 Algebra

$P+i\cdot{P}\cdot{n}=A$

We have:
$i=0.07$
$n=5$
$A=8775$

So:

$P+0.07\cdot{P}\cdot{5}=8775$

So:

$P+0.35\cdot{P}=8775$

$1.35P=8775$

5. ## Re: Grade 8 Algebra

So 1.35 is the amount we started with?

6. ## Re: Grade 8 Algebra

No, we have to solve this equation for P, by dividing both sides by 1.35

7. ## Re: Grade 8 Algebra

But we haven't solved P.

We have A and i and n
We are trying to solve P.

8. ## Re: Grade 8 Algebra

Originally Posted by JesseElFantasma
But we haven't solved P.

We have A and i and n
We are trying to solve P.
I agree. In order to solve an equation for P, you need to isolate P (get P by itself). This is what I've done here.

Originally Posted by Quacky
$P+i\cdot{P}\cdot{n}=A$

We have:
$i=0.07$
$n=5$
$A=8775$

So:

$P+0.07\cdot{P}\cdot{5}=8775$ Here I've just substituted the known values into the formula.

So:

$P+0.35\cdot{P}=8775$ I've simplified, but otherwise haven't changed anything.

$1.35P=8775$ I've collected like terms together. In the step above, I had 1P and 0.35P, so these combine to make 1.35P
Then, if we divide both sides by $1.35$, we get:

$P=\frac{8775}{1.35}=6500$

9. ## Re: Grade 8 Algebra

But in
P + 0.35 x P = 8775
what happens to the 1st P?

10. ## Re: Grade 8 Algebra

The first is $1\times{P}$
The second is $0.35\times{P}$

Because they are both amounts of P that are being added together, we can combine them as $1.35P$ in the same way that $3+3+3+3= 4\times{3}$

11. ## Re: Grade 8 Algebra

I get it!
Thank you.

So for another I would go:

P + 0.085 . P.7
P + 0.595 x P
P + 1.595
1.595P = 34866.70
34866.70/ 1.595P
= 21860

12. ## Re: Grade 8 Algebra

If $34866.70$ is your final amount, and the question is asking you to solve for P, then yes - although there are some very tiny notation errors, the method is fine. I'd set it out like this:

$P + 0.085 \cdot{P}\cdot{7}=34866.70$

$P + 0.595 \times P=34866.70$

$1.595P = 34866.70$

$34866.70/ 1.595=P$

$= 21860$