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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 algebraic word problems are as follows.

STEP 1 :

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

STEP 2:

Choose a letter or letters such as x (and y) to represent the desired unknown number(s).

STEP 3:

Represent the problem statements step by step in symbolic language.

STEP 4:

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

STEP 5:

Solve the sentence(s) obtained in step 4.

STEP 6:

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

EXAMPLE 1 (on linear equations in one variable)

Problem Statement:

A fifth of butterflies in a garden are on jasmine and a third on roses. Three times the difference between butterflies on jasmine and roses on lilies. If the rest are flying freely, find the number of butterflies in the garden.

problem solution:

Let x be the number of butterflies in the garden.

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

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

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

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

Therefore, the 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.

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

This is a Linear Equation that is created by converting verbal expressions into symbolic language.

Now we have to solve this equation.

x = x/5 + x/3 + x – 3x/5 + 1

Canceling x on both sides, we get

0 = x/5 + x/3 – 3x/5 + 1

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

Multiply both sides of the equation by 15, we get

15(0) = 15(x/5) + 15(x/3) – 15(3x/5) + 15(1) i.e. 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.

The number of butterflies in the garden = x = 15. Ans.

Considerations:

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 number of butterflies = 3 + 5 + 6 + 1. = 15. As Answer. (confirmation.)

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 will keep you twice as much.” B replies “if you give me 10, I’ll leave you three times as much.” How many marbles are there in each one?

Solution to the problem:

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

According to the data, when this happens, A has twice as much left as B.

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

ie x – 2y = -90 ……….(i)

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

According to the data, when this happened, B stayed with A three times.

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

ie 3x – y = 40 ………..(ii)

Sentences (i) and (ii) are Linear Equations formed by converting verbal expressions into symbolic language.

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

(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. Using this in equation (ii), we get 3(34) – y = 40.

ie 102 – y = 40 ie – y = 40 – 102 = -62 ie y = 62.

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

Considerations:

If B gives A 30 of his 62, then A’s 34 + 30 = 64 and B’s 62 – 30 = 32. Twice 32 becomes 64. (edit.)

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

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 that distance one hour earlier. Find the original speed of the bicycle.

Problem Solving:

Let the original speed of the bicycle be x km/h.

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

If he increases his speed by 2 km/h, time taken to stop = 60/(x + 2)

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

Therefore, 60/(x + 2) = 60/x – 1

Multiply 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 the 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)]/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. Therefore, x = 10.

Therefore, Original speed of the bicycle = x kmph. = 10 km/h. Ans.

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