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

Solve this rational equation.

AdamTaurus
Sep 12, 2017

#1**0 **

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.

Add 7 to both sides:

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**+1 **

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

Factor the x^{2} - 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 - 2x^{2} = 5x - 20

-2x^{2} + 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