# MODELING WITH LINEAR EQUATIONS WORD PROBLEMS

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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