PDF Format springboard algebra 2 teacher edition

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 down

what is given and what is required.

 

STEP 2 : 

 

Select a letter or letters say x (and y ) to represent

the unknown quantity(ies) asked for.

 

STEP 3 :

 

Represent the word statements of the problem 

in the symbolic language step by step.

 

STEP 4 : 

 

Look for quantities which are equal as per

conditions given and form an equation or equations.

 

STEP 5 : 

 

Solve the equation(s) obtained in step 4.

 

STEP 6 :

 

Check the result for making sure that your answer

satisfies the requirements of the problem.

 

EXAMPLE 1  (on Linear equations in one variable)

 

 Statement of the Problem :

 

One fifth of a number of butterflies in a garden are on jasmines

and one third of them are on roses. Three times the difference of

the butterflies on jasmines and roses are on lilys.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.

As per data,

Number of butterflies on jasmines = x/5. 

Number of butterflies on roses = x/3.

Then

difference of the butterflies on jasmines and roses

= x/3 -x/5

As per data

Number of butterflies on lilys

=  Three times the difference of the butterflies on jasmines and roses

= 3(x/3 – x/5)

As per data,

Number of butterflies flying freely = 1.

So, number of butterflies in the garden = x

= Number of butterflies on jasmines + Number of butterflies on roses

 + Number of butterflies on lilys + 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 to the symbolic language.

Now we have to solve this equation. 

So, x =  x/5 + x/3 + x – 3x/5 + 1

Cancelling x which is present on both sides, we get

0 = x/5 + x/3  – 3x/5 + 1

L.C.M. of the denominators 3, 5 is (3)(5) = 15.

Multiplying both sides of the equation with 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. Ans.

 

Check:

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

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

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

Number of butterflies flying freely = 1.

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

 

  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 30 to me, I will have twice as many as left with you.”

B replies “if you give me 10, I will have thrice as many as left with you.”

How many marbles does each have?

 

 Solution to the problem :

 

Let x be the number of marbles A has.

And Let y be the number of marbles B has.

If B gives 30 to A, then A has x + 30

and B has y – 30.

By data, When this happens, A has twice as many as left with 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 10 to B, then A has x – 10

and B has y + 10.

By data,

When this happens, B has thrice as many as left with 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 the Linear Equations formed

by converting  the given word statements to the symbolic language.

 

Now we have to solve these simultaneous equations. 

To solve (i) and (ii), Let us make y coefficients same.

 

(ii)(2) gives 6x – 2y = 80 ………..(iii)

x – 2y = -90 ……….(i)

Subtracting  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.

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

 

Check:

If B gives 30 to A from his 62, then A has 34 + 30 = 64

and B has 62 – 30 = 32. Twice 32 is 64. (verified.)

If A gives 10 to B from his 34, then A has 34 – 10 = 24

and B has 62 + 10 = 72. Thrice 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 kmph,

he will cover the distance one hour before.

Find the original speed of the cyclist.

 

 Solution to the Problem :

 

Let the original speed of the cyclist be x kmph.

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

If he increases his speed by 2 kmph, the time taken

= 60/(x + 2)

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

So, 60/(x + 2) = 60/x – 1

Multiplying both sides with (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 ± ?(b^2 – 4ac)}/2a

Applying this Quadratic Formula here, we get

x = {- b ± ?(b^2 – 4ac)}/2a

=  [-2 ± ?{ (2)^2 – 4(1)( -120)}]/2(1)

=  [-2 ± ?{ 4 + 4(1)(120)}]/2

= [-2 ± ?{4(1 + 120)}]/2

= [-2 ± ?{4(121)}]/2

= [-2 ± 2(11)}]/2 = -1±11 = -1+11 or -1-11 = 10 or -12

But x can not be negative. So, x = 10.

So, The original speed of the cyclist = x kmph. = 10 kmph. Ans.

 

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