Question

how would I graph 3y+1=7

Answer 1

having issue graphing 3x-4y-5=-17

Answer 2

as well

Answer 3

solve for "y" on both cases, then pick a couple of random "x" values and get "y", and plot it

Answer 4

3y+1=7 =3y=6 y=2

Answer 5

so would that make x =1

Answer 6

yeap so y =2 that means \(\large \begin{array}{rrllll}
x&y
\\\hline\\
\pm 5&2\\
\pm 50&2\\
\pm 5,000&2\\
\pm 5,000,000&2\\
\pm \infty&2
\end{array}\) thus

Answer 7

so the 3x-4y-5=-17 would that have be made into the y=mx+b form 4y=-3x+4y=-17
so then for the 3y+1=7 eqation do not have to find the other two points.

Answer 8

right.. the \(\bf 3y+1=7 \) is just a horizontal line... you could look for two points, look above... there are more than two.. then again.. they'll all end up over the y =2 line anyhow

as far as the 2nd one \(\bf 3x-4y-5=-17\implies -4y-5=-17-3x\implies -4y=-17-3x+5
\\ \quad \\
-4y=-12-3x\implies y=\cfrac{-12-3x}{-4}\implies y=\cfrac{\cancel{ -12 }}{\cancel{ -4 }}-\cfrac{-3x}{-4}
\\ \quad \\
y=3+\cfrac{3x}{4}\)

Answer 9

ok so it got a bit truncated... so \(\bf 3x-4y-5=-17\implies -4y-5=-17-3x
\\ \quad \\
-4y=-17-3x+5 \implies-4y=-12-3x\implies y=\cfrac{-12-3x}{-4}
\\ \quad \\
y=\cfrac{\cancel{ -12 }}{\cancel{ -4 }}-\cfrac{-3x}{-4}\implies y=3+\cfrac{3x}{4}\)

Answer 10

and then just substitute then into one of the equations to find x

Answer 11

well... then just pick 2 random "x" and get "y" that would give you 2 coordinate points and since is a LINEar equation, its graph is a line, and to graph a line all you need is two points... .draw a line through those points

Answer 12

got it thanks a lot for your help

Answer 13

yw