# SYSTEM OF EQUATIONS - SOLVING, WORD PROBLEMS MADE EASY

Please study

Linear Equations in Two Variables before System Of Equations

if you have not already done so.

There, we explained the method of solving the equations, with examples.
That knowledge is a prerequisite here.

Here, we apply that knowledge to solve word problems.

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### Example 1 of System Of Equations

Solve the following Word Problem on System Of Equations.

The area of a rectangle gets reduced by 8 m2, if its length
is reduced by 5 m and breadth increased by 3 m. If we increase
the length by 3 m and breadth by 2 m, the area is increased
by 74 m2. Find the length and breadth of the rectangle.

Solution to Example 1 of System Of Equations :

Let l m and b m be the length and breadth of the rectangle respectively.
Then its area is lb m2.

Reducing length by 5 m and increasing breadth by 3 m, the area becomes
(l - 5) x (b + 3) m2 = l(b + 3) - 5(b + 3) = lb + 3l - 5b - 15.
By data this is 8 m2 less than the original area, lb m2
lb + 3l - 5b - 15 = lb - 8
Cancelling lb which is present on both sides, we get
3l - 5b - 15 = -8 ⇒ 3l - 5b = -8 + 15 = 7.
3l - 5b = 7............................................................(i)

Increasing length by 3 m and breadth by 2 m, the area becomes
(l + 3) x (b + 2) m2 = l(b + 2) +3(b + 2) = lb + 2l + 3b + 6.
By data this is 74 m2 more than the original area, lb m2
lb + 2l + 3b + 6 = lb + 74
Cancelling lb which is present on both sides, we get
2l + 3b + 6 = 74 ⇒ 2l + 3b = 74 - 6 = 68.
2l + 3b = 68............................................................(ii)

Equations (i) and (ii) are the Linear Equations
in two variables formed by converting the given
word statements to the symbolic language.

Now we have to solve these simultaneous equations.
To solve (i) and (ii), let us make the coefficients of b the same.
(i) x 3 gives 9l - 15b = 21............................................................(iii)
(ii) x 5 gives 10l + 15b = 340.....................................................(iv)
(iii) + (iv) gives 9l + 10l = 21 + 340 ⇒ 19l = 361
l = 361⁄19 = 19.
Using this in (ii), we get
2(19) + 3b = 68 ⇒ 38 + 3b = 68 ⇒ 3b = 68 - 38 = 30
b = 30⁄3 = 10.
Thus Length and Breadth of the Rectangle
are 19 m and 10 m respectively. Ans.

Check to Example 1 of System Of Equations:
Area = Length x Breadth = 19 m x 10 m = 190 m2
If Length is reduced by 5 m and Breadth is increased by 3 m,
Area becomes (19 - 5) x ( 10 + 3) = 14 x 13 = 182 m2
which is 8 m2 less than 190 m2. (verified.)

If Length is increased by 3 m and Breadth is increased by 2 m,
Area becomes (19 + 3) x ( 10 + 2) = 22 x 12 = 264 m2
which is 74 m2 more than 190 m2. (verified.)

Thus the Solution to Example 1 of
System Of Equations is verified.

### Example 2 of System Of Equations

Solve the following Word Problem on System Of Equations.

A 90% acid solution is mixed with a 97% acid solution to obtain
21 litres of a 95% solution. Find the quantity of each of the
solutions to get the resultant mixture.

Solution to Example 2 of System Of Equations :

Let x litres be 90% acid solution and y litres be 97% acid solution.
By data, x + y = 21...................................(i)
Also by data, 90%of x + 97% of y = 95% of 21
⇒ 0.90x + 0.97y = (0.95) x 21
As there are 2 digits ( maximum) after the decimal place,
we multiply both sides by 102 = 100 to get
90x + 97y = 95 x 21...................................(ii)

Equations (i) and (ii) are the Linear Equations
in two variables ( System Of Equations )
formed by converting the given
word statements to the symbolic language.

Now we have to solve these simultaneous equations.
To solve Equations (i) and (ii), Let us make the coefficients of y, same.
(i) x 97 gives 97x + 97y = 97 x 21...................................(iii)
(iii) - (ii) gives 97x - 90x = 97 x 21 - 95 x 21 ⇒ 7x = (97 - 95) x 21
7x = 2 x 21 = 42 ⇒ x = 42⁄7 = 6.
Using this in (i), we get
6 + y = 21 ⇒ y = 21 - 6 = 15.
Required 90% acid solution = 6 litres. Ans.
and required 97% acid solution = 15 litres. Ans.

Check to Example 2 of System Of Equations:
6 + 15 = 21 (verified.)
90%of 6 + 97% of 15 = 0.9 x 6 + 0.97 x 15 = 5.4 + 14.55 = 19.95
95% of 21 = 0.95 x 21 = 19.95
90%of 6 + 97% of 15 = 19.95 = 95% of 21 (verified.)

Thus the Solution to Example 2 of
System Of Equations is verified.

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### Example 3 of System Of Equations

Solve the following Word Problem on System Of Equations.

6 men and 8 boys can finish a piece of work in 14 days while
8 men and 12 boys can do it in 10 days. Find the time taken by
one man alone and that by one boy alone to finish the work.

Solution to Example 3 of System Of Equations :

Let the time taken by one man alone to finish the work be x days.
and the time taken by one boy alone to finish the work be y days.
⇒ In one day, the work that can be done by one man is 1⁄x
and in one day, the work that can be done by one boy is 1⁄y
The work that can be done by 6 men and 8 boys in 14 days
= 6 x (1⁄x) x 14 + 8 x (1⁄y) x 14 = (2 x 14)(3⁄x + 4⁄y)
By data, this = 1. (Since 6 men and 8 boys in 14 days finish the work.)
∴ 28(3⁄x + 4⁄y) = 1 ⇒ 3⁄x + 4⁄y = 1⁄28
Let us denote (1⁄x) by p and (1⁄y) by q.
The above equation becomes 3p + 4q = 1⁄28...........................(i)

The work that can be done by 8 men and 12 boys in 10 days
= 8 x (1⁄x) x 10 + 12 x (1⁄y) x 10 = (4 x10)(2⁄x + 3⁄y)
By data, this = 1. (Since 8 men and 12 boys in 10 days finish the work.)
∴ 40(2⁄x + 3⁄y) = 1 ⇒ 2⁄x + 3⁄y = 1⁄40
Since we denoted (1⁄x) by p and (1⁄y) by q,
the above equation becomes 2p + 3q = 1⁄40...........................(ii)

Equations (i) and (ii) are the Linear Equations
in two variables formed by converting the given
word statements to the symbolic language.

Now we have to solve these simultaneous equations.
To solve Equations (i) and (ii), Let us make the coefficients of q, same.
(i) x 3 gives 9p + 12q = 3⁄28..........................(iii)
(ii) x 4 gives 8p + 12q = 4⁄40 = 1⁄10..........................(iv)
(iii) - (iv) gives 9p - 8p = 3⁄28 - 1⁄10 ⇒ p = 3⁄28 - 1⁄10
Multiplying both sides with 140 (L.C.M. of denominators 28 and 10), we get
140p = 140(3⁄28) - 140(1⁄10) = 5(3) - 14 = 15 - 14 = 1 ⇒ p = 1⁄140
Using this in (ii), we get
2(1⁄140) + 3q = 1⁄40 ⇒ 3q = 1⁄40 - 1⁄70
Multiplying both sides with 280 (L.C.M. of denominators 40 and 70), we get
280(3q) = 280(1⁄40) - 280(1⁄70) = 7 - 4 = 3 ⇒ 280q = 1 ⇒ q = 1⁄280
But p and q stand for 1⁄x and 1⁄y respectively.
p = 1⁄x = 1⁄140 ⇒ x = 140
and q = 1⁄y = 1⁄280 ⇒ y = 280
Thus time taken by one man alone to finish the work = 140 days.Ans.
and time taken by one boy alone to finish the work = 280 days.Ans.

Check to Example 3 of System Of Equations:
Work by 6 men and 8 boys in 14 days
= 6(1⁄140) x 14 + 8(1⁄280) x 14 = 3⁄5 + 2⁄5 = 5⁄5 = 1 (verified.)
Work by 8 men and 12 boys in 10 days
= 8(1⁄140) x 10 + 12(1⁄280) x 10 = 4⁄7 + 3⁄7 = 7⁄7 = 1 (verified.)

Thus the Solution to Example 3 of
System Of Equations is verified.

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### Exercise on System Of Equations

Solve the following Word Problems on System Of Equations.

1. A jeweller has bars of 18-carat gold and 12-carat gold. How much
of each must be melted together to obtain a bar of 16-carat gold,
weighing 120 gm ? It is given that pure gold is 24 carat.
2. 4 men and 4 boys can do a piece of work in 3 days; while 2 men
and 5 boys can finish it in 4 days. How long would it take
to do it by (i) 1 boy ? (ii) 1 man ?

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### Answers to Exercise on System Of Equations

The Answers to the Word Problems on
System Of Equations given
in the exercise above, are given below.

1. 80 gm of 18-carat gold and 40 gm of 12-carat gold
2. 36 days, 18 days