Precalculus

M

Maxine Carr

New member
Joined
Jun 14, 2005
Messages
5
I'm having a hard time with this problem: The Board of Governors at the Federal Reserve Bank set the prime interest rate. These rates tickle down through the economy at various rates and affect consumer spending and saving habits. The FR sample households. They found in one household the family investement for 1 year earned $7.50 simple interest. If the principal had been $25 more and the interest rate 1% less, the interest would have been the same. Find the principal and the rate.

Systems of Equations for finding principal and rate


Equation one: t=1
I= 7.50


Equation two: I=7.50
7.50= (p+25)(r-.01)

I do not know what to do now!
 
You fooled yourself. Try a different Equation 1.

p*r = 7.50

Now do you know what to do?
 
Equation 1 is
7.50 = p*r
equation 2 is correct
7.50= (p+25)(r-.01)
You have two equations in two unknowns. Solve for one unknown in Equation 1 and substitute it in Equation 2
 
Gene said:
Equation 1 is
7.50 = p*r
equation 2 is correct
7.50= (p+25)(r-.01)
You have two equations in two unknowns. Solve for one unknown in Equation 1 and substitute it in Equation 2
Click to expand...

okay so

7.50=(p+25)(7.50/p - .01)

7.50 = 7.50p?p^2-.01p+187.50/25p - 2.20p

I do not know what I am doing!
 
Gene said:
Equation 1 is 7.50 = p*r
Equation 2 7.50= (p+25)(r-.01)
You have two equations in two unknowns.
Click to expand...

Your task is to find values for p and r that work simultansously in both equations. One way to do that is to transform the two equations in two variables into a single meaningful equation in only one variable.

The "Substitution Method" is like this:

Pick one and solve for one of the variables

7.50 = p*r
r = 7.50/p

Use this value of 'r' and substitute into the other equation

7.50= (p+25)(r-.01)
7.50= (p+25)((7.50/p)-.01)

The task, then, is to solve this equation for the single remaining variable. You may have to draw on ALL you have learned about manipulating algebraic symbols up to this point.

7.50= (p+25)((7.50/p)-.01)

Solve this equation for 'p'. (You should get p = 125.)

Then use this relationship, r = 7.50/p, to find 'r'.
 
tkhunny said:
Gene said:
Equation 1 is 7.50 = p*r
Equation 2 7.50= (p+25)(r-.01)
You have two equations in two unknowns.
Click to expand...

Your task is to find values for p and r that work simultansously in both equations. One way to do that is to transform the two equations in two variables into a single meaningful equation in only one variable.

The "Substitution Method" is like this:

Pick one and solve for one of the variables

7.50 = p*r
r = 7.50/p

Use this value of 'r' and substitute into the other equation

7.50= (p+25)(r-.01)
7.50= (p+25)((7.50/p)-.01)

The task, then, is to solve this equation for the single remaining variable. You may have to draw on ALL you have learned about manipulating algebraic symbols up to this point.

7.50= (p+25)((7.50/p)-.01)

Solve this equation for 'p'. (You should get p = 125.)

Then use this relationship, r = 7.50/p, to find 'r'.
Click to expand...
 
You are having trouble with the multiplication. Have they taught you about FOIL?
Thatis a a way to remember how to multiply binomials.
(p + 25)((7.50/p) -.01)

F is for first. Take the first from each and multiply them.
p * 7.50/p Hint: there is no no p^2
here.
O is for outside.
p * -.01
I is for inside
25 * 7.75/p
L is for last.
25 * -.01
Then 7.50 = the sum of those four terms.

See what you can do with that
 
7.50= (p+25)((7.50/p)-.01)

I might be tempted tp multiply by 100 and 'p'

7.50*100= (p+25)((7.50/p)-.01) *100
750= (p+25)[((7.50/p)-.01) *100]
750= (p+25)((7.50/p)*100-.01*100)
750= (p+25)((7.50*100/p)-1)
750= (p+25)((750/p)-1)

Since we're pretty sure p != 0

750*p= (p+25)((750/p)-1)*p
750*p= (p+25)[((750/p)-1)*p]
750*p= (p+25)((750/p)*p-1*p)
750*p= (p+25)((750*p/p)-p)
750*p= (p+25)((750*(p/p))-p)
750*p= (p+25)((750*(1))-p)
750*p= (p+25)(750-p)

ONE STEP AT A TIME. DON'T HURRY.
Carefully and deliberately.

Now FOIL

750*p= (p+25)(750-p)
750*p= p*750 - p^2 + 25*750 - 25*p

750*p= p*750 - p^2 + 25*750 - 25*p

That's just begging us to subtract 750*p from both sides.

0 = - p^2 + 25*750 - 25*p

This is looking a lot like a quadratic equation, so let's rearrange it and make it look even more like one.

p^2 + 25*p - 25*750 = 0

Factor

(p - 125)(p + 150) = 0

So, p = 125 or p = -150

Well, we are talking about real money, so p > 0. This discards p = -150 as silly.

p = 125 is all we've left.

OK, so what is 'r'?
 
You must log in or register to reply here.
Top