# How To Add Fractions With Different Denominators Math Is Fun Solving Algebra Word Problems Made Easy With a Lucid Explanation of the Method With Examples

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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 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.

STEP 2:

Select one or more letters that say x (iy ) 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 one 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:

A fifth of a number of butterflies in a garden are found on jasmines and a third of them on roses. Three times the difference of butterflies to jasmines and roses are to lilies. If the rest fly 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 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 on lilies = Three times the difference of butterflies on jasmines and roses = 3 (x/3 – x/5)

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

Therefore, number of butterflies in the garden = x = Number of butterflies on jasmines + Number of butterflies on roses + Number of butterflies on lilies + Number of free-flying butterflies = 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 sentences into symbolic language.

Now we need 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

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) 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. Ans.

To 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 the 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)

Statement of the problem:

A and B each have a certain number of balls. A says to B: “if you give me 30, I’ll have twice as many as yours.” B replies “if you give me 10, I will have three times as many as you will have”. How many marbles does each one have?

Solution to the problem:

Let x be the number of balls A has. AND Let y be the number of balls B has. If B gives 30 to A, then A has x + 30 and B has y – 30.

By data, when this happens, AA has twice as much 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 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 i.e. y – 3x = -30 -10

that is, 3x – y = 40 ………..(ii)

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

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

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

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

To check:

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

Statement of the problem.

A cyclist covers a distance of 60 km in a given time. If you increase your speed by 2 km/h, you will cover the distance one hour earlier. Find the cyclist’s original speed.

Solution to the problem:

Let the cyclist’s original speed be x km/h.

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

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

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

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

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. So x = 10.

Thus, the cyclist’s original speed = x km/h. = 10 km/h. Ans.

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