Beforehand let me apologize for my sloppy handwriting. I'm a lefty, so please deal with me. . Anyway, the equation you listed didn't really have a x, because a linear equation is y= mx+b, and anything with x is the slope. But I did see y and a number on one side, so I'm like maybe I can get these two alone. So here's what I did:

2y + 4=0
-4 -4 I subtracted 4 on both sides. Why because you would want "y" alone.
Next,
2y= -4
Now to get "y" alone you want to divide on both sides.
2y= -4/ 2
y= -2
Now you're probably thinking "how do you graph it? There's no "x" in the equation." Well, you just graph it. Since the answer is y= -2, you go to the y axis, look for -2, and place a line to indicate that is the equation. And to make it clear, remember y is the y-intercept.

I really do hope this helps you, if not message me. I'll be happy to help, and again I'm sorry for my handwriting!

4 0

3 years ago

Read 2 more answers

What is the equation of the line that passes through the points (-3,5) (6,8)

Ratling [72]

$(\stackrel{x_1}{-3}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{8}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{8}-\stackrel{y1}{5}}}{\underset{run} {\underset{x_2}{6}-\underset{x_1}{(-3)}}}\implies \cfrac{3}{6+3}\implies \cfrac{3}{9}\implies \cfrac{1}{3}$

$\begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{5}=\stackrel{m}{\cfrac{1}{3}}(x-\stackrel{x_1}{(-3)}) \\\\\\ y-5=\cfrac{1}{3}(x+3)\implies y-5=\cfrac{1}{3}x+1\implies y=\cfrac{1}{3}x+6$

6 0

2 years ago