SOLUTION: 1. Graph 3x-1=5x-2 2. Graph y=|x| +3 3. Graph x+2=2x-1 4. Graph y=5x-2 My answers so far: For Number 4, I substituted x for 0 I got (0,-2) I graphed a line straight up and d

Algebra ->  Graphs -> SOLUTION: 1. Graph 3x-1=5x-2 2. Graph y=|x| +3 3. Graph x+2=2x-1 4. Graph y=5x-2 My answers so far: For Number 4, I substituted x for 0 I got (0,-2) I graphed a line straight up and d      Log On


   



Question 178154: 1. Graph 3x-1=5x-2
2. Graph y=|x| +3
3. Graph x+2=2x-1
4. Graph y=5x-2
My answers so far:
For Number 4, I substituted x for 0
I got (0,-2) I graphed a line straight up and down.
If x=1, then y=3 give my points at (1,3) This line is sloped now. Why is it changed? Which answer is right?
For Number 2, I subbed x for 0
I got (0,-1) and again got a straight line up and down.
If I made x=1, then y=4. This gets changed again to a slated line. Is this right?
For Number 3, I subbed x for 1.
I'm confused by this bc I came up with 3=1. Are these my points?
For Number 1, I subbed x for 0
I got 3(0)-1=5(0)-2
Then -1=-2
Now what do I do?
I really don't understand this. What I really don't understand is how to put these points on a graph.

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!

Your first problem is that you are making one substitution to get a single ordered pair representing a point and then trying to characterize a line based solely on that information. You can't do that. You must have two points to define a straight line.

Now let's take a look at your specific questions.

1.

Step 1 is to collect like terms:





Now you have an equation that has a solution set

Meaning that y can be anything you like as long as , in other words, a vertical line that intercepts the x-axis at . Now you should be able to see why you got an absurd result when you tried to substitute 0 for x in this case. If you take a look at your problem #3, you have the same difficulty. Solve problem 3 using the process just shown.

The graph for #1:



2. This problem contains an absolute value. In order to characterize a graph of an absolute value equation, you need to define three ordered pairs. The first one must be the ordered pair that results from setting x equal to a value that makes the expression inside of the absolute value bars be equal to zero.

In the case of the given problem: , this is trivial (if you had a more complex expression inside of the absolute value bars, you would first need to set this expression equal to zero and then solve for x to find the appropriate value). We need to solve for



Giving us our first ordered pair:

Next we need a value of x that makes the value inside of the absolute value bars be < 0; Let's choose -1.



Giving us our second ordered pair:

Lastly, we need a value of x that makes the value inside of the absolute value bars be > 0; Let's choose 1.



Giving us our third ordered pair:

Then plot the three points:



Then connect the dots...



4. This one is straight-forward, except for trying to define the line after you had only plotted one point. The two ordered pairs you calculated are correct. Just plot them and draw the line.