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# Solving Rational Equations

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$${1\over x(x-4)}+{5-2x\over x^2-3x-4}={5\over x(x+1)}$$

Solve this rational equation.

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#1
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Solve for x:
1/(x (x - 4)) + (5 - 2 x)/(x^2 - 3 x - 4) = 5/(x (x + 1))

Hint: | Write the left hand side as a single fraction.
Bring 1/((x - 4) x) + (5 - 2 x)/(x^2 - 3 x - 4) together using the common denominator x (x - 4) (x + 1):
(-2 x^2 + 6 x + 1)/(x (x - 4) (x + 1)) = 5/(x (x + 1))

Hint: | Multiply both sides by a polynomial to clear fractions.
Cross multiply:
x (x + 1) (-2 x^2 + 6 x + 1) = 5 x (x - 4) (x + 1)

Hint: | Write the quartic polynomial on the left hand side in standard form.
Expand out terms of the left hand side:
-2 x^4 + 4 x^3 + 7 x^2 + x = 5 x (x - 4) (x + 1)

Hint: | Write the cubic polynomial on the right hand side in standard form.
Expand out terms of the right hand side:
-2 x^4 + 4 x^3 + 7 x^2 + x = 5 x^3 - 15 x^2 - 20 x

Hint: | Move everything to the left hand side.
Subtract 5 x^3 - 15 x^2 - 20 x from both sides:
-2 x^4 - x^3 + 22 x^2 + 21 x = 0

Hint: | Factor the left hand side.
The left hand side factors into a product with five terms:
-x (x + 1) (x + 3) (2 x - 7) = 0

Hint: | Multiply both sides by a constant to simplify the equation.
Multiply both sides by -1:
x (x + 1) (x + 3) (2 x - 7) = 0

Hint: | Find the roots of each term in the product separately.
Split into four equations:
x = 0 or x + 1 = 0 or x + 3 = 0 or 2 x - 7 = 0

Hint: | Look at the second equation: Solve for x.
Subtract 1 from both sides:
x = 0 or x = -1 or x + 3 = 0 or 2 x - 7 = 0

Hint: | Look at the third equation: Solve for x.
Subtract 3 from both sides:
x = 0 or x = -1 or x = -3 or 2 x - 7 = 0

Hint: | Look at the fourth equation: Isolate terms with x to the left hand side.
x = 0 or x = -1 or x = -3 or 2 x = 7

Hint: | Solve for x.
Divide both sides by 2:
x = 0 or x = -1 or x = -3 or x = 7/2

Hint: | Now test that these solutions are correct by substituting into the original equation.
Check the solution x = -3.
1/((x - 4) x) + (5 - 2 x)/(x^2 - 3 x - 4) ⇒ (5 - 2 (-3))/(-4 - 3 (-3) + (-3)^2) + 1/((-4 - 3) (-3)) = 5/6
5/(x (x + 1)) ⇒ -5/(3 (1 - 3)) = 5/6:
So this solution is correct

Hint: | Check the solution x = -1.
1/((x - 4) x) + (5 - 2 x)/(x^2 - 3 x - 4) ⇒ (5 - 2 (-1))/(-4 - 3 (-1) + (-1)^2) + 1/((-4 - 1) (-1)) = ∞^~
5/(x (x + 1)) ⇒ -5/(1 - 1) = ∞^~:
So this solution is incorrect

Hint: | Check the solution x = 0.
1/((x - 4) x) + (5 - 2 x)/(x^2 - 3 x - 4) ⇒ 1/((0 - 4) 0) + (5 - 2 0)/(-4 - 3 0 + 0^2) = ∞^~
5/(x (x + 1)) ⇒ 5/(0 (1 + 0)) = ∞^~:
So this solution is incorrect

Hint: | Check the solution x = 7/2.
1/((x - 4) x) + (5 - 2 x)/(x^2 - 3 x - 4) ⇒ 1/(1/2 (7/2 - 4) 7) + (5 - (2 7)/2)/(-4 - (3 7)/2 + (7/2)^2) = 20/63
5/(x (x + 1)) ⇒ 5/(7/2 (1 + 7/2)) = 20/63:
So this solution is correct

Hint: | Gather any correct solutions.
The solutions are:
x = -3          or          x = 7/2

Guest Sep 13, 2017
#2
+5905
+1

$$\frac{1}{x(x-4)}+\frac{5-2x}{x^2-3x-4}=\frac{5}{x(x+1)}$$

Factor the  x2 - 3x - 4 . What two numbers add to  -3  and multiply to  -4 ?       +1  and  -4  .

$$\frac{1}{x(x-4)}+\frac{5-2x}{(x+1)(x-4)}=\frac{5}{x(x+1)}$$

Multiply the first fraction by  $$\frac{x+1}{x+1}$$ .

$$\frac{1(x+1)}{x(x-4)(x+1)}+\frac{5-2x}{(x+1)(x-4)}=\frac{5}{x(x+1)}$$

Multiply the middle fraction by  $$\frac{x}{x}$$ .

$$\frac{1(x+1)}{x(x-4)(x+1)}+\frac{(5-2x)x}{(x+1)(x-4)x}=\frac{5}{x(x+1)}$$

Multiply the last fraction by  $$\frac{x-4}{x-4}$$ .

$$\frac{1(x+1)}{x(x-4)(x+1)}+\frac{(5-2x)x}{(x+1)(x-4)x}=\frac{5(x-4)}{x(x+1)(x-4)}$$

Now we have a common denominator. Let's multiply both sides by this denominator...but first we must say that   x ≠ 0 ,  x ≠ 4 ,  and  x ≠ -1 , because these values cause a zero in the denominator of the original equation.

1(x + 1) + (5 - 2x)x  =  5(x - 4)          When    x ≠ 0 ,  x ≠ 4 ,  and  x ≠ -1  .

Multiply out the parenthesees and simplify.

x + 1 + 5x - 2x2  =  5x - 20

-2x2 + x + 21  =  0

Solve for  x  with the quadratic formula.

$$x = {-1 \pm \sqrt{1^2-4(-2)(21)} \over 2(-2)} \\~\\ x = {-1 \pm 13\over -4} \\~\\ x=-\frac{12}{4}=-3\qquad\text{ or }\qquad x=\frac{-14}{-4}=\frac{7}{2}$$

hectictar  Sep 13, 2017

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