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

2 Answers
May 10, 2018

#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#

May 10, 2018

#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#