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-25 < 5(w + 3)

-9q + 18 -3 > 6q

19 < 7 + 1/3k + 2 -2/3k

I hate algebra. XD Thanks!!

2006-11-15 02:28:22 · 10 answers · asked by BadRomance 2 in Science & Mathematics Mathematics

10 answers

-25 < 5(w + 3)
-25< 5w +15
-40< 5w
-8
-9q + 18 -3 > 6q
15 > 15q
1> q


19 < 7 + 1/3k + 2 -2/3k
multiply by 3 to get rid of denominators
57 < 21 +k +6 -2k
30< -1k
divide by -1
-30

2006-11-15 02:35:07 · answer #1 · answered by ignoramus 7 · 1 0

- 25 < 5(w + 3)

- 25 < 5w + 15

- 25 - 15 < 5w + 15 - 15

- 40 < 5w

- 40/ 5 < 5w/5

- 8 < w

- - - - - - - -

9q + 18 - 3 > 6q

9q + 15 > 6q

9q + 15 - 9q > 6q - 9q

15 > -3q

15 / - 3 < -3q/-3

- 5 < q

- - - - - - - - - - - - - -

19 < 7 + 1/3k + 2 - 1/3k

19 < 9 + 1/3k - 2/3k

19 - 9 < 9 + 1/3k - 2/3k - 9

10 < 1/3k - 2/3k

10 < - 1/3k

-3(10) > - 3(- 1/3) k

- 30 > k

- - - - - - -s-

2006-11-15 11:31:22 · answer #2 · answered by SAMUEL D 7 · 1 0

-25<5(w+3)
=5(w+3)>-25 =5w+15>-25
=5w>-25-15 =5w>-40 =w>-40/5
=w>-8


-9q+18-3>6q
=6q<-9q+18-3 =6q<-9q+15
=6q+9q<15 =15q<15 q<15/15 =q<1


19 < 7 + 1/3k + 2 -2/3k
7 + 1/3k + 2 -2/3k >19 =7 + 1+ 2 -2>19*3k*3k
=19*3k*3k<8 =19*k*k < 8 / 3 / 3 =k<0.216

2006-11-15 10:47:28 · answer #3 · answered by Nemesis 1 · 1 0

-25 < 5(w + 3)
-25<5w+15
-40<5w
w>-8

-9q + 18 -3 > 6q
-9q+15>6q
15>15q
q<1

19 < 7 + 1/3k + 2 -2/3k
19<9-1/3 k
10<-1/3 k
30<-k
k<-30

2006-11-15 11:25:22 · answer #4 · answered by yupchagee 7 · 1 0

Hmph, inequalities are fairly easy to solve. They act pretty much like an equality, except when you divide by a negative number they swap the inequality sign. Also, doing things like squaring and square rooting does not work as nicely as you would like.

The first is simple- you divide through by 5 to get -5
The second is incredibly easy, simply add 9q to both sides, tidy it up, then divide through by the number equal to q.

The final one is slightly more complex. Tidying it up somewhat we have

10<-1/3k

We really don't want to mulyiply through by k here, as that might change the sign of the inequality (actually unless you meant 1/3*k, in which case this is trivial)

the solution is instead to multiply through by k^2, which is positive gives 10k^2<-1/3K

so we have 3K^2+K<0

so K(3K+1)<0 So either K<0 or 3K+1<0.......

Algebra involves applying simple rules that you would apply to numbers normally.....

2006-11-15 10:42:40 · answer #5 · answered by mister k 1 · 1 0

-25 < 5(w + 3)
-5 -5-3 -8
-9q + 18 -3 > 6q
18-3>6q+9q
15>15q
1>q

19 < 7 + 1/3k + 2 -2/3k
19-7-2<1/3k-2/3k
10<-1/3k
30k<-1
k<-1/30

2006-11-15 10:39:03 · answer #6 · answered by Dupinder jeet kaur k 2 · 1 0

1)-25<5w+15
5w>-40
w>-8

2)-9q+15>6q
15q<15
q<1

3)Combine like terms:
19<-1/3k+9

Multiply both sides both 3 to get rid of the fractions:
3(19)<3(-1/3k+9)
57<-k+27

Subtract both sides by 27:
-k>30
k<-30

I hope that helps!

2006-11-15 10:41:51 · answer #7 · answered by Anonymous · 0 0

-8 < w (or w > -8 if your teacher wants the variable first)

q > 1

-30 > k (or k < -30)

2006-11-15 10:36:55 · answer #8 · answered by jatass 2 · 1 0

w+3>-5
w>-8

15q<15
q<1

-1/3k>10
k<-30

2006-11-15 10:36:42 · answer #9 · answered by . 3 · 1 0

I've never needed it in my life. I hate it too! Now go study.

2006-11-15 10:33:45 · answer #10 · answered by Anonymous · 1 1

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