Determine whether the origin is included in the shaded region and whether the shaded region is above or below the line for the graph of the following inequality:
y > three fifthsx − 2
The origin is not included in the shaded region and the shaded area is above the line.
The origin is not included in the shaded region and the shaded area is below the line.
The origin is included in the shaded region and the shaded area is above the line.
The origin is included in the shaded region and the shaded area is below the line.

- anonymous

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

@queen-of-tokyo

- anonymous

@WhateverYouSay

- anonymous

@sweetburger

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

@zepdrix

- anonymous

@yomamabf

- zepdrix

Hey there :)
Let's answer the second part of the question first, because it's really simple.

- zepdrix

When you have something like this:
\(\large\rm y\lt \color{orangered}{2x+3}\)
Think of it like this:
\(\large\rm y\lt \color{orangered}{\text{the line}}\)
The orange part represents the line.
So in this example,
we want all of the `y values` which are `less than` `the line`.
So in this example we would shade `below` the line.

- zepdrix

How bout in your problem?
y is GREATER THAN some line.
So where we gonna shade?

- anonymous

above the line correct? @zepdrix

- zepdrix

k good :) that already narrows down our choices!

- zepdrix

\[\large\rm y\gt\frac{3}{5}x-2\]Recall that the origin is the point (0,0).
So let's just plug it in...
and find out if the inequality holds true!

- zepdrix

\[\large\rm 0\gt\frac{3}{5}\cdot0-2\]After you simplify this a bit,
does the inequality hold true?

- anonymous

no

- anonymous

its the same right?

- zepdrix

So after simplifying, we have, 0 > -2

- zepdrix

Hmm that appears to hold true, ya?
0 is larger than -2

- anonymous

oh yes my bad i was looking at the wrong thing

- zepdrix

Ok good!
So we figured out that the point (0,0) DOES in fact satisfy our inequality.
So it is included in the shaded area.

- anonymous

thanks could you help mw with one more

- zepdrix

sure, i can try :)

- anonymous

Given the system of equations presented here:
3x + 5y = 29
x + 4y = 16
Which of the following actions creates an equivalent system such that, when combined with the other equation, one of the variables is eliminated?
Multiply the second equation by −1 to get −x − 4y = −16
Multiply the second equation by −3 to get −3x − 12y = −48
Multiply the first equation by −1 to get −3x − 5y = −29
Multiply the first equation by −3 to get −9x − 15y = −87

- zepdrix

You're trying to match up the coefficients on either the x's or the y's, but not both.
If we wanted to match up the x's, maybe we would need some kind of 3.
If we wanted to match up the y's, it would be a little more difficult.
We would multiply the first equation by 4,
and the second equation by -5.
They would then both have coefficients of 20, one of them being negative.
So they would combine nicely.
They didn't do that here though.
They went the easier route.
They're matching up the x's somehow.

- zepdrix

We want them to match.
Notice the first equation already has a 3, so it doesn't need anything more.

- zepdrix

So what do you wanna do to the second equation to match them up?

- anonymous

add a 1 to the x?

- zepdrix

The first equation has 3x.
The second equation has x.
If we add 1 to x, hmm I'm not really sure what that will do for us :d
We want the same number of x in both equations.

- zepdrix

We don't want to add.
We want to multiply that second equation by some number that will turn it from x to some kind of 3x.

- anonymous

you would multiply it by 3 then

- anonymous

@zepdrix

- zepdrix

Good good good, we would multiply the second equation 3!
That would give us 3x.
But in order for them to combine, and disappear, we want one of them to be negative.
So let's instead multiply the second equation by negative 3.

- zepdrix

boom bam! yay team!

- anonymous

so the answer is D yes? @zepdrix

- zepdrix

Option D talks about changes to equation 1.
We were making changes to equation 2.

- anonymous

oh ok i wasn't reading carefully but thank you for your help :) @zepdrix

Looking for something else?

Not the answer you are looking for? Search for more explanations.