Example 1 :
The scale reading, R(mm) when weights w (grams) are placed on a spring balance is given by the formula R = 0.4w + 5.
Find w when :
(a) R = 27 (b) R = 42
Solution :
(a) Reading R = 27 mm
R
= 0.4w + 5
27 = 0.4w + 5
27 – 5 = 0.4w
22/0.4 = w
w = 55 g
So, the weight (w) is 55 g
(b) Reading R = 42 mm
R
= 0.4w + 5
42 = 0.4w + 5
42 – 5 = 0.4w
37/0.4 = w
w = 92.5 g
So, the weight (w) is 92.5 g
Example 2 :
In the united states of America temperature is measured in degrees Fahrenheit (°F) rather than in degrees Celsius (°C). The rule showing the relationship between these two temperature scales is given by
F = 1.8C + 32
What temperature in °C corresponds to a temperature of :
(a) 40°F (b) 0°F (c) 200°F
Solution :
(a) F = 40°F
F
= 1.8C + 32
40° = 1.8C + 32
40 – 32 = 1.8C
8/1.8 = C
C = 4.44°C
So, temperature is 4.44˚C
(b) F = 0°F
F
= 1.8C + 32
0° = 1.8C + 32
0 – 32 = 1.8C
-32/1.8 = C
C = -17.77°C
So, temperature is -17.8°C
(c) F = 200°F
F
= 1.8C + 32
200° = 1.8C + 32
200 – 32 = 1.8C
168/1.8 = C
C = 93.33°C
So, temperature is 93.3°C.
Example 3 :
The total cost $C, to sink a bore to given by the rule
C = 15d + 350
where d is the depth in metres. How deep a bore can a farmer obtain for a cost of :
a) $2000 b) $3200
Solution :
(a) C = $2000
Depth (d) = ?
By given rule,
C = 15d + 350
2000 = 15d + 350
2000 – 350 = 15d
1650/15 = d
d = 110 m
So, the depth is 110 m.
(b) C = $3200
Depth (d) = ?
By given rule,
C = 15d + 350
3200 = 15d + 350
3200 – 350 = 15d
2850/15 = d
d = 190 m
So, the depth is 190 m.
Example 4 :
Ben walks at a rate of 3 miles per hour. He run at a rate of 6 miles per hour. In one week, the combined distance that he walks and runs is 210 miles.
a) Write a linear model that relates the number of hours that Ben walks to the number of hours Ben runs.
b) Ben runs for 25 hours. For how many hours does he run ?
Solution :
(a) x -----> be the number of hours Ben walks at 3 miles.
y -----> be the number of hours Ben run at 6 miles.
Total distance = 210 miles
By using linear equation,
3x + 6y = 210 -----(1)
(b) y -----> be the number of hours Ben run at 6 miles
y = 25 hours
By applying y = 25 hrs in equation (1), we get
3x + 6y = 210
3x + 6(25) = 210
3x + 150 = 210
3x = 210 – 150
3x = 60
x = 20 hrs
So, he run 20 hours.
Example 5 :
A salesperson receives a base salary of $35000 and a commission of 10% of the total sales for the year.
a) Write a linear model that shows the salesperson’s total income based on total sales for k dollars.
b) If the salesperson sells $250000 worth of merchandise, what is her total income for the year, including her base salary ?
Solution :
(a) k----> be the total sales for the year.
y ----> be the total income.
10 % of total sales for year
= 10/100k
k = 0.1k
Base salary = $35000
By linear model,
y = 0.1k + 35000 -----(1)
(b) k = $250000
By applying k = 250000 in equation (1), we get
y = 0.1k + 35000
y = 0.1(250000) + 35000
y = 25000 + 35000
y = $60000
So, total income for the year is $60,000
Example 6 :
Amery has x books that weigh 2 pounds each and y books that weigh 3 pounds each. The total weight of his books is 60 pounds.
a) Write a linear model that relates the number of 2 pound books to the number of 3 pound books Amery has.
b) If Amery has 10 3-pound books, how many 2 pound books does he have ?
Solution :
(a)
Given,
x is the number of books weigh 2-pound.
y is the number of books weigh 3-pound.
Total weigh of books = 60 pounds.
Then,
2x + 3y = 60 -----(1)
(b) y is the number of books weigh 3 pound.
y = 10 books
By applying y = 10 in equation (1), we get
2x + 3y = 60
2x + 3(10) = 60
2x + 30 = 60
2x = 30
x = 15
So, 15 2-pound books.
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