Question 178154
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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. *[tex \Large 3x - 1 = 5x - 2]


Step 1 is to collect like terms:


*[tex \Large 3x - 5x = 1 - 2]


*[tex \Large -2x = -1 \text { } \Rightarrow \text{ } \math x = {1 \over 2}]


Now you have an equation that has a solution set *[tex \Large \{(x, y) | y \in \R, x = {1 \over 2}\}]


Meaning that <i>y</i> can be anything you like as long as *[tex \Large x = {1 \over 2}], in other words, a vertical line that intercepts the <i>x</i>-axis at *[tex \Large ({1 \over 2},0)].  Now you should be able to see why you got an absurd result when you tried to substitute 0 for <i>x</i> 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:


{{{drawing(
400,400,-5,5,-5,5,
grid(1),
line(.5,-5,.5,5)
)}}}


2. This problem contains an absolute value.  In order to characterize a graph of an absolute value equation, you need to define <i><b>three</b></i> ordered pairs.  The first one must be the ordered pair that results from setting <i>x</i> 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:  *[tex \Large y=|x| +3], 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 <i>x</i> to find the appropriate value).  We need to solve for *[tex \Large x = 0]


*[tex \Large y=|0| +3 \text { } \Rightarrow \text{ }\math y = 3]


Giving us our first ordered pair:  *[tex \Large P_1:(0, 3)]


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


*[tex \Large y=|-1| + 3 \text { } \Rightarrow \text{ }\math y = 1 + 3 = 4]


Giving us our second ordered pair:  *[tex \Large P_2:(-1, 4)]


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


*[tex \Large y=|1| + 3 \text { } \Rightarrow \text{ }\math y = 1 + 3 = 4]


Giving us our third ordered pair:  *[tex \Large P_3:(1, 4)]


Then plot the three points:


{{{drawing(
400,400,-3,3,-1,5,
grid(1),
circle(0, 3, .05),
locate(.2, 3, P1(0, 3)),
circle(-1, 4, .05),
locate(-.8, 4, P2(-1, 4)),
circle(1, 4, .05),
locate(1.2, 4, P3(1, 4))
)}}}


Then connect the dots...


{{{drawing(
400,400,-3,3,-1,5,
grid(1),
circle(0, 3, .05),
locate(.2, 3, P1(0, 3)),
circle(-1, 4, .05),
locate(-.8, 4, P2(-1, 4)),
circle(1, 4, .05),
locate(1.2, 4, P3(1, 4)),
graph(
400,400,-3,3,-1,5,
y=abs(x) + 3
))}}}


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.


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