Algebra Problems? Help?
1) Where does y= 1/3x -12 cross the y-axis?
2) What is the x-intercept of the line defined by 4x-7y=24
3) Does the point (1, -1/4) lie on the line defined by 4x+8y=-2
Please answer any of these questions. You don't have to answer all of them; at least just one. I would prefer all of them answered. Please explain how you got your answer too. Thank you. ^.^
Comments
Hi,
1) Where does y= 1/3x -12 cross the y-axis?
Let x = 0. The equation becomes:
y= 1/3x -12
y= 1/3(0) -12
y = 0 - 12
y = -12
It crosses at (0,-12). <==ANSWER
2) What is the x-intercept of the line defined by 4x-7y=24
To find the x intercept, let y = 0, and solve.
4x - 7y = 24
4x - 7(0) = 24
4x - 0 = 24
4x = 24
x = 6
The x intercept is (6,0). <==ANSWER
3) Does the point (1, -1/4) lie on the line defined by 4x+8y=-2
4x + 8y = -2
4(1) + 8(-1/4) = -2
4 + -2 = -2
2 = -2 <==not true, so (1, -1/4) is not a solution.
I hope that helps!! :-)
1)y=(1/3)x-12
(1/3)x-12=0 (set equation equal to 0)
(1/3)x=12 (isolate the x)
x=36
2)4x-7y=24
4x-7(0)=24 (substitute y for 0)
4x=24
x=6
3)4x+8y=-2
4(1)+8(-1/4)=-2 (Substitute values)
4-2=-2 (false statement because 4-2=2)
1) -12
2) 6
3) 4 - 2 = 2
No it doesn't
for the first one you take x = 0
y = -12
so the line will cross y axis at ( 0, -12)
number 1 is in y= mx + b (slope intercept form). m is the slope and b would be the y- intercept. So wouldnt -12 be the y-intercept? and 1/3 be the slope? I think thats what you're asking, im not sure if i answered it?