Solve for all values of x.

Question

x, minus, start fraction, 3, x, plus, 5, divided by, x, plus, 2, end fraction, equals, start fraction, 1, divided by, x, plus, 2, end fraction
x−
x+2
3x+5
​
=
x+2
1
​

Answer 1

To solve for all values of x, we first need to set up the equation:

x - (3x + 5)/(x + 2) = 1/(x + 2)

Now we simplify the left side of the equation:

x - (3x + 5)/(x + 2) = 1/(x + 2)

x - (3x + 5)/(x + 2) = 1/(x + 2)

Multiplying through by (x + 2):

(x)(x + 2) - (3x + 5) = 1

Expanding:

x^2 + 2x - 3x - 5 = 1

x^2 - x - 5 = 1

x^2 - x - 6 = 0

Now we have a quadratic equation that we can solve for x:

(x - 3)(x + 2) = 0

x - 3 = 0 or x + 2 = 0

x = 3 or x = -2

Therefore, the values of x that satisfy the equation are x = 3 and x = -2.

Answer 2

answer that will resalt in division by zero should be excluded

Answer 3

Checking for potential division by zero:

When x = -2, the denominator of both fractions becomes zero.

Therefore, x = -2 is not a valid solution and it should be excluded.

The only valid solution is x = 3.