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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 note what is given and what is required.

Step 2:

Select the letter or letters x (and y) to represent the unknown quantity(ies) being asked.

Step 3:

Represent the problem’s word statements in step-by-step symbolic language.

Step 4:

Find the quantities that are equivalent 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 requirements of the problem.

Example 1 (On Linear Equations in One Variable)

Statement of the problem:

One fifth of the many butterflies in the garden are on jasmine and one third of them on roses. The threefold difference of the butterfly to the jasmine and the rose is to the lily. If the remaining one is flying 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 jasmine = x/5. Number of butterflies in rose = x/3.

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

According to the data number of butterflies in lily = triple difference of butterflies in jasmine and rose = 3(x/3 – x/5)

Number of butterflies flying freely = 1 according to statistics.

So, number of butterflies in garden = x = number of butterflies in jasmine + 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.

It is a linear equation created 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 that exists on both sides, we get

0 = x/5 + x/3 – 3x/5 + 1

LCM of 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.

ie 0 = 8x – 9x + 15 ie 0 = -x + 15 ie 0 + x = 15 ie x = 15.

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

Check out:

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

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

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

Number of butterflies flying freely = 1.

Total dolls = 3 + 5 + 6 + 1. = 15. Like the answer. (Attested.)

Example 2 (On Linear Equations in Two Variables)

Statement of the problem:

A and B each have a certain number of marbles. A says to B, “If you give me 30, I will have double your amount.” B replies “If you give me 10, I will have three times what you have left.” How many marbles does each have?

Solution to the problem:

Let X be the number of marbles A. and y is the number of marbles B. If B gives 30 to A then A gives x + 30 and B gives y – 30.

By data, when this happens, A has more than twice as much as B.

So, x + 30 = 2(y – 30) = 2y – 2(30) = 2y – 60. That is, 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 as much left as A.

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

ie 3x – y = 40 ………..(ii)

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

Now we have to solve these simultaneous equations. To solve (i) and (ii), make the y coefficient equal.

(ii)(2) gives 6x – 2y = 80 ………..(iii)

x – 2y = -90 ……….(i)

Subtracting (i) from (i), 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.

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

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

Check out:

If B gives 30 out of his 62 to A then A has 34 + 30 = 64 and B has 62 – 30 = 32. 32 twice is 64. (Attested.)

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

Example 3 (In 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 per hour, he will cover the distance one hour earlier. Find the original speed of the cyclist.

Solution to the problem:

The original speed of the cyclist is x kmph.

Then, the cyclist takes 60 km = 60/x to cover the distance

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

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

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)]/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.

So, original speed of cyclist = x kmph. = 10 km per hour. Answer.

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