How do you solve #1/(2x) + 3/(x+7) = -1/x# and find any extraneous solutions?

Question

Ethan Adkins · Accepted Answer

#x=-7/3#

You can first write the corresponding equation:

#1/(2x)+3/(x+7)+1/x=0#

and finally, by multiplying each term by 2x(x+7), obtain a single fraction.

if #x!=0 and x!=-7#:

#1/cancel(2x)cancel(2x) (x+7)+3/cancel(x+7)2x(cancel(x+7))+1/cancelx2cancelx(x+7)=0#

#x+7+6x+2x+14=0#

#9x+21=0#

#x=-21/9=-7/3#

Jameson Allen · Answer

undefined To solve the equation 1/(2x) + 3/(x+7) = -1/x and find any extraneous solutions, we can follow these steps:

1. Find a common denominator for the fractions on the left side of the equation. In this case, the common denominator is 2x(x+7).

2. Multiply each term by the common denominator to eliminate the fractions. This gives us: (x+7) + 6x = -2(x+7).

3. Simplify the equation by expanding and combining like terms: x + 7 + 6x = -2x - 14.

4. Combine like terms again: 7x + 7 = -2x - 14.

5. Move all the terms to one side of the equation: 7x + 2x = -14 - 7.

6. Simplify: 9x = -21.

7. Divide both sides of the equation by 9 to solve for x: x = -21/9.

8. Simplify the fraction: x = -7/3.

To check for extraneous solutions, substitute the found value of x back into the original equation:

1/(2*(-7/3)) + 3/((-7/3)+7) = -1/(-7/3).

Simplifying this equation, we find that both sides are equal to -3/7.

Therefore, the solution x = -7/3 is valid and there are no extraneous solutions.