Re: Basic algebra problems

Note that a/(x-b) + b = a/(x - b) + b(x-b)/(x-b) = (a + b(x - b)) / (x - b). Then (a + bx - b^2) / (x - b).

Cross multiply.

(1 - 2x)(x - b) = (3x - 2)(a + bx - b^2)

x - b - 2x^2 + 2bx = 3ax + 3bx^2 - 3b^2 x + 2b^2 - 2a - 2bx

-2x^2 + (2b + 1)x - b = 3bx^2 + (3a - 3b^2 - 2b)x + 2b^2 - 2a

Equate coefficients

(*) -2 = 3b

(**) 2b + 1 = 3a - 3b^2 - 2b

(***) 2b^2 - 2a = -b

(*) => b = -2/3. Plug this into (**) and (***).

(**) => 2(-2/3) + 1 = 3a - 3(-2/3)^2 - 2(-2/3)

=> -4/3 + 1 = 3a - 3(4/9) + 4/3

=> -1/3 = 3a - 4/3 + 4/3

=> -1/3 = 3a

=> a = -1/9

(***) => 2(-2/3)^2 - 2a = -(-2/3)

=> 2(4/9) - 2a = 4/3

=> 8/9 - 2a = 12/9

=> -2a = 12/9 - 8/9

=> -2a = 4/9

=> a = -2/9

We see that a = -1/9 and a = -2/9 which is impossible. There must be some typo.

Re: Basic algebra problems

Hello, deSitter!

There is a typo in the problem . . .

Long division:

. .

We have: .

Divide numerator and denominator by 3:

. . . .

Look at what we have: .

Re: Basic algebra problems

Thanks guys, you have both been very helpful. Any thoughts on the second problem? I'm still unsure what the question is asking. I'm sure it's something simple but the wording has me confused.

Quote:

Find all linear functions f(x) = ax + b such that f(x) = f^{ -1} (x), where f(x) = (x - 3)^{2}(4 - x)(2x + 1)

Re: Basic algebra problems

I think I have made an error with the last question. It should be -

"Find all linear functions f(x) = ax + b such that f(x) = f ^{-1} (x)"

(x-3)^{2}(4-x)(2x+1) was from a previous question and I has assumed it carried over, but this is probably incorrect. I had supposed that f(x) = ax + b was just the form of the linear functions I was to attempt to find. Does anyone have any clues I where I should start with the above question?

PS. yes there was a typo in the denominator, it should have been 3x + 2. Thank you again for you assistance.

Re: Basic algebra problems

Yes, the added "f(x) = (x - 3)^{2}(4 - x)(2x + 1)" makes no sense because that is NOT a linear function.

You are correct that a linear function is of the form y= ax+ b. Now can you find the inverse of that?

Since inverting y= f(x) gives x= f^{-1}(y), start by swapping x and y: x= ay+ b and solve for y. Then use f^{-1}(x)= f(x) to solve for a and b.

(You cannot solve for a and b as specific numbers because there are an infinite number of such functions.)

Re: Basic algebra problems

Remember, given f(x) we can determine f^(-1) (x) by the following steps

1. Set y = f(x).

2. Interchange x and y.

3. Solve for new y.

4. Set new y = f^(-1) (x)

Using these steps will yied f^(-1) (x) = x/a - b/a

Set f(x) = f^(-1) (x).

Then ax + b = x/a - b/a

Now equate coefficients. Then a = 1/a and b = -a/b. Then a^2 = 1 and b^2 = -a. Since a^2 = 1, we see that either a = 1 or a = -1. If a = 1, then -a = -1. Then b^2 = -1 which yields no real solution for b. Thus, a = -1. Then b^2 = -(-1) = 1. Then b = 1 or b = -1.

Solution: f(x) = -x + 1 and f(x) = -x - 1.

Re: Basic algebra problems

EDIT: Cross multiply should then be (1 - 2x)(x - b) = (3x + 2)(a + bx - b^2)

x - b - 2x^2 + 2bx = 3ax + 3bx^2 - 3b^2 x + 2a + 2bx - 2b^2

3ax + 3bx^2 - 3b^2 x + 2a + 2bx - 2b^2 - x + b + 2x^2 - 2bx = 0

x^2(3b + 2) + x(3a - 3b^2 + 2b - 1 - 2b) + (2a - 2b^2 + b) = 0

Then 3b + 2 = 0 and 3a - 3b^2 - 1 = 0 and 2a - 2b^2 + b = 0

3b + 2 = 0 => b = -2/3

3a - 3(-2/3)^2 - 1 = 0 and 2a - 2(-2/3)^2 + (-2/3) = 0

3a - 3(4/9) - 1 = 0 and 2a - 2(4/9) + -2/3 = 0

3a - 12/9 - 9/9 = 0 and 2a - 8/9 - 6/9 = 0

3a = 21/9 and 2a = 14/9

a = 21/27 and a = 14/18

a = 7/9

Answer: a = 7/9 and b = -2/3

Re: Basic algebra problems

Thank you all for your assistance. You have been very helpful.

mathguy25, you equate b = -a/b, would this not be b = -b/a?

Which leads to -

a^{2} = 1

a = 1

b = -b/a

ba = -b

if a = 1, b = -b, which makes no sense (unless I guess b = 0), therefore a = -1 (I don't know if I can just assume this).

ax + b = (x - b)/a

a^{2}x + ba + b = x

(-1)^{2}x + b(-1) + b = x

b = 1

Solution f(x) = -x + 1

Re: Basic algebra problems

Yes, I made a mistake there. It should be b = -b/a. Remember a = +/- 1. Thus,

a = +/- 1 and b = -b/a

a = +/- 1 and ab = -b

a = +/- 1 and ab + b = 0

a = +/- 1 and b(a + 1) = 0

a = +/- 1 and b = 0 or a + 1 = 0

If b = 0, then a = +/- 1 so f(x) = x and f(x) = -x would be solution.

If b =/= 0, then a = -1. Then f(x) = -x + b is a solution for any number b =/= 0.

Thus, either f(x) = x or f(x) = -x + b for any b (since f(x) = -x is the case where b = 0).

Soln: {f(x) = x, f(x) = -x + b where b is any real number}

Re: Basic algebra problems

Great stuff, thanks again.