How do you find the intercepts for #-3x-4y=18#?

1 Answer
Nov 6, 2015

Answer:

#x=-6#

#y=-9/2#

Explanation:

To find the #x# and #y# intercepts for an equation, you need to put the equation in Standard Form (#y=mx+b#).

Thus, we need to solve #-3x -4y = 18# for #y#.

First, we can add #3x# to both sides. Doing so yields:

=#-4y=18+3x#

or

=#-4y=3x+18#

Then, we need to divide both sides by #-4# to isolate #y#.

#y=(3x+18)/(-4)#

Simplifying slightly gets you:

#y=-(3x)/4-18/4#

which equates to

#y=-(3x)/4-9/2#

Now that we have the equation in Standard form, we can look for the intercepts. Conceptually, an #x#-intercept will occur when #y=0# and a #y#-intercept will occur when #x=0#.

So, simply plug in those values of #0# seperately to solve for each intercept.

Solving for the #x#-intercept:

#(0)=-(3x)/4-9/2#

Adding #(3x)/4# to both sides results in:

#(3x)/4=-9/2#

Multiplying both sides by #4/3# gets:

#x=(-9/2)(4/3)#

or:

#x=-6#

Solving for the #y#-intercept:

#y=-(3(0))/4-9/2#

Canceling out the term with the #0# being multiplied to it gives:

#y=-9/2#

It is worthy to notice that the y-intercept of the equation, when put in Standard Form, is simply the term without the #x# in it (or the #b# term when written as #y=mx+b#).