Modulus inequality: abs(2x - 5) ? 2 - x

[Monkeyseat]

Solve the following inequality: abs(2x - 5) ? 2 - x

I sketched a graph:

Because they don't intercept, is there no solution?

Thanks.

 

[stapel]

Because they don't intercept, is there no solution?

Intersection points would be the solutions to equalities; that is, where the lines cross (occupy the same coordinates), the expressions have the same value.

You are solving an inequality. You are being asked to find the x-values for which the red line (the absolute-value expression) is higher than (has a greater value than) the green line (the linear expression).

For what x-values is the red line not higher than the green line? :wink:

Eliz.

 

[Monkeyseat]

Re:

Because they don't intercept, is there no solution?

Intersection points would be the solutions to equalities; that is, where the lines cross (occupy the same coordinates), the expressions have the same value.

You are solving an inequality. You are being asked to find the x-values for which the red line (the absolute-value expression) is higher than (has a greater value than) the green line (the linear expression).

For what x-values is the red line not higher than the green line? :wink:

Eliz.

Isn't the red line always greater than the green line?

Sorry stapel, I don't quite understand what you're saying. I was taught to draw a graph, look where they intercept and solve them algebraically.

 

[galactus]

If we solve each equation, we see all the bases are covered.

\(\displaystyle -2x+5>2-x\)

\(\displaystyle 2x-5>2-x\)

Solving the first one we get \(\displaystyle x<3\)

The second one yields \(\displaystyle x>\frac{7}{3}\)

That pretty much covers it.

 

[Monkeyseat]

If we solve each equation, we see all the bases are covered.

\(\displaystyle -2x+5>2-x\)

\(\displaystyle 2x-5>2-x\)

Solving the first one we get \(\displaystyle x<3\)

The second one yields \(\displaystyle x>\frac{7}{3}\)

That pretty much covers it.

Thanks for the reply. I've been doing it a different way and I'm not quite sure about this. For example, if a line cut a curve between two points, I would substitute a value in (between those two values) to see if the modulus part for each point was positive or negative and then solve it like you have.

How did you jump straight to that step and get:

\(\displaystyle -2x+5>2-x\)

\(\displaystyle 2x-5>2-x\)

I'm not sure how you did that galactus as today was the first time I have done this topic.

Also, why is the range just 7/3 < x < 3? Can't x be greater than that and abs(2x - 5) will still be greater than 2 - x? E.g. 2(5) - 5 = 10, whereas 2 - 5 = -3. Furthermore, from looking at the graph abs(2x - 5) always seems greater than 2 - x as isn't the red line always greater than the green line?

I know I am wrong, but I don't quite understand how you got to that step (and why the answer is restricted to 7/3 < x < 3). It's greater than/equal to as well I think btw. Thanks.

 

[galactus]

It is not \(\displaystyle \frac{7}{3}<x<3\)

It is \(\displaystyle x<3, \;\ and \;\ x>\frac{7}{3}\)

What is less than 3 and greater than 7/3?. Everything. Your graph shows it. It certainly appears that the absolute value graph is always higher than the line graph. I just showed you algebraically.

When you have absolute values, you solve two separate equations.

Afterall, what is the absolute value except where the two lines come together and form a V.

The two lines are y=2x-5 and y=-2x+5

In this case, \(\displaystyle -2x+5>2-x\)

and

\(\displaystyle 2x-5>2-x\)

Solve each for x and you'll see.

 

[Monkeyseat]

It is not \(\displaystyle \frac{7}{3}<x<3\)

It is \(\displaystyle x<3, \;\ and \;\ x>\frac{7}{3}\)

What is less than 3 and greater than 7/3?. Everything. Your graph shows it. It certainly appears that the absolute value graph is always higher than the line graph. I just showed you algebraically.

When you have absolute values, you solve two separate equations.

Afterall, what is the absolute value except where the two lines come together and form a V.

The two lines are y=2x-5 and y=-2x+5

In this case, \(\displaystyle -2x+5>2-x\)

and

\(\displaystyle 2x-5>2-x\)

Solve each for x and you'll see.

I saw how you solved the two inequalities to get the values, but I don't know how you got the inequalities to begin with. I know that if you put x < 5/2 and x > 5/2, one is positive and one is negative.

Okay so is this what you are saying:

Baisically, x can be anything less than 3 and x can be anything greater than 2.3 (but not just within that region of 0.7). Therefore, x can be anything? Does this go back to what I said in my first post where I said "Isn't the red line always greater than the green line?"

I was a bit confused why it wasn't 7/3 < x < 3. For example, In the first question I did, abs(4 - x) ? 1:

I got x ? 3 and x ? 5 and the answer was 3 ? x ? 5. What makes this different? Is that because the two lines intercepted there and they don't here?

Thanks.

 

[galactus]

Because 7/3<x<3 would mean the only place it is larger is when x is between 7/3 and 3. That is certainly not the case. It is everywhere.

Basically, x can be anything less than 3 and x can be anything greater than 2.3 (but not just within that region of 0.7). Therefore, x can be anything? Does this go back to what I said in my first post where I said "Isn't the red line always greater than the green line?"

Yes.

 

[Monkeyseat]

Because 7/3<x<3 would mean the only place it is larger is when x is between 7/3 and 3. That is certainly not the case. It is everywhere.

Basically, x can be anything less than 3 and x can be anything greater than 2.3 (but not just within that region of 0.7). Therefore, x can be anything? Does this go back to what I said in my first post where I said "Isn't the red line always greater than the green line?"

Yes.

Thanks, just a few things to clarify. Bear with me, first day I've done these.

Okay, so I think I'm getting it now. x can be ANY value because the red line is ALWAYS greater than the green line. So x < 3 and x > 7/3 (but not the two combined) is the same as the image below which encompasses all values?

Although it can include 3 and 7/3 because it is less than/equal to and greater than/equal to. I take it you used the not equal to sign for simplicity.

Also, for the example I gave my previous post, was that 3 ? x ? 5 because the modulus line and the linear line intercepted (and that is not the case in my original question)?

Finally, for:

-2x + 5 > 2 - x
2x - 5 > 2 - x

If you sub values for x that are greater than 2.5 (x intercept) and less than 2.5 into 2x - 5 does that show if it the line at that point is positive or negative? E.g. if x < 2.5, abs(2x - 5) = 5 - 2x. If x > 2.5, abs (2x - 5) = 2x - 5. Is that correct?

Many thanks.