solve (x-3)/(x+1) > 5

I get x < -2 every time, but it doesnt work when I plug it back in. (I plugged in -5)

Multiply both sides times x + 1

x-3 > 5x + 5

-4x > 2

x<-2 You are correct

You choose -5 to check which is correct.

(-5 -3)/(-5 + 1) > 5
-8/-4 >5
2 > 5
That is correct 2 is greater than 5.

Two is greater than five? That's what I do not understand.. Thank you

-4x >8 excuse my typo

x < -2

Lets pretend this is an equation for one minute

(x-3)/(x+1) = 5
Put in the -2
(-2-3/(-2-1) =5
-5/-1 =5
We are correct there...

so.. Since we have an inequality, we have two choices for the final answer.

x> -2 or x > -2

If we choose something larger than -2 will that work?
Lets pick a 2.. ( 2-3)/(2+3) = -1/5 so that is not greater than 5
If we pick a 10 we would have 7/11 which is not greater than 5. We are sure we shouldn't go to the right.

Going to the left..we are having the same problem. You are right 2>5 is wrong.

Try x = -2.5 see how that works.

To solve the inequality (x-3)/(x+1) > 5, we need to follow a systematic approach. Let's go step by step:

Step 1: Get rid of the fraction by multiplying both sides of the inequality by (x+1):
(x+1) * (x-3)/(x+1) > 5 * (x+1)

Simplifying the left side:
(x-3) > 5(x+1)

Step 2: Distribute the 5 on the right side:
(x-3) > 5x + 5

Step 3: Move all terms with x to one side of the inequality:
(x - 5x) - 3 > 5

Simplifying:
-4x - 3 > 5

Step 4: Move the constant term to the other side:
-4x > 5 + 3

Simplifying:
-4x > 8

Step 5: Finally, divide both sides of the inequality by -4, remembering that we need to reverse the inequality sign when dividing by a negative number:
x < 8/-4
x < -2

So, the solution to the inequality (x-3)/(x+1) > 5 is x < -2.

Now, to verify if -5 is a valid solution, we can plug it back into the original inequality and check:

(x-3)/(x+1) > 5
(-5-3)/(-5+1) > 5
-8/-4 > 5
2 > 5

Since 2 is not greater than 5, -5 is not a valid solution. Therefore, it seems that the solution x < -2 is correct, and you might have made a computation mistake when plugging it back in.