weredumbmaster22

Could u rewrite it so that it is not in latex form

Thank you!

The solution can be written in multiple ways. Find the best way that you need.

 \begin{bmatrix}\mathrm{Solution:}\:&\:x<-2\quad \mathrm{or}\quad \:x>\frac{3}{2}\:\\ \:\mathrm{Decimal:}&\:x<-2\quad \mathrm{or}\quad \:x>1.5\\ \:\mathrm{Interval\:Notation:}&\:\left(-\infty \:,\:-2\right)\cup \left(\frac{3}{2},\:\infty \:\right)\end{bmatrix}

I am sorry.

I have seen people doing that in the past so I thought it would be okay. If you would like I could repost the questions one by one.

I did distribute and got this:

3x-(-3)^2=12

3x+9=12

3x=3

x=1

3(1)-y^2=12

3-y^2=12

-y^2=9

What do I do after??????

Find the area of the region bounded by the y-axis,y=6 the line , y=1/2x and the line .

Could you do this?

Oh, that wasy. i wonder why I got stuck lol.

Thanks :)

Thank you EP. I will surely "SUStitute" the y for the x. XD :)

I did but this part:

Multiply the first equation by some constant, so that in the resulting equation, the coefficient of x matches the coefficient of x in the second equation.

Enter the entire equation that results from multiplying the first equation by the constant.

I get when it asks to multiply the first question by constant but after that its just a blur in my head.

Thank you, it helps. But I dont understnad how the answer could help me find my second question, it tells me to substitute but it doesn't give deep details about the question.

Oh ok, I understand now. Thank you. 

a^2+b^2 would be 10^2+10^2 which is 100+100 and the answer is 200. Its wrong.