Inequalities Lesson Plan

Inequalities are the main concepts in mathematical algebra.Usually inequalities are the combination of variables and constants  greater than and less than sign.  For example 2y -10 < 20 is a less than inequality. In this article we are going to study about inequalities lesson plan through suitable example problems.
Brief Summary about Inequalities Lesson Plan


General form of an inequality:
Inequalities take the following form,

     ax +by <= c

     Here x and y are variables and then ‘a’ and ‘b’ are coefficients of x and y and then c is the constant term.  <= is denote the inequality sign.

     The following signs are used to represent the inequalities.  They are

    < = less than
    > = Greater than
    <= less than or equal to
    >= Greater than or equal to

Examples on Solving Inequalities Lesson Plan

Example 1:

     Solve the following inequality 5p + 15 > 45

Solution:

     The given inequality is 5p + 15 > 45

     Now we have to subtract the value 15 on both sides, then we have to get the following inequality,

    5p + 15 -15 > 45 - 15

     That is 5p > 30

     Dividing the value 5 on both sides, then we have to the following inequality,

     That is p>6

     That is p value is greater than 6.

Example 2:

     Solving the following inequality 2r + 18 `<=` 36 and represent it on the number line

Solution:

     The given inequality is 2r + 18 `<=` 36

     Now we have to subtract the value 18 on both sides, then we have to get the following inequality,

     2r + 18 -18 `<=` 36 -18

     That is 2r `<= ` 18

     Dividing the value 2 on both sides, then we have to the following inequality,

     That is r`<=` 9

     That is r value is less than or equal to 9.

     That is r=9 and r is less than 9.


Representation of number line:

 Inequalities on a number line

     First we have to draw the number line and mark the values.  Here r is less than or equal to 9 so we can first shade the value 9 and then draw the arrow in the direction of negative direction.  Because it is the less then (<) value.

    These are the important problems in the inequality lesson plan.

Equation Of A Line

Equation of a line is the equation satisfied by all the points which lie on the line.

 So, we can call every point on the line as a solution of the equation.

 As there are infinitely many points on a line, the equation of a line has infinitely many solutions.

 For every value of x, you can find a value for y by substituting the x value in the equation of the line.

 That is, consider the line whose equation is 2x + y = 4.

 We can find the value of y, when x equals 2.

 Substituting x = 2 in 2x + y = 3, we have

 2(2) + y = 3

 ? 4 + y = 3

 ? y = -1

 So, (2, -1) is a point on the line 2x + y = 4.

 There are various approaches to derive equation of a line.
Point Slope Form of Equation of a Line

We can draw a line if we know its slope and a point through which it passes. And the line we get is unique. Is it? Try.

 Hence we can find the equation of a line when its slope m and a point P(x1, y1) through which passes are given.

 The equation of the line is given as (y – y1) = m(x – x1)
Slope Intercept Form of Equation of a Line :

Suppose if c is the y-intercept of the line, then (0, c) is a point on the line.

 So, the equation (y – y1) = m(x – x1) can be written as (y – c) = m(x – 0)

 That is, y = mx + c

 So, we can find the equation of the line if its slope and y-intercept are known.
Equation of a Line Passing through Two Given Points:

Equation of a  line passing through two given points is also unique.

 So, we can find the equation of a line, if we know any two points on it.

 Suppose line l passes through the points A(x1, y1) and B(x2, y2), then slope of the line l is

 m = y2 - y1 / x2 -x1

Now we know the slope and a point on the line (considering either A or B) and hence we can find the equation of the line using the formula (y – y1) = m(x – x1).

So, the equation of the line is

(y - y1) = y2 - y1 / x2 -x1  (x - x1)

OR

(y - y2) = y2 - y1 / x2 -x1  (x - x2)

That is y - y1/ x - x1 = y2 - y1 / x2 -x1

In general, we can rewrite the equation of a line in the form ax + by + c = 0, where a, b and c are real numbers.

LCM Using Prime Factorization

The expanded form of LCM is Lease common multiple which means find the common multiple with least value for a given set of numbers. This can be done by using either prime numbers or by using composite numbers or with a combination of prime numbers and composite numbers. The LCM can also be written with exponents.
Steps to Find the Lcm Using Prime Factorization:

Step 1: Write the given set of numbers.

Step 2: Take the first number from the given set.

Step 3: Write the number with the multiples of prime numbers.

Step 4: Repeat step 2 and Step 3 for all the numbers in a given set.

Step 5: Choose the highest exponent term form all and multiply it to get LCM of given numbers.
Example Problems – Lcm Using Prime Factorization:

Example 1- LCM using prime factorization:

Find the LCM of 30 and 10.

Solution:

The given numbers are 30 and 10.

First it needs to find the prime factors of 30.

30 = 2*15

     = 2 * 3  * 5

The prime factors of 30 are 21, 31 and 51.

Next it needs to find the prime factors of 10.

10 = 2 * 5.

The prime factors of 10 are 21 and 51.

The highest exponent of all terms is 21, 31 and 51.

The LCM of 30 and 10 is 2 * 3* 5 `=>` 30.

My forthcoming post is on prime factorization algorithm, adding and subtracting rational expressions with unlike denominators will give you more understanding about Algebra

Example 2- LCM using prime factorization:

Find the LCM of 42 and 60.

Solution:

The given numbers are 42 and 60.

First it needs to find the prime factors of 42.

42 = 2*21

     = 2 * 3  * 7

The prime factors of 30 are 21, 31 and 71.

Next it needs to find the prime factors of 60.

60 = 2 * 30

     = 2 * 2 * 15

     = 2* 2 * 3* 5

The prime factors of 10 is 22 , 31 and 51.

The highest exponent of all terms is 22, 31, 51 and 71.

The LCM of 30 and 10 is 22 * 3* 5* 7 `=>` 420.

Example 3- LCM using prime factorization:

Find the LCM of 50 and 100.

Solution:

The given numbers are 50 and 100.

First it needs to find the prime factors of 50.

50 = 2*25

     = 2 * 5  * 5

The prime factors of 50 are 21 and 52

Next it needs to find the prime factors of 60.

100 = 2 * 50

       = 2 * 2 * 25

       = 2* 2 * 5* 5

The prime factors of 10 is 22 and 52

The highest exponent of all terms is 22 and 52

The LCM of 30 and 10 is 22 * 52 `=>` 100

Equivalent Percentage

Percentage is a number represents fraction of 100. In percentage, numerator is any number but denominator is only 100. So that it is called as Percentage. We can denote percentage of a number by using symbol %.

Example of Percentage: `50/100` (or) 50%

Equivalent percentage is a percentage for equivalent fraction or decimal.

In this article, we shall see about how to find equivalent percentage for given values.
Example Problems – Equivalent Percentage:

Problem 1:

What is the equivalent percentage of 5 out of 10.

Solution:

Step 1:

First, we divide numerator by denominator.

So `5/10` = 0.5

Step 2:

Multiply and divide the above decimal value by 100.

 `(0.5*100)/100`

Step 3:

So we get, `50/100` is equal to 50%.

Answer:

Equivalent percentage of `5/6` is 50%.

Problem 2:

What is the equivalent percentage of 6 out of 8.

Solution:

Step 1:

First, we divide numerator by denominator.

So `6/8` = 0.75

Step 2:

Multiply and divide the above decimal value by 100.

 `(0.75*100)/100`

Step 3:

So we get, `75/100` is equal to 75%.

Answer:

Equivalent percentage of `6/8` is 75%.

Problem 3:

What is the equivalent percentage of fraction `4/5` .

Solution:

Step 1:

First, we divide numerator by denominator.

So `4/5` = 0.8

Step 2:


Multiply and divide the above decimal value by 100.

`(0.8*100)/100 `

Step 3:

So we get, `80/100` is equal to 80%.

Answer:

Equivalent percentage of `4/5` is 80%.

Problem 4:

What is the equivalent percentage of 0.06?

Solution:

Multiply and divide the given decimal value by 100

`(0.06*100)/100 `

So we get, `6/100` is equal to 6%.

Answer:

Equivalent percentage of 0.06 is 6%.

Problem 5:

What is the equivalent percentage of 0.25?

Solution:

Multiply and divide the given decimal value by 100

`(0.25*100)/100 `

So we get, `25/100` is equal to 25%.

Answer:

Equivalent percentage of 0.25 is 25%.
Practice Problems: Equivalent Percentage

Problem 1:

What is the equivalent percentage of `8/5` ?

Answer:

160%

 Problem 2:

What is the equivalent percentage of `10/1000` ?

Answer:

1%

Problem 3:

What is the equivalent percentage of 0.81?

Answer:

81%

Problem 4:

What is the equivalent percentage of 0.025?

Answer:

2.5%

How to Solve polynomials

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An algebraic expression of the form a₀+a₁x+a₂x²+….+a_nxⁿ where a₀, a₁, a₂,….a_n are real numbers, n is a positive integer is called a polynomial in x.

Polynomial consists of:-

Constants: Like -8, 3/5, 20

Variables: Like a and b or x and y

Exponents: Like the 2 in a²

Purpose of Polynomials:

Listed below is the purpose of polynomials.

• Polynomials show in a wide mixture of area of mathematics and science.
• They are used to form polynomial equations, which instruct a wide range of troubles, from simple word tribulations to complex harms in the sciences.
• The purpose of polynomials is to describe polynomial functions, which emerge in setting range from fundamental chemistry and physics to economics and social science.
• Polynomials are used in calculus and arithmetic examination to estimate extra functions.
• In advanced mathematics, polynomials are used to build polynomial rings, a mid concept in intangible algebra and reckoning geometry.

Introduction of distributive property problems

An operation is distributive if the result of applying it to a sum of terms equals the sum of the results of applying it to the terms individually.
a ( b + c ) = ( a x b ) + ( a x c )
Here, ‘a’ is multiplied with the sum of two terms ‘b and c’ in the left hand side which, gives the same answer when ‘a’ is multiplied individually with ‘b’ and ‘c’ and then added.

Distributive Property Problems Example Part - 1:

1) Solve the problem using distributive property 9(9 + x).

Solution:

=9(9 + x)

=(9 * 9 + x * 9)

=(81 + 9x)


2) Solve the problem using distributive property 2(4 + 9x)

Solution:

2(4 + 9x)

(4 * 2 + 9x * 2)

(8 + 18x)


3) Solve the problem using distributive property 7(-1 + x)

Solution:

=7(-1 + x)

=(-1 * 7 + x * 7)

=(-7 + 7x)


4) Solve the problem using distributive property 12(a + b + c)

Solution:

=12(a + b + c)

=(a * 12 + b * 12 + c * 12)

=(12a + 12b + 12c)


5) Solve the problem using distributive property 7(a + c + b)

Solution:

=7(a + b + c)

=(a * 7 + b * 7 + c * 7)

=(7a + 7b + 7c)

Distributive Property Example Problems Part - 2:


6) Solve the problem using distributive property -10(3 + 2 + 7x)

Solution:

-10(3 + 2 + 7x)
Combine like terms: 3 + 2 = 5
-10(5 + 7x)
(5 * -10 + 7x * -10)
(-50 + -70x)


7) Solve the problem using distributive property -1(3w + 3x + -2z)

Solution:

=-1(3w + 3x + -2z)
=(3w * -1 + 3x * -1 + -2z * -1)
=(-3w + -3x + 2z)


8) Solve the problem using distributive property 1(-2 + 2x2y3 + 3y2)

Solution:

=1(-2 + 2x2y3 + 3y2)

=(-2 * 1 + 2x2y3 * 1 + 3y2 * 1)

=(-2 + 2x2 y3 + 3y2)


9) Solve the problem using distributive property 5(5 + 5x)

Solution:

=5(5 + 5x)

=(5 * 5 + 5x * 5)

=(25 + 25x)


 10) Solve the problem using distributive property y(1 + x)

Solution:

=(y + yx)

=(y + xy)

11) Solve the problem using distributive property 5(x + 10).

Solution:

=5( x + 10)
=(5* x + 5 * 10)
=(5x + 50)

Calculus problems solver

All the algebra, functions, relations etc that we study in math can be termed as pre-calculus. The next huge topic in math is calculus. Calculus is distinct from pre-calculus. In calculus there is less of static properties and more of dynamic properties. We talk of one quantity tending to another in calculus.
Calculus is a wide topic that encompasses various subtopics. The most common of which are: Limits, continuity, differential calculus and integral calculus.

The origin of calculus goes back at least 2500 years to the ancient Greeks. They use to find help with calculus problems of how to find area of any polygon.  The method they used was like this: the polygon was divided into triangles and the area of the triangles was added to find the area of the entire polygon, which is very similar to what we do in integration.

In solving calculus problems like finding the area of closed region bounded by curves, it is not possible to split the region to triangles as the boundaries are curved. The Greek method was to inscribe polygons in the figure the circumscribe polygons about the figure and then let the number of sides of the polygon increase. The limiting position of these areas was the area under consideration. Eudoxus used this method to solve calculus problems pertaining to area of circle and to prove the famous formula of A = pi r^2.

The problem of trying to find the slope of the tangent line to a curve gave rise to the branch of calculus called differential calculus. In problems of finding rate of increase of quantities, velocity, acceleration etc, the idea of limits is useful.

Sample calculus problems:

1. The price of a box of pencils is as follows: 5x/(x-1). Where x is the number of boxes one buys. So that means the more then number boxes that a person buys, the lower is the per box cost. What would the price tend to if a person buys lesser and lesser number of boxes?
Solution: The above problem is same as solving the limit: lim(x->1)(5x/(x-1)). We can see that x tends to 1, the term tends to infinity. So if someone wants to buy just one box, the cost would be pretty high.

2. Radius of a circle increases at a constant rate of 4 cm/s. Find the rate of change of area of circle when radius is 6cm.
Solution: This is application of derivatives.
A = pir^2
dA/dt = 2pir.dr/dt
dr/dt=4 and r=6
so, dA/dt = 2*pi*6*4 = 48pi (cm)^2/s