# How To Solve An Equation By Using The Quadratic Formula 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 in the algebra word problem solving method are as follows.

STEP 1 :

Read the problem carefully and write down what is given and what is required.

STEP 2:

Choose the letter or letters that say x (and y ) to represent the unknown quantity(s) asked.

STEP 3:

Give the word statements of the problem step by step in symbolic language.

STEP 4:

Find the 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:

EXAMPLE 1 (linear equations in one variable)

Problem statement:

A fifth of the butterflies in the garden are on jasmines and a third of them on roses. Jasmines and roses have triple the difference of butterflies to lilies. If the rest fly 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 the difference between jasmine and rose butterflies = x/3 -x/5

According to the data, the number of butterflies on lilies = three times the difference between the number of butterflies on jasmine and roses = 3 (x/3 – x/5)

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

So, number of butterflies in 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.

It is a linear equation formed by converting the given word sentences into symbolic language.

Now we need to solve this equation.

x = x/5 + x/3 + x – 3x/5 + 1

Canceling x, which is 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) or 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:

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 free flying butterflies = 1.

Total number of butterflies = 3 + 5 + 6 + 1. = 15. Same as in answer (confirmed)

EXAMPLE 2 (for linear equations in two variables)

Problem statement:

Al and Bl each have a certain number of marbles. A says to B, “If you give me 30, I will have twice as much as you have left.” B replies, “If you give me 10, I’ll have three times what you have left.” How many marbles does each have?

Solution to the problem:

Let x be the number of balls in Al. And let y be the number of balls in Bl. If B gives A 30, then Al is x + 30 and Bl is y-30.

According to the data, when this happens, Al is twice as likely as B.

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 Al is x – 10 and Bl is y + 10.

According to the data, when this happens, Bl is three times more than A.

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 linear equations formed by converting the given word sentences into symbolic language.

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

(ii)(2) gives 6x – 2y = 80 ………..(iii)

x – 2a = -90 …….(i)

Subtracting (iii) from (i), 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 Al is 34 and Bl is 62 marbles. Ans.

Check:

If B gives 30 of his 62 to A, then Al is 34 + 30 = 64 and Bl 62-30 = 32. Twice 32 is 64. (verified.)

If A gives B 10 of his 34, then Al is 34-10 = 24 and Bl is 62 + 10 = 72. Three times 24 is 72. (verified.)

Problem statement.

A cyclist covers a distance of 60 km in a given time. If he increases his speed by 2 km/h, he covers 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 km/h.

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

If he increases his speed by 2 km/h, the 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 of (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

From the quadratic formula, we know that 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, initial speed of the cyclist = x km/h. = 10 km/h. Ans.

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