Pre-calculus help please- algebra worksheet problem

Find the solution set of 5/x-3 < 3/1-x in interval notation.

If anyone could please show me how to solve this problem i would really apreciate it <3

5/(x - 3) < 3/(1 - x)

It helps to first state the domain of the two functions. No denominator can equal 0, ever. So:

5/(x - 3) < 3/(1 - x) | x ≠ 1 and x ≠ 3

Multiply both sides by (1 - x)(x - 3). The constraints remain.

5(1 - x) < 3(x - 3) | x ≠ 1 and x ≠ 3

5 - 5x < 3x - 3 | x ≠ 1 and x ≠ 3

-8x < -8 | x ≠ 1 and x ≠ 3

8x > 8 | x ≠ 1 and x ≠ 3 ... The rule of opposites for inequalities.

x > 1 | x ≠ 1 and x ≠ 3

x > 1 | x ≠ 3 ... Non-redundance.

The variable x can equal anything greater than 1 except for 3. The answer will use nothing but open intervals because the bounds 1 and 3 are not included.

5/(x - 3) < 3/(1 - x)

x ϵ (1, 3)U(3, ∞)

"Variable x is an element of the set of all values between 1 and 3 in union with the set of all values between 3 and infinity."

I know what I did wrong! The sixth line should read 5 - 5x < 3x - 9 | x ≠ 1 and x ≠ 3 .

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if x>3 then: 5(1-x)>3(x-3) which yields x<14/8 : contradiction
if 114/8 or x>1.75 which yields the interval (1.75,3)
if x<1 then: 5(1-x)>3(x-3) which yields x<14/8 considering both of them, x<1 holds:

solution set interval is (-infinity,1) and (1.75,3)