Can someone please break down how to solve this equation. I am so confused.
(x - 2)/(x2 + 3x - 10) – x/(x + 5) = 4
Show how you get the answers {-3.8, 2}, yet only one of them is a true solution.
(x - 2)/(x^2 + 3x - 10) – x/(x + 5) = 4
(x-2)/[(x+5)(x-2)] - x/(x+5) = 4
1/(x+5) -x/(x+5) = 4
1-x = 4x+20
5 x = -19
x = 3.8
x = 2 does not satisfy the original equation and I do not know where you got it. If x = 2 then the denominator of the first term x^2+3x-10 = 0 so it is undefined.
Thanks a bunch. It makes sense. I was adding too many steps in and confusing myself.
To solve the given equation:
1. Start by finding the LCD (Least Common Denominator) of the fractions in the equation. In this case, the LCD is (x + 5)(x - 2).
2. Multiply both sides of the equation by the LCD to clear the fractions. This will give you:
(x - 2)(x + 5)(x - 2)/(x2 + 3x - 10) – x(x - 2)(x + 5)/(x + 5)(x - 2) = 4(x + 5)(x - 2)
Simplifying this equation gives:
(x - 2)(x + 5) - x(x - 2) = 4(x + 5)(x - 2)
3. Expand and simplify the equation:
(x^2 + 3x - 10) - (x^2 - 2x) = 4(x^2 + 3x -10)
x^2 + 3x - 10 - x^2 + 2x = 4x^2 + 12x - 40
4. Combine like terms:
5x - 10 = 4x^2 + 12x - 40
5. Move all the terms to one side of the equation to create a quadratic equation:
0 = 4x^2 + 7x - 30
6. Factor the quadratic equation or use the quadratic formula to find the possible values of x. Factoring in this case gives:
0 = (2x + 15)(2x - 2)
7. Set each factor equal to zero and solve for x:
2x + 15 = 0 or 2x - 2 = 0
8. Solve for x in each equation:
2x = -15 or 2x = 2
x = -15/2 or x = 1
9. Check if each solution is a true solution by substituting it back into the original equation:
For x = -15/2:
[(x - 2)/(x^2 + 3x - 10)] – [x/(x + 5)] = 4
[(-15/2 - 2)/((-15/2)^2 + 3(-15/2) - 10)] – [(-15/2)/(-15/2 + 5)] = 4
(-19/2)/[(225/4) - 45/2 - 10] + (15/2)/(5/2) = 4
(-19/2)/(225/4 - 90/4 - 40/4) + (15/2)/(5/2) = 4
(-19/2)/(95/4) + (15/2)/(5/2) = 4
(-19/2)/(95/4) + (15/2)(2/5) = 4
(-19/2)/(95/4) + 30/10 = 4
(-19/2)(4/95) + 3 = 4
(-19/47) + 3 = 4
-19/47 + 141/47 = 4
122/47 = 4
The equation is not satisfied when x = -15/2.
For x = 1:
[(x - 2)/(x^2 + 3x - 10)] – [x/(x + 5)] = 4
[(1 - 2)/(1^2 + 3(1) - 10)] – [(1)/(1 + 5)] = 4
(-1)/(-6) - (1)/(6) = 4
1/6 - 1/6 = 4
0 = 4
The equation is not satisfied when x = 1.
Therefore, there are no true solutions to the given equation. The solutions {-3.8, 2} are extraneous or invalid solutions.