Someone please solve this for me:
2(x+3)-5(2x+1)
and explain how you solved it.
Thank you!
2007-08-19 12:38:16 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
2(x+3)-5(2x+1)
2x+6-10x-5 = -8x+1
multiply the numbers through and group like terms
2007-08-19 12:45:02 · answer #1 · answered by bradlelf 2 · 2⤊ 0⤋
You first start by multiplying out the bracketed parts of equation:
from 2(x+3)-5(2x+1) =2x +6 -10x -5
then add/subtract like numerals/expressions
so then 2x +6 -10x -5
= 6 -5 +2x -10x
= 1 -8x
and to solve for x; 1 -8x
8x = 1
x = 1/8 = 0.125
2007-08-19 13:18:33 · answer #2 · answered by no_einstein 4 · 0⤊ 0⤋
2 (x + 3) - 5 (2X + 1)
2x + 6 - 10x -5
-8x + 1
2007-08-19 12:47:34 · answer #3 · answered by Sweet Pea 3 · 0⤊ 0⤋
2(x + 3) - 5(2x + 1)
2x + 6 - 10x - 5
-8x + 1
2007-08-19 12:45:01 · answer #4 · answered by Anonymous · 0⤊ 0⤋
= 2x + 6 - 10x - 5
= 6 - 5 + 2x - 10x
= 1 - 8x
2007-08-22 07:40:52 · answer #5 · answered by Como 7 · 0⤊ 0⤋
I guess you forgot the = sign.
2(x+3) = 5(2x+1)
2x + 6 = 10x + 5
so 8x = 1
x = 1/8
2007-08-19 12:45:52 · answer #6 · answered by Steve 2 · 0⤊ 0⤋
2(x+3)-5(2x+1) You need to distribute and then add like terms.
2x+6-10x-5 Now add like terms.
2007-08-19 12:44:01 · answer #7 · answered by Ed S 4 · 0⤊ 0⤋
to discover which of A,B C,D,E might desire to be actually shall we glance at the three opportunities given above. in accordance to the reality that then M being equidistant from X and Y determination i might desire to be actually i.e. If M is the middle of the cirlce the XM and YM regularly is the radi and equivalent lenght so i might desire to be actually. on condition that M is placed so as that XM and YM are equivalent lengths exhibits that M could be a element on the arc XY such that it quite is indoors the midst of the arc XY and that doesn't violate determination II and so it would desire to be actually. If M is exterior the cirlcle then it would desire to regardless of the reality that be equidistant from the climate X and Y and so determination III is frequently actually. on condition that opportunities I & II & III might desire to all be in all likelihood actually ( not simulataneously regardless of the reality that) so E is the wonderful answer.
2016-11-12 22:44:39 · answer #8 · answered by Anonymous · 0⤊ 0⤋