How would I clear the fractions in the equation in the attached picture, in order to get the following?:

4(x+3)+4x(x−2) = 9(x−2)(x+1)

4x+12+4x^2−8x = 9x^2−9x−18

How would I clear the fractions in the equation in the attached picture in order to get the following 4x34xx2 9x2x1 4x124x28x 9x29x18 class=

Respuesta :

Answer:

  You clear fractions by the usual method: multiply by the least common denominator.

Step-by-step explanation:

The denominators of the three fractions are ...

  • x² -x -2 = (x -2)(x +1)
  • x +1
  • 4

The "least common denominator" is the least common multiple of these expressions, which will be the product that contains all of the unique factors: 4(x -2)(x +1). This is what the equation is multiplied by in order to clear fractions.

[tex]\dfrac{x+3}{x^2-x-2}+\dfrac{x}{x+1}=\dfrac{9}{4}\\\\\dfrac{(x+3)(4)(x-2)(x+1)}{(x-2)(x+1)}+\dfrac{x(4)(x-2)(x+1)}{(x+1)}=\dfrac{9(4)(x-2)(x+1)}{4}\\\\4(x+3)+4x(x-2)=9(x-2)(x+1) \qquad\text{your first form with fractions cleared}\\\\4x+12+4x^2-8x=9x^2-9x-18 \qquad\text{eliminate parentheses}\\\\5x^2-5x-30=0 \qquad\text{subtract the terms on the left}\\\\x^2-x-6=0 \qquad\text{divide by 5}\\\\(x-3)(x+2)=0 \qquad\text{factor}[/tex]

The solutions are x=-2, x=3.

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When we say "multiply" or "subtract" or "divide", we mean do it to both sides of the equation. Anything you do to an equation must be done to both sides in order to maintain the truth of the equal sign.

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