# find and graph the solution set of (3 x - 5)/(x + 1) < 2 Express solution set in set builder notation.

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To find the solution of (3x-5)/x+1) <2.

We know that an inequation is unaffected by subtaction of equals from both sides.

We subtract 2 from both side: of the inequation:

(3x-5)/(x+1) -2 < 0

(3x-5)/(x+1) - 2(x+1)/(x+1) < 0

(3x-5 -2x-2)/(x+1) < 0

(x- 7)/(x+1) < 0.

Let f(x) = (x-4)/(x+1) < 0 .

f(x) < 0 if x lies between -1 and 7. But x can not take the boundary value of x= -1 and x = 7 also.

Therefore such a solution set S is given by: S = {x: -1 < x < 7} for which (3x-5)/(x+1) < 2.

We'll start by imposing contraints of existence of the ratio. For this reason, we'll calculate the values of x for the denominator is cancelling.

x + 1 = 0

We'll subtract 1:

x = -1

So, for x = -1, the ratio is indefinite.

We'll re-write the expression, after performing the cross multiplying:

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

We'll remove the brackets and we'll get:

3x - 5 < 2x + 2

We'll subtract 2x + 2 both sides:

3x - 2x - 5 - 2 < 0

We'll combine like terms:

x - 7 < 0

We'll add 7 both sides:

**x < 7**

The ratio (3 x - 5)/(x + 1) < 2 if and only if the values of x belong to the interval (- infinite , 7).

**Note: Since the ratio is indefinite for x = -1, we'll reject the value -1 from the interval (- infinite , 7).**

**The solution of the inequality is the interval:**

**(- infinite , 7) - {-1}.**

We have the inequation (3 x - 5)/(x + 1) < 2

Now (3 x - 5)/(x + 1) < 2 can be rewritten as

=> 3x- 5 < 2*(x+1) [assuming x+1 is positive]

=> 3x - 2x - 5 -2 < 0

=> x - 7 <0

=> x < 7

=> x < 7

Now we had assumed that x+1>0 => x > -1

Therefore x= (-1 , 7)

Now assume x+1 is negative

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

=> 3x - 2x - 5 -2 > 0

=> x - 7 >0

=> x > 7

As x+ 1 is negative x< -1

We don't get valid values here as x cannot be less than -1 and greater than 3.

**Therefore x can take values (-1 , 7)**