SOLUTION: Tell what you must do the first inequality in order to get second. Be sure to list all the steps.show me how to Solve it 1.)4j+5>=23;j>=4.5 2.) 2(q-3)<8;

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Question 27881: Tell what you must do the first inequality in order to get second. Be sure to list all the steps.show me how to Solve it
1.)4j+5>=23;j>=4.5

2.) 2(q-3)<8;q<7

3.) 8-4s>16;-4s>8

4.) -8>z/-5-2;30

5.) 2y-5>9+y;y>14

6.) 2/3g+7>=9;2/3g>=2]

7.) 6<12-s;s<6

8.) 3+5t>=6(t-1)-t;3>=-6

9.) 6.2<-r;-6.2>r

Solve it and graph it

10.) 2x-2>4

11.) 2-2x>4

12.) 2x+2>4

13.) 2x+2>4x

14.) -2x-2>4

15.) -2(x-2)>4

Solve each inequality. Graph the solutions on a number line.
17.) 5<=11+3h

18.) 3(y-5)>6

19.) -4x-2<8

20.) r+6+3r>=15-2r

21.) 5-2n<=3-n

22.) 3(2v-4)<=2(3v-6)

23.) 2(m-8)-3m<-8

24.) -(6b-2)>0

25.) 7a-(9a+1)>5

pls send me the answer and how to do it really quickly
Answer by sdmmadam@yahoo.com(530) (Show Source):
You can put this solution on YOUR website!
Graphs/27881 (2006-02-20 01:56:31): Tell what you must do the first inequality in order to get second. Be sure to list all the steps.show me how to Solve it
1.)4j+5>=23;j>=4.5
4j+5>=23
4j>=(23-5) (subtracting 5 from both the sides)
4j>=18
j>=18/4 (dividing by 4>0, multip or dividing by a positive quantity does not alter the inequality sign)
j>=(2X9)/(2X2) = 9/2 = 4.5
Therefore j>=4.5
2.) 2(q-3)<8;q<7
2(q-3)<8
2q-6<8
2q<8+6 (adding 6 to both the sides of the inequality)
2q<14
q<14/2
That is q <7

3.) 8-4s>16;-4s>8
8-4s>16
8-4s+4s> 16+4s (adding 4s to both the sides)
8+0 > 16+4s
8>16+4s
8-16 > +16+4s -16 (subtracting 16 from both the sides)
-8> 4s+16-16 (using additive commutativity on the RHS)
-8>4s+0
-8>4s
4s <-8 (Adam older than Tom is the same as Tom younger to Adam)
Dividing by 4>0
(and multiplying or dividing an inequality by a postitive quantity keeps the inequality unaltered.)
s<(-2)

4.) -8>z/(-5)-2
-8>(z/-5)-2
-8+2 >(z/-5) -2+2 (adding 2 to both the sides)
-6>z/(-5) +0
-6>z/(-5)
Multiplying both the sides by (-5)
(-6)X(-5) < z
(why change in the inequality?
It is because mulltiplication or division by a negative quantity alters the inequality)
That is 30 or z>30

5.) 2y-5>9+y;y>14
2y-5>9+y
2y-5-y >9+y-y (subtracting y from both the sides or adding -y to both the sides)
2y-y-5> 9+0 (using additive associativity,commutativity on the LHS)
y-5>9
y -5+5 >9+5 (adding 5 to both the sides)
y+0 >14
y>14

6.) 2/3g+7>=9;2/3g>=2]
2/3g+7>=9
2/3g+7-7 >=9-7 (adding (-7) to both the sides)
2/3g+0 >=2
2/3g >=2
[g>=2X3/2 (multiplying by 3/2>0)
g>=3]

7.) 6<12-s;s<6
6<12-s
6+s<12-s+s (adding s to both the sides)
6+s<12+0
6+s<12
-6+6+s< -6+12 (adding (-6) to both the sides)
0+s <6
s<6
8.) 3+5t>=6(t-1)-t;3>=-6
Is the problem 1)3+5t>=6(t-1) OR 2)3+5t>=6(t-1)-t
1) 3+5t>=6(t-1)
3+5t>=6t-6
3+6 >=6t-5t (you can simultaneously add 6 to both the sides and subtract 5t from both the sides, in fact it is like things being transfered from one side to another , changing sign while changing side)
9>=t
2) 3+5t>=6(t-1)-t
3+5t>=6t-6-t
3+5t>=6t-t-6
3+5t>=5t-6
3+5t-5t>= 5t-6-5t
3+0>=-6+5t-5t
3>=-6+0
3>=-6
Note as 3 cannot be equal to -6
we have 3>-6

9.) 6.2<-r;-6.2>r
6.2<-r
Multiplying by (-1) on both the sides, we have
-6.2 > (-1)X(-r) [multiplication by a negative quantity alters the inequality]
-6.2>+r
-6.2>r
Solve it and graph it
I do not know todo the technique of doing graphs on your answer sheet.
My graphs are not getting copied to the answer sheet.
10.) 2x-2>4
Dividing by 2>0 (dividing by a positive quantity does not alter the inequality)
x-1>2
x-1+1>2+1 (adding 1 on both the sides)
x+0>3
x>3
11.) 2-2x>4
Dividing by 2>0 (dividing by a positive quantity does not alter the inequality)
1-x>2
-2+(1-x)+x>-2+(2)+x
(as you need to transfer 2 from the right to left and (-x ) from the left to right,we add-2 from the left on both the sides and +x from the right on both the sides)
(-2+1)+(-x+x) >[-2+2] +x (by additive associativity on both the sides)
-1+0>0+x
-1>x
x<-1
Note: If you are asked to every step with this particular problem in focus, then do as above.
But otherwise if such an inequality is a portion of a bigger problem then you may do it quickly as follows:
2-2x>4
1-x>2 (Dividing by2>0)
1-2>+x
(you may transfer things from one side to another changing sign for the quantity
everytime while changing side)
-1>x
x<-1 (using a>b implying b 12.) 2x+2>4
x+1>2 (dividing by 2)
x>2-1
x>1
13.) 2x+2>4x
x+1>2x (dividing by 2)
1>2x-x
1>x
x<1
14.) -2x-2>4
-x-1>2
-x>2+1
-x>3
x<-3 (Multiplying by (-1) and multiplication by a negative quantity alters the inequality)
15.) -2(x-2)>4
-1(x-2)>2
-x+2>2
-x>2-2
-x>0
(-1)X (-x)<(-1)X0 (multiplying by (-1) and multiplication by a negative quantity alters the inequality)
x<0
Solve each inequality. Graph the solutions on a number line.
17.) 5<=11+3h
5-11<=3h
-6<=3h
-2<=h (dividing by 3>0)
h>=-2 (using a<=b means b>=a)
18.) 3(y-5)>6
(y-5)>2 (dividing by3>0)
y>2+5
y>7
19.) -4x-2<8
-2x-1<4 (dividing by 2>0)
-2x<4+1
-2x<5
x>5/(-2) (dividing by (-2) and dividing by a negative quantity alters the inequality)
x>(-5/2)
20.) r+6+3r>=15-2r
r +3r +2r >=15-6
(transferring (-2r) from the right to the left and transferring 6 from the left to the right)
6r>=9
r>=9/6
r>=3/2
21.) 5-2n<=3-n
5-3<=2n-n (transferring 3 from the right to the left and (-2n) from the left to the right )
2<=n
That is n>=2
22.) 3(2v-4)<=2(3v-6)
6v-12<=6v-12
Note: (6v-12) cannot be less than itself
Therefore 6v-12 = 6v-12 alone holds and this is true for all values of v
23.) 2(m-8)-3m<-8
2m-16 -3m<-8
2m-3m-16<-8 (additive commutativity and associativity)
-m<-8+16
-m<8
m>-8 (multiplying by (-1) multiplication by a negative quantity alters the inequality)
24.) -(6b-2)>0
-6b +2 >0
2>6b (transferring (-6b) from the left to the right )
Dividing by 6
2/6>b
1/3>b
That is b<1/3

25.) 7a-(9a+1)>5
7a-9a-1>5
-2a>5+1
-2a>6
a<6/(-2) ( dividing by (-2) amd dividing by a negative quantity alters the inequality)
a<-3