^{2}-2.5x+1=0

^{2}+3x+0.5=0

1. (3/2x)+(5/x)=(13/6)

2. X^{2}-2.5x+1=0

3. (17/2x+1)=(17/5)

4. 2.5x^{2}+3x+0.5=0

5.(x/x-2)=(7/x-2)

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Kyle S. | Music Theory & Algebra TutorMusic Theory & Algebra Tutor

For number 1, you want to get rid of the fractions without changing the value of your equation. In order to do so, you need to find the LCM (least common multiple) of each denominator. To do so,

1. Write each denominator as a product of prime numbers.

2x = (2)(x). x = (1)(x). 6 = (2)(3).

2. Multiply all unique factors. If a factor repeats, multiply it as many times as it appears in the number in which it appears the most.

All the unique factors are 1, 2, 3, and x. 2 appears twice, but only appears one time with each denominator, so we will only multiply it once. If the last denominator were twelve, we would have two 2s and so would our LCM. So, our LCM is (1)(2)(3)(x) or 6x.

Now, you will multiply everything by 6x. Since both sides are equal, if we multiply them by the same number they remain equal. Remember, when you multiply fractions by non fractions, you only multiply the numerator of the fraction. Now we should have the equation:

(18x/2x) + (30x/x) = (78x/6)

Simplified, it is

(9) + (30) = 13x

Combine like terms:

39 = 13x

divide both sides by 13:

(39/13) = (13x/13)

Simplify:

3 = x

The same process applies for numbers 3 & 5.

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