+0

# graphs, slopes, and stuffs

+3
86
11

Find a linear inequality with the following solution set. Each grid line represents one unit.

click to see image!

Give your answer in "standard form" or ax+by+c>0 where  a, b and c are integers with no common factor greater than 1. Aug 17, 2020

#3
+1

Find two good points   -3 -2    and   1, 0

find slope      (-2 -0) / (-3-1) = 1/2

(y+2) = 1/2 (x+3)     is the equation of the line   re-arrange to   y = 1/2 x - 1/2

The LINE is dashed and not included in the area above the line      so    y > 1/2 x - 1/2

Aug 17, 2020
#4
+2

Thanks EP.  I will just add my 2c worth

The line appears to go through   (1,0)  and (-1,-1)

$$gradient=\frac{rise}{run}\\ gradient=\frac{\text{distance between the y values}}{\text{distance between the x vlaues}}\\ gradient=\frac{y_2-y_1}{x_2-x_1}\\ gradient=\frac{-1-0}{-1-1}\\ gradient=\frac{-1}{-2}\\ gradient=\frac{1}{2}\\$$

now to get the equation I usually just think to myself

$$gradient=\frac{y_2-y_1}{x_2-x_1}\\ \text{keep the same }(x_1,y_1) \text{point but just let the other point be (x,y)}\\ \frac{1}{2}=\frac{y-0}{x-1}\\ 1(x-1)=2(y-0)\\ x-1=2y\\ x-2y-1=0$$

(0,0) is in the wanted part of the graph so you have to sub in (0,0) and choose the correct inequality

(it is a dashed line, not a solid one, so there will be no equal sign)

$$LHS=x-2y-1 \qquad sub\;\;in\;\;(0,0)\\ LHS=0-0-1\\ LHS=-1\\ LHS<0\\ so\\ x-2y-1<0$$   LaTex:

gradient=\frac{\text{distance between the y values}}{\text{distance between the x vlaues}}\\

\text{keep the same }(x_1,y_1) \text{point but just let the other point be (x,y)}\\
\frac{1}{2}=\frac{y-0}{x-1}\\
1(x-1)=2(y-0)\\
x-1=2y\\
x-2y-1=0

LHS=0-0-1\\
LHS=-1\\
LHS<0\\
so\\
x-2y-1<0

Aug 18, 2020
#5
0

ok thank you!!!! I also got the answer eventually this way:

-x + 2y + 1 > 0 is the answer. :)

Let's find the equation of the dashed line first. From the graph, we can see that it has a slope of 1/2 and a y-intercept of -1/2 so using slope-intercept form, the equation of the line is y = 1/2x - 1/2. Since the line is dashed and not solid, we know that we will use either < or > instead of ≤ or ≥, and because we can see that the shaded region is above the dashed line, we know that the linear equation is y > 1/2x - 1/2. However, we want the answer to be in the form ax + by + c > 0 where a, b, and c are integers so therefore:

y > 1/2x - 1/2

= 2y > x - 1

= -x + 2y + 1 > 0

iamhappy  Aug 18, 2020
#6
+1

I have done that.

I hope you have looked at my answer properly.

Do you understand that my answer and your answer are the same only mine is slightly better presented?

Anyway, I am glad you understand the question.

Melody  Aug 19, 2020
#7
+2

Sorry, Melody. :( I only read the beginning of your answer, and then decided to come up with my own to see if I could start off of what you did. :) I then read your whole answer and realized they were very similar and yes, your's was explained way more thoroughly. :) I just wanted to see if I could do it by starting off of your well explained answer, since I thought I would learn best that way~ Sorry again, I shouldn't have made another answer. 😢

iamhappy  Aug 19, 2020
#8
+1

I was just a bit put out because you said you were really struggling with the whole topic, not just this question, and I gave you a much more thorough teaching answer then I would normally bother to do.

But in reality, you were not struggling with the whole topic and were capable of answering by yourself.

The time I took to give you a full explanation could have been better used.

However, you do not need to feel bad about it, I know you didn't do it on purpose, you were just being a little dramatic.  (in your private message)

Cause that is how you felt at the time.

(it would have been nice to get a point from you though, not that the point itself is significant to me, but the show of gratitude always is)

Melody  Aug 19, 2020
edited by Melody  Aug 19, 2020
#9
+3

Melody, thank you for the advice. :) My first instinct was to give you and EP a thumbs up, but as I reported before, I cannot. I think it may be a glitch, but I just am not able to give any hearts. SORRY. :(

iamhappy  Aug 20, 2020
#10
+1

That is ok, many people are having this problem.  :)

Melody  Aug 21, 2020
#11
+3

Yeah, thanks for understanding though!

iamhappy  Aug 21, 2020