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3 years ago

Can someone plz help me I’m tryna go to sleep .

Mathematics

1 answer:

schepotkina [342]3 years ago

6 0

Take a warm shower, however you want, dont eat too much before hand, relax your face and don’t think too much about it, dont think much just think “im sleepy im going to sleep”
Incase you don’t want to read all of this

It seems like you’re stressed especially since you’re thinking about it too much to the point you have to use brainly. You’re stressed and need to relax and not think too much, you might be having too much things going around you and or many possibilities. For me I just take a warm shower and just stay and think, probably cry if i have to, without sleep or little you’re going to be anxious and things bother you more and makes sleeping even more harder on top of that

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3 years ago

2x+3y=1 and y=-2 - 9 solve using linear combination method. Please explain steps.

kaheart [24]

Answer:

$x=17$

Step-by-step explanation:

$2x+3y=1$

$y=-2-9$

In order to combine these two equations, an idea you need to keep in mind is finding a way of setting these equations as equal to each other. I saw that each equation shared a common value, $y$ . In this case, we need to isolate $y$ in the first equation so that both equations $=y$ .

$2x+3y=1$

$3y=-2x+1$

$\frac{3y}{3}=\frac{-2x+1}{3}$

$y=-\frac{2}{3}x+\frac{1}{3}$

With this, we now know that both $-2-9$ and $-\frac{2}{3}x+\frac{1}{3}$ are equal to $y$ , so we can set them equal to each other.

$y=-\frac{2}{3}x+\frac{1}{3}$

$y=-11$

$-\frac{2}{3}x+\frac{1}{3}=-11$

Reply to this if anything I'm saying or doing is confusing in any way, or if you find a mistake. :) Solve for $x$ .

$-\frac{2}{3}x+\frac{1}{3}=-11$

$-\frac{2}{3}x=-11-\frac{1}{3}$

$-\frac{2}{3}x =-\frac{33}{3}-\frac{1}{3}$

$-\frac{2}{3}x =-\frac{34}{3}$

$-\frac{2}{3}x(-\frac{3}{2})=-\frac{34}{3}(-\frac{3}{2})$

$x=\frac{102}{6}=17$

$x=17$

Hopefully this answer is correct AND makes sense in terms of how I achieved it. Again, reply to this with any questions or mistakes I made and I'll do my best to answer or fix them.

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