Question

In the equation 5y = 3x - 15 if I plug 0(zero) in for x what would y be? (answer is just the number like 15) *

Answer 1

So if the plug in 0 for x we would have 5y=(3)(0)−15
Yes?

Answer 2

\(\color{#0cbb34}{\text{Originally Posted by}}\) @supie
So if the plug in 0 for x we would have 5y=(3)(0)−15
Yes?
\(\color{#0cbb34}{\text{End of Quote}}\)
yes I think so

Answer 3

So do you know how we would solve that?

Answer 4

would it be - 12

Answer 5

y=-12?

Answer 6

\(\color{#0cbb34}{\text{Originally Posted by}}\) @supie
y=-12?
\(\color{#0cbb34}{\text{End of Quote}}\)
maybe I'm confused

Answer 7

hm

well the first thing we would to is multiply 3 and 0 which equals 0
So we would have \(5y=0+−15\)
thne we combine like terms
you think you would do that?

Answer 8

then*

Answer 9

\([5y=(0+−15)]\)
Solve what it inside of the parenthesis `()`

Answer 10

5y- 15

Answer 11

Yes, do you know the next step?

Answer 12

then add 0

Answer 13

\(\color{#0cbb34}{\text{Originally Posted by}}\) @b3lla2025
5y- 15
\(\color{#0cbb34}{\text{End of Quote}}\)
Actually it's \(5y=−15\)

Answer 14

\(\color{#0cbb34}{\text{Originally Posted by}}\) @b3lla2025
then add 0
\(\color{#0cbb34}{\text{End of Quote}}\)
No-

Answer 15

What we do is ÷5 to both sides

Answer 16

so -15 divided by 5 =-3

Answer 17

so y=-3

Answer 18

So that should look like
\(\LARGE \frac{5y}{5}=\frac{-15}{5}\)
So then 5y/5 cancels
\(\cancel{\LARGE \frac{5y}{5}}\LARGE=\frac{-15}{5}\)
So all you have left is \(\frac{-15}{5}\)
so we simplify that
then we get \(\frac{-15}{5}=-3\)
So y=-3

Answer 19

so I was right