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miababygirl1987
Group Title
Anyone Know how to do this kind of work that will help me through these problems would be really helpful!!!!!!!! 7( x  1 ) ≥ 14
 2 years ago
 2 years ago
miababygirl1987 Group Title
Anyone Know how to do this kind of work that will help me through these problems would be really helpful!!!!!!!! 7( x  1 ) ≥ 14
 2 years ago
 2 years ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.1
divide by 7 get \[x1\geq 2\] then add 1 to both sides to get \[x\geq 1\]
 2 years ago

saifoo.khan Group TitleBest ResponseYou've already chosen the best response.0
\[x1 \ge  \frac{14}{7}\] \[x1 \ge  2\] Add 1 on both sides, \[x \ge 2 +1\] \[x \ge Answer\]
 2 years ago

miababygirl1987 Group TitleBest ResponseYou've already chosen the best response.0
3( x  4 )  2( x + 4) ≤ 9
 2 years ago

viniterranova Group TitleBest ResponseYou've already chosen the best response.1
7x7>=14; 7x>=14+7; x>=7/7; x>=1
 2 years ago

miababygirl1987 Group TitleBest ResponseYou've already chosen the best response.0
3( x  4 )  2( x + 4) ≤ 9
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[x\geq 1\] as before
 2 years ago

miababygirl1987 Group TitleBest ResponseYou've already chosen the best response.0
5( x + 1)  3( x  1) > 9
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[x\geq \frac{17}{2}\]
 2 years ago

viniterranova Group TitleBest ResponseYou've already chosen the best response.1
3x+122x8=<9; 5x+4=<9; 5x=<49; 5x=<13; (x1) x>=13/5
 2 years ago

miababygirl1987 Group TitleBest ResponseYou've already chosen the best response.0
4(x + 3) > 6(x  5)
 2 years ago

viniterranova Group TitleBest ResponseYou've already chosen the best response.1
4x+12>6x30; 4x6x>3012; 2x>42 (x1) x<42/2; x<21
 2 years ago

Chlorophyll Group TitleBest ResponseYou've already chosen the best response.0
x 1 >= 2 >x >= 1
 2 years ago

miababygirl1987 Group TitleBest ResponseYou've already chosen the best response.0
3x + 5 < 5
 2 years ago

Wolf77 Group TitleBest ResponseYou've already chosen the best response.0
In general you want to approach inequalities the same way that you would a simple equality however when you multiply or divide by a negative then you must change the direction of the inequality. i.e.) x > 5 becomes x < 5.
 2 years ago

miababygirl1987 Group TitleBest ResponseYou've already chosen the best response.0
ok for 3x+ 5< 5 I got x ≥ 3 is that right ?
 2 years ago

miababygirl1987 Group TitleBest ResponseYou've already chosen the best response.0
sorry 5(x + 3) ≥ 0 I got x ≥ 3
 2 years ago
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