6. One variable only so pretty straightforward.
length-x+4
width-x
x+x+4=80
2x=76
x=38
x+4=42
answer: length 42cm and width 38cm
7. Another money problem!
n-# of nickels
q-# of quarters
n=3+q
0.05n+0.25q=2.85
Substitution works like a charm!
0.05(3+q)+0.25q=2.85
0.15+0.05q+0.25q=2.85
0.3q=2.7
q=9
n=3+q
n=3+9
n=12
answer: 9 nickels and 12 quarters
8. One variable situation again.
Ann's money-2b+9
Betty's money-b
b+2b+9=60
3b=51
b=17
2b+9=43
answer: Ann has $43 and Betty has $17.
9. # of red m&m's-x+1
# of blue m&m's-x
x+1+x=13
x=6
x+1=7
answer: 6 blue and 7 red m&m's
10. a-number of adult tickets
s-number of student tickets
a+s=785 ----> a=785-s
5a+2s=3280
5(785-s)+2s=3280
-3s=-645
s=215
a+s=785
a+215=785
a=570
answer: 215 children tickets and 570 adult tickets
Answer:
1300 and 700 respectively.
Step-by-step explanation:
Let x be invested in first account and y be invested in second account.
ATQ, x+y=2000 and 101=(4)*x/100+7*y/100. Solving it will give us x=1300 and y=700
1. 5%
2. That’s the only answer I have, sorry.
the second one because the exddddsedfffdffdggf
