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## Solving Algebra Word Problems Made Easy With a Lucid Explanation of the Method With Examples

Equations are often used to solve practical problems.

The steps involved in solving an algebraic word problem are as follows.

STEP 1 :

Read the problem carefully and note what is given and what is required.

STEP 2 :

Choose the letter or letters x (and y ) to represent the unknown number(s) requested.

STEP 3 :

Show verbal explanations of the problem in symbolic language step by step.

STEP 4 :

Find the equivalent quantities in terms of the given conditions and create an equation or equations.

STEP 5 :

Solve the problem(s) found in step 4.

STEP 6 :

Check the output to make sure your answer meets the requirements of the problem.

EXAMPLE 1 (for linear equations in one variable)

Problem Statement :

One fifth of the number of butterflies in the garden are on jasmines and one third of them are on roses. The triple division of butterflies on jasmines and roses on lilies. If the remaining one moves freely, find the number of butterflies in the garden.

solution to the problem :

Let x be the number of butterflies in the garden.

According to the data, Number of butterflies in jasmines = x/5. Number of butterflies on roses = x/3.

Then the difference of butterflies in jasmines and roses = x/3 -x/5

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

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

So, the number of butterflies in the garden = x = Number of butterflies in jasmines + Number of butterflies in roses + Number of butterflies in 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 a Linear Equation created by converting the given word definitions into symbolic language.

Now we need to solve this equation.

x = x/5 + x/3 + x – 3x/5 + 1

Canceling x present on both sides, we get

0 = x/5 + x/3 – 3x/5 + 1

The 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) ie 0 = 3x + 5x – 3(3x) + 15.

i.e. 0 = 8x – 9x + 15 i.e. 0 = -x + 15 i.e. 0 + x = 15 i.e. x = 15.

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

Check up:

Number of butterflies in jasmines = x/5 = 15/5 = 3.

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

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

Number of butterflies flying freely = 1.

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

EXAMPLE 2 (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 will have twice as much.” B replies “if you give me 10, I will have three times the amount left with you.” How many marbles does one have?

Solution to the problem:

Let x be the number of marbles A has. And Let be the number of marbles 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 the number of B.

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

ie x – 2y = -90 ……….(i)

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

By data, When this happens, B has three times more than A.

So, y + 10 = 3(x – 10) = 3x – 3(10) = 3x – 30 i.e. y – 3x = -30 -10

that is 3x -y = 40 ………..(ii)

Equations (i) and (ii) are Linear Equations created by converting the given word definitions into symbolic language.

Now we have to solve these equations simultaneously. To solve (i) and (ii), let us make u equal.

(ii)(2) gives 6x – 2y = 80 ………..(iii)

x – 2y = -90 ……….(i)

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

ie x = 170/5 = 34. Applying this to Equation (ii), we get 3(34) – y = 40

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

Thus A has 34 marbles and B has 62 marbles. Ans.

Check up:

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

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

EXAMPLE 3 (Quadratic Equations)

Problem Statement.

A cyclist travels a distance of 60 km in a certain time. If he increases his speed by 2 km/h, he will cover the distance one hour earlier. Find the initial speed of the cyclist.

Solution to the Problem :

Let the initial speed of the cyclist be x kmph.

Then, the cycling time taken to cover a distance of 60 km = 60/x

If he increases his speed by 2 kmph, time taken = 60/(x + 2)

According to the data, the second period is less than the first by 1 hour.

Therefore, 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

ie 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 the Quadratic Formula, x = – b ± square root(b^2 – 4ac)/2a

Using 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)]/2

= [-2 ± square root4(1 + 120)]/2 = [-2 ± square root4(121)]/2

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

But x cannot be negative. So, x = 10.

So, Initial speed of the cyclist = x kmph. = 10 kmph. Ans.

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