Thread: Half right, missed the other half...

"Find all values of such that the slope of the line through the points and is less than 5."

I solved it by first finding the slope which simplifies to . Then I used the point-slope form to derive the equation . Then, I got lazy, and plugged in a couple of test numbers into the equation, found that numbers above 2 seemed to qualify, and decided that the answer was the set . Unfortunately, the answer should have included also the set .

Algebraically, what should I have done to get to that answer? What part did I skip?

Hello, earachefl!

Find all values of such that the slope of the line
through the points and is less than 5.

You should have solved the inequality . . . tricky, but safe.


You found the slope of . . . good!

. . Then we have: . . [1]


I'd like to multiply both sides by
. . but the result depends on whether is positive or negative.

Recall: when multiplying or dividing an inequality by a negative quantity,
. . . . . .reverse the inequality.


Suppose is positive . . . that is: .

Multiply [1] by by positive

. .

Divide both sides by -3: .

We have two inequalities to satisfy: . and

We take the "stronger" of the two: .


Suppose is negative . . . that is: .

Multiply [1] by negative

. .

Divide both sides by -3: .

We have two inequalities to satisfy: . and

We take the "stronger" of the two: .


Therefore: . .. . . .or: .

Originally Posted by Soroban

Hello, earachefl!


We have two inequalities to satisfy: . and

We take the "stronger" of the two: .

Interesting approach. I don't think my book ever put it in quite these terms.

Thanks!