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

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 note what is given and what is required.

2ND STEP :

Select a letter or letters called x (and y ) to represent the unknown quantity or quantities requested.

STEP 3:

Represent the verbal problem statements step by step 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 result to make sure your answer meets the problem requirements.

EXAMPLE 1 (on linear equations in one variable)

Problem statement:

A fifth of a number of butterflies in a garden are on jasmines and a third of them on roses. Three times the difference of butterflies on jasmines and roses on lilies. If the remaining one flies 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 on jasmines = x/5. Number of butterflies on roses = x/3.

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

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

According to the data, number of free-flying butterflies = 1.

So number of butterflies in the garden = x = Number of butterflies on jasmines + 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 into 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 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) 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.

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

Check:

Number of butterflies on the jasmines = x/5 = 15/5 = 3.

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

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

Number of free-flying butterflies = 1.

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

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 double what you have left.” B replies “if you give me 10, I will have three times as many as 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 ax + 30 and B ay – 30.

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

So x + 30 = 2(y – 30) = 2y – 2(30) = 2y – 60. i.e. x – 2y = -60 – 30

i.e. x – 2y = -90 ……….(i)

If A gives B 10, then A ax – 10 and B ay + 10.

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

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

i.e. 3x – y = 40………..(ii)

Equations (i) and (ii) are the linear equations formed by converting the statements of given words into symbolic language.

Now we need to solve these simultaneous equations. To solve (i) and (ii), let y be the same coefficients.

(ii)(2) gives 6x – 2y = 80 ………..(iii)

x – 2y = -90 ……….(i)

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

i.e. 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. Rep.

Check:

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

If A gives B 10 from 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)

Statement of the problem.

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 one hour earlier. Find the original speed of the cyclist.

Solution to the problem:

Let the initial speed of the cyclist be equal to x km/h.

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

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

According to the data, the second time is less than the first by one 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

i.e. x^2 + 2x – 120 = 0

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

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

We know by 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. So x = 10.

So, The original speed of the cyclist = x km/h. = 10 km/h. Rep.

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