SOLUTION: how does the graph of the line y=2/3x=1 differ from the graph of the line y=4x+1? a.) y=2/3x+1 is steeper than y=3x-1 b.)y=3x+1 is steeper than y=2/3x+1 c.)y=2/3x+1 and y=3x+1 d

Algebra ->  Linear-equations -> SOLUTION: how does the graph of the line y=2/3x=1 differ from the graph of the line y=4x+1? a.) y=2/3x+1 is steeper than y=3x-1 b.)y=3x+1 is steeper than y=2/3x+1 c.)y=2/3x+1 and y=3x+1 d      Log On


   

Question 746811: how does the graph of the line y=2/3x=1 differ from the graph of the line y=4x+1?
a.) y=2/3x+1 is steeper than y=3x-1
b.)y=3x+1 is steeper than y=2/3x+1
c.)y=2/3x+1 and y=3x+1 do not differ
d.)the y-intercept of y= ex+1 is 2(1/3) unites higher than the y-intercept of y=2/3x+1
e.)the x-intercept of y=3x+1 is 2(1/3) units to the rigfht of the x-intercept of y=2/3x+1
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
first find two points for each line and graph them:
y=%282%2F3%29x%2B1
set x=0 and find y-intercept
y=%282%2F3%290%2B1 => y=1...so, one point is (0,1)
set y=0 and find x-intercept
0=%282%2F3%29x%2B1 => x=-1%2F%282%2F3%29=-3%2F2...so, other point is (-3%2F2,0)
y=4x%2B1
set x=0 and find y-intercept
y=4%2A0%2B1 => y=1...so, one point is (0,1)
set y=0 and find x-intercept
0=4x%2B1 => x=-1%2F4...so, other point is (-1%2F4,0)
now plot these points and draw lines through

compare lines: A higher slope value indicates a steeper incline.
y=%282%2F3%29x%2B1 => a slope is m=2%2F3
y=4x%2B1=> a slope is m%5B1%5D=4
m%5B1%5D=4%29 a higher slope value than m=2%2F3

so, the line y=4x%2B1 is steeper than y=%282%2F3%29x%2B1
answer: b.)