(4/3)(x + 1) <_ (1/2)(x - 3) I am trying to figure out whether or not the problem is less than or equal to.
2006-09-10 17:06:12 · 3 answers · asked by prettygirl7641 1 in Science & Mathematics ➔ Mathematics
Multiply everything by 6 to get rid of the fractions.
6(4/3)(x+1) <_ 6(1/2)(x-3)
8(x+1) <_ 3(x-3)
Distribute the 8 on the left side and then distribute the 3 on the right side
8x + 8 <_ 3x - 9
Subtract 3x from each side and subtract 8 from each side
5x <_ -17
Divide both sides by 5
x <_ -17/5
Answer: (-infinity, -17/5]
Note: A bracket is used this time because the sign is less than or equal to.
2006-09-10 17:25:38 · answer #1 · answered by MsMath 7 · 1⤊ 1⤋
(4/3)(x + 1) <= (1/2)(x - 3)
(4/3)x + (4/3) <= (1/2)x - (3/2)
Multiply everything by 6
8x + 8 <= 3x - 9
5x <= -17
x <= (-17/5)
In interval notation
(-infinity,-3.4]
2006-09-11 01:55:10 · answer #2 · answered by Sherman81 6 · 0⤊ 0⤋
x<1/11 is answer
2006-09-11 00:13:52 · answer #3 · answered by free aung san su kyi forthwith 2 · 0⤊ 1⤋