Write an inequality for the phrase

one third of the opposite of an number is less than 12

Let's first return to the original definition of absolute value: "| x | is the distance of x from zero." For instance, since both –2 and 2 are two units from zero, we have | –2 | = | 2 | = 2:

With this definition and picture in mind, let's look at some absolute value inequalities.

Suppose you're asked to graph the solution to | x | < 3. The solution is going to be all the points that are less than three units away from zero. Look at the number line:

The number 1 will work, as will –1; the number 2 will work, as will –2. But 4 will not work, and neither will –4, because they are too far away. Even 3 and –3 won't work (though they're right on the edge). But 2.99 will work, as will –2.99. In other words, all the points between –3 and 3, but not actually including –3 or 3, will work in this inequality. Then the solution looks like this:

The open circles at the ends of the blue line indicate "up to, but not including, these points." Your book might use parentheses instead of circles. Translating this picture into algebraic symbols, you find that the solution is –3 < x < 3.

The other case for absolute value inequalities is the "greater than" case. Let's first return to the number line, and consider the inequality | x | > 2.

The solution will be all points that are more than two units away from zero. For instance, –3 will work, as will 3; –4 will work, as will 4. But –1 will not work, and neither will 1, because they're too close. Even –2 will not work, and neither will 2 (although they're right on the edge). But 2.01 will work, as will –2.01. In other words, the solution will be two separate sections: one being all the points more than two units from zero off to the left, and the other being all the points more than two units from zero off to the right. The solution looks like this:

Translating this solution into symbols, we get "x < –2 or x > 2". That is, the solution is TWO inequalities, not one. DO NOT try to write this as one inequality. If you try to write this solution as
"–2 > x > 2", you will probably be counted wrong: if you take out the x in the middle, you'll see that you would be saying "–2 > 2", which certainly isn't true. Take the extra half a second, and write the solution correctly.

only 10/60 isn't equivalent

hope this : )

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!