Algebra II Final Part 2: Rationals, Systems, Absolute Value, and Inequalities

Includes some definitions, solving inequalities, and probably more

Define: Solving an equation

To make it ture. For example, using x+2=6, the solution would be x=4, because 4+2=6, making the statement true.

Solve: I2x+3I<6

-4.5<x<1.5

Using the inequalities y≤x²-2 and y≥2x²-2x-5, find a point which makes them both false. Additionally, find a solution that has whole number coordinates and a solutions that does not.

There are many options, but mine were:

(-2,5)

(-1,-1)

(2.4,1.72)

Pickle Peter was looking at his stupid notes and he noticed that he had written an inequality, but his dumb ass smeared the ink, so all he had was ?x²-1>0. All his foolish mind could remember was that it had no solutions. What would the smudge(?) have to had been?

A negative number. For example, if he used -1, plugging in any x value would make the left side come out negative, making the inequality false (so, -1x²>0). Stupid Pickle Peter.

Solve algebraically: x²+11x+18>0

x<−9 or x>−2

1. Factor and set to 0 to get (x+9)(x+2)=0

2. Set each factor to 0 (x+9=0, x+2=0) and solve

3. Write your answers (view initial answer)

NOTE: inequalities using more than are written as to answers as all valid inputs exclude a given domain, but include everything on either side of it.

Solve algebraically: x²+5x−36<0

−9<x<4

1. Factor and set to 0 to get (x+9)(x−4)=0

2. Set each factor to 0 (x+9=0,x-4=0) and solve

3. Write you answer (view initial answer)

NOTE: inequalities using less than are written as one answer as all valid inputs are within a single given domain.