1. Joe is 5 years older than Sally. Represent Joe's age in terms of Sally's age.

2. The length of a rectangle is 3 more than twice its width. The perimeter is 48 feet.

3. Fifteen more than four times a number is 6 more than five times the number.

4. Martha and Sally had lunch together on Tuesday. Sally paid $2 more than Martha. The total bill was $5.64.

5. Bob paid $3 less than Tim at a local pizza parlor. Together they spent $7.80.

6. There were six more women on a committee than men. The committee was made up of 30 people.

7. Three consecutive integers have the sum of 153.

Question

1.	Joe is 5 years older than Sally. Represent Joe's age in terms of Sally's age.
 
2.	The length of a rectangle is 3 more than twice its width. The perimeter is 48 feet.
 
3.	Fifteen more than four times a number is 6 more than five times the number.
 
4.	Martha and Sally had lunch together on Tuesday. Sally paid $2 more than Martha. The total bill was $5.64.
 
5.	Bob paid $3 less than Tim at a local pizza parlor. Together they spent $7.80.
 
6.	There were six more women on a committee than men. The committee was made up of 30 people.
 
7.	Three consecutive integers have the sum of 153.

anonymous · Answer

1. sally age =x

joe age=x+5

anonymous · Answer

there up i put the answers for these

angela210793 · Answer

1-      joe's age   y=x+5 where x=sally's age
2-      l=3+2w
3-      15+4x=6++5x
4-       2x+x=5.64 where X=amount in $
5-       (x-3)+x=7.80 
6-       x no. of men so : 6x+x=30
7-       x+(x+1)+(x+2)=153

Now these r too easy I guess u can solve them by urself :)