Question

How can you solve for `b` in `y=mx+b`?

Answer 1

`b` tends to be the y intercept which can be worked out given that you have the gradient and any coordinate of a point on that line. Just substitute them values in, rearrange to make b the subject which will give you the y-intercept (`b`).

Explanation:

For example:

`y = 2x + b`

and you know that the coordinate (5, 2) lies on this function. Simple substitute them in and work out your value of b.

`2 = 2(5) + b`
`b = 2-10`
`b = -8`

Therefore your final equation will be: `y = 2x - 8`

Answer 2

`b=y-mx`

Explanation:

Given: `y=mx+b`

The objective is to have just 1 of `b` and for this to be on one side of the equals and everything else on the other side.

To get `b` on its own subtract `color(red)(mx)` from both sides

`color(green)(ycolor(white)("d")=color(white)("d")mx+b color(white)("dddd")-> color(white)("dddd")ycolor(red)(-mx)color(white)("d")=color(white)("d")ubrace(mxcolor(red)(-mx))+b)`
`color(white)("dddddddddddddddddddddddddddddddddd.")darr`

`color(green)(color(white)("ddddddddddddddd")->color(white)("dddd")y-mxcolor(white)("d")=color(white)("dddd")0color(white)("ddd")+b)`

`b=y-mx`