Equations are frequently used to solve practical problems.

The steps involved in the method of solving an algebra word problem are as follows.

STEP 1:

Read the problem carefully and write down what is given and what is required.

STEP 2:

Select a letter or letters, say x (yy) to represent the unknown quantity(s) requested.

STEP 3:

Represent step by step the verbal statements of the problem in symbolic language.

STEP 4:

Find quantities that are equal under the given conditions and form an equation or equations.

STEP 5:

Solve the equation(s) obtained in step 4.

STEP 6:

Check the output to make sure that your answer satisfies the problem’s requirements.

EXAMPLE 1 (on linear equations in one variable)

Problem Statement:

One fifth of a number of butterflies in a garden are on jasmines and one third of them are on roses. Three times the difference of butterflies in jasmines and roses are in lilies.

solution to the problem:

Let x be the number of butterflies in the garden.

According to data, Number of butterflies on jasmine = x/5. Number of butterflies on roses = x/3.

Then difference of butterflies on jasmine and roses = x/3 -x/5

According to data Number of butterflies in lilies = Three times the difference of butterflies in jasmine and roses = 3(x/3 – x/5)

According to the data, Number of butterflies flying freely = 1.

So, number of butterflies in the garden = x = Number of butterflies on jasmine + Number of butterflies on roses + Number of butterflies on lilies + Number of butterflies flying freely = x/5 + x/3 + 3(x/3 – x /5) + 1.

So, x = x/5 + x/3 + 3(x/3) – 3(x/5) + 1.

This is the linear equation formed by converting the given word statements to symbolic language.

Now we have to solve this equation.

x = x/5 + x/3 + x – 3x/5 + 1

Canceling x which is present on both sides, we get

0 = x/5 + x/3 – 3x/5 + 1

LCM of the denominators 3.5 is (3)(5) = 15.

Multiplying both sides of the equation by 15, we get

15(0) = 15(x/5) + 15(x/3) – 15(3x/5) + 15(1) that is, 0 = 3x + 5x – 3(3x) + 15.

that is, 0 = 8x – 9x + 15, that is, 0 = -x + 15, that is, 0 + x = 15, that is, x = 15.

Number of butterflies in the garden = x = 15. Years.

Check:

Number of butterflies on jasmine = x/5 = 15/5 = 3.

Number of butterflies on roses = x/3 = 15/3 = 5.

Number of butterflies on lilies = 3(5 – 3) = 3(2) = 6.

Number of butterflies flying freely = 1.

Total butterflies = 3 + 5 + 6 + 1. = 15. Same as answer (verified).

EXAMPLE 2 (on linear equations in two variables)

Problem Statement:

A and B each have a certain number of marbles. A says to B, “if you give me 30, I’ll have twice what you have left.” B replies “if you give me 10, I will have triple what you have left”. How many marbles does each have?

Solution to the problem:

Let x be the number of marbles that A has. And let y be the number of marbles that B has. If B gives 30 to A, then A has x + 30 and B has y – 30.

By data, when this happens, A has twice as many left as B.

So, x + 30 = 2(y – 30) = 2y – 2(30) = 2y – 60. That is, x – 2y = -60 – 30

that is, x – 2y = -90 ……….(i)

If A gives B 10, then A has x – 10 and B has y + 10.

By data, when this happens, B has triple what A has left.

So, y + 10 = 3(x – 10) = 3x – 3(10) = 3x – 30 that is, y – 3x = -30 -10

that is, 3x – y = 40 ………..(ii)

Equations (i) and (ii) are linear equations formed by converting the given word statements to symbolic language.

Now we have to solve these simultaneous equations. To solve (i) and (ii), let’s make the coefficients and equal.

(ii)(2) gives 6x – 2y = 80 ………..(iii)

x – 2y = -90 ……….(i)

Subtracting (i) from (iii), we get 5x = 80 – (-90) = 80 + 90 = 170

that is, x = 170/5 = 34. Using this in Equation (ii), we get 3(34) – y = 40

i.e. 102 – y = 40 i.e. – y = 40 – 102 = -62 i.e. y = 62.

So A has 34 marbles and B has 62 marbles. Year.

Check:

If B gives A 30 of his 62, then A has 34 + 30 = 64 and B has 62 – 30 = 32. Double 32 is 64. (verified).

If A gives B 10 of his 34, then A has 34 – 10 = 24 and B has 62 + 10 = 72. Three times 24 is 72. (verified).

EXAMPLE 3 (on quadratic equations)

Problem Statement.

A cyclist covers a distance of 60 km in a given time. If he increases his speed by 2 km/h, he will cover the distance an hour earlier. Find the original velocity of the cyclist.

Solution to the problem:

Let the original speed of the cyclist be x kmph.

So the time it takes the cyclist to travel a distance of 60 km = 60/x

If you increase your speed by 2 kmph, the time taken = 60/(x + 2)

By data, the second time is less than the first in 1 hour.

So, 60/(x + 2) = 60/x – 1

Multiplying both sides by (x + 2)(x), we get

60x = 60(x + 2) – 1(x+ 2)x = 60x + 120 – x^2 – 2x

that is, x^2 + 2x – 120 = 0

Comparing this equation with ax^2 + bx + c = 0, we get

a = 1, b = 2 and c = -120

We know by quadratic formula, x = {- b ± square root (b ^ 2 – 4ac)}/2a

Applying this quadratic formula here, we get

x = {- b ± square root (b^2 – 4ac)}/2a

= [-2 ± square root{ (2)^2 – 4(1)( -120)}]/2(1)

= [-2 ± square root{ 4 + 4(1)(120)}]/two

= [-2 ± square root{4(1 + 120)}]/2 = [-2 ± square root{4(121)}]/two

= [-2 ± 2(11)}]/2 = -1±11 = -1+11 or -1-11 = 10 or -12

But x cannot be negative. So x = 10.

So the original speed of the cyclist = x kmph. = 10 km/h. Year.