A = P + Prt
Solve for P
It would go, according to what I see
(1):
A = P + Prt
(2):
A = P(1+rt)
And that's where I get lost. Wouldn't combining P and P be 2P at that point, so where did the second P go and where did this 1 come from and why is it being added to rt and then multiplied by the P? Can someone show exactly what's happening when we go from (1) to (2) so I may better understand?
Solve for P
It would go, according to what I see
(1):
A = P + Prt
(2):
A = P(1+rt)
And that's where I get lost. Wouldn't combining P and P be 2P at that point, so where did the second P go and where did this 1 come from and why is it being added to rt and then multiplied by the P? Can someone show exactly what's happening when we go from (1) to (2) so I may better understand?
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Going from step 1 to step 2 uses the distributive property of multiplication over addition. That is, you're factoring a P out of both terms, P and Prt.
The distributive property goes something like this:
ab + ac = a(b+c)
You can see that this works if you substitute numbers for a, b, and c.
If a = 2, b = 3, c = 4, for example:
2*3 + 2*4 = 2*(3+4)
6 + 8 = 2*7
14 = 14
So, you're just taknig something of the form ab + ac and putting it into the form a(b+c).
In the case of P + Prt, let a = P, b = 1, and c = rt.
P + Prt =
P * 1 + P * rt =
P * (1 + rt)
And of course, writing the multiplication symbol is optional, so this is the same as P(1+rt).
The distributive property goes something like this:
ab + ac = a(b+c)
You can see that this works if you substitute numbers for a, b, and c.
If a = 2, b = 3, c = 4, for example:
2*3 + 2*4 = 2*(3+4)
6 + 8 = 2*7
14 = 14
So, you're just taknig something of the form ab + ac and putting it into the form a(b+c).
In the case of P + Prt, let a = P, b = 1, and c = rt.
P + Prt =
P * 1 + P * rt =
P * (1 + rt)
And of course, writing the multiplication symbol is optional, so this is the same as P(1+rt).