# SOLVING LINEAR EQUATIONS WORD PROBLEMS

Problem 1 :

On selling the product x at 5% gain and the product y at 10% gain, a store owner gains \$2000. But if he sells the product x at 10% gain and the product y at 5% loss, he gains \$1500 on the transaction. Find the actual prices of the product x and the product y.

Solution :

Let x and y be the selling prices of the products x and y respectively.

Given : On selling the product x at 5% gain and the product y at 10% gain, the store owner gains \$2000.

Then, we have

0.05x +  0.1y  =  2,000

To get rid of the decimal, multiply each side by 100.

5x + 10y  =  200,000

Divide each side by 5.

x + 2y  =  40,000 -----(1)

Given : If x is sold at 10% gain and y at 5% loss, the gain is \$1500.

0.1x - 0.05y  =  1,500

To get rid of the decimal, multiply each side by 100.

10x - 5y  =  150,000

Divide each side by 5.

2x - y  =  30,000 -----(2)

In order to eliminate y in (1) and (2), add (1) and 2 times of (2).

(1) + 2(2) :

5x  =  100,000

Divide each side by 5.

x  =  20,000

Substitute 20,000 for x in (2).

(2)-----> 2(20,000) - y  =  30,000

40,000 - y  =  30,000

Subtract 40,000 from each side.

- y  =  - 10,000

Multiply each side by (-1).

y  =  10,000

So, the actual prices of x and y are \$20,000 and \$10,000 respectively.

Problem 2 :

Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5. Find the numbers.

Solution :

Let x and y be the two numbers.

Given : Two number are in the ratio 5 : 6.

Then, we have

x / y  =  5 / 6

6x  =  5y

6x - 5y  =  0 -----(1)

Given : If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5.

Then, we have

(x - 8) / (y - 8)  =  4 / 5

5(x - 8)  =  4(y - 8)

5 x - 40  =  4y - 32

5x - 4y  =  -32 + 40

5x - 4y  =  8 ----(2)

In order to eliminate y in (1) and (2), subtract 5 times (2) from 4 times (1).

4(1) - 5(2) :

x  =  40

Substitute 40 for x in (2).

(2)-----> 5(40) - 4y  =  8

200 - 4y  =  8

Subtract 200 from each side.

- 4y  =  - 192

Divide each side by (-4).

y  =  48

So, the required numbers are 40 and 48.

Problem 3 :

4 Indians and 4 Chinese can do a piece of work in 3 days. While 2 Indians and 5 Chinese can finish it in 4 days. How long will it take for 1 Indian to do it? How long will it take for 1 Chinese to do it ?

Solution :

Let x and y be the number of days taken by each Indian and each Chinese.

Then,

Work done by 1 Indian in 1 day  =  1 / x

Work done by 1 Chinese in 1 day  =  1 / y

Given : 4 Indians and 4 Chinese can complete the work in 3 days.

Then,

Work done by 4 Indians and 4 Chinese in 1 day  =  1 / 3

That is,

4/x + 4/y  =  1/3

Let a  =  1/x and b  =  1/y.

4a + 4b  =  1/3 ----(1)

Given : 2 Indians and 5 Chinese can finish it in 4 days

Then,

Work done by 2 Indians and 5 Chinese in 1 day  =  1 / 4

That is,

2/x + 5/y  =  1/4

2a + 5b  =  1/4 ----(2)

In order to eliminate a in (1) and (2), subtract 2 times (2) from (1).

(1) - 2(2) :

- 6b  =  1/3 - 1/2

- 6b  =  2/6 - 3/6

- 6b  =  (2 - 3)/6

- 6b  =  -1 / 6

b  =  1 / 36

Substitute 1/36 for b in (1).

(1)-----> 4a + 4(1/36)  =  1/3

4a + 1/9  =  1/3

4a  =  (1/3) - (1/9)

4a  =  (3/9) - (1/9)

4a  =  (3 - 1)/9

4a  =  2 / 9

a  =  1 / 18

Find the values of x and y from the values of a and b.

 a  =  1/181/x  =  1/18x  =  18 b  =  1/361/y  =  1/36y  =  36

So, one Indian will take 18 days to complete the work and one Chinese will take 36 days to complete the same work. Apart from the stuff given in this section if you need any other stuff in math, please use our google custom search here.

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