Posted by Jeff on Tuesday, February 17, 2009 at 10:49pm.
Uhh im not sure if this is right but i think you should first put it in slope intercept form.
then make an X and Y chart to find solutions.
y=4x-7
(0,-7)
(1, -3)
(2, 1)
Would (2,1) (3,5) and (4,9) work as well? Thanks!
yea i got (2,1) in my first post. Yeah, they should all work.
Even though you switched it to the y = 4x - 7? Is this all the same or is the -7 and 7 the constants changed? Thanks
no, when i put in slope intercept form i got -7
the original equation was
4x-y=7
subtract 4x to both sides
-y=-4x+7
you cant have a -y so move the negative over, this will change the sign of everything
y=4x-7
if you had 4x+y=7
then the slope intercept form of that would be
y=-4x+7
So which is the correct method? Switching the original, or just using variables as i did to make the equation true
4x-y = 7
4(2) - 1 = 7 (2,1)
4(3) - 5 = 7 (3,5)
4(4) - 9 = 7 (4,9)
Which method is correct? Or does it not matter because the equation has infinate solutions? Thanks!
Both method is fine.
Are you a math teacher? Just wondering?
no im not. but im in alg 2
That makes sense..High School?
yeah high school.
Ok! Have a good night and thanks for the help!