Question

Which of the following coordinates are intercepts of the linear relation
2x- 3y + 30 = 0 ?

I. (0 , 10)
II.(0 , 2/3)
III.(-10 , 0)
IV.(-15, 0)


A. I only
B. I and IV only
C. II and III only
D. II and IV only

This was what I was trying to do:

2(0)-3(10) + 30 = 0
-30y +30 = 0
subtract 30 frm both sides

-30y/-30y = -30/-30

= 0

Im utterly confused.

Answer 1

2x- 3y + 30 = 0 ?

to find x and y intercepts, put equation in the form
ax + by = c
x intercept = (c/a, 0)
y intercept = (0, c/b)

to find x-intercept, y = 0
2x - 3(0) = -30
2x = -30
x = -15
x intercept = (c/a, 0)
x intercept = (-15, 0)

to find y-intercept, x = 0
2(0) - 3y = -30
-3y = -30
y = 10
y intercept = (0, c/b)
y intercept = (0, 10)

B. I, IV

do you understand?

Answer 2

it would be easier to actually find the intercepts, rather than testing the answers

2x - 3y = -30

here is an easy way:
to let x=0, put your "pinkie" over the x term, what do you see ?
In your head you should be able to solve -3y=-30
or y = 10
to let y=0, put your "pinkie over the y term, what do you see?
can you in your head solve 2x = -30 to get
x = -15

so my intercepts are (0,10) and (-15,0)
do we see those?

Answer 3

Oh that makes sense now. thank you both helper and Reiny so much =]