Study Substitution

Introduction to Substitution:-

In the substitution of solve equation, through a particular variable, another variable can be solve if any one equation is solve. A linear equation is the grouping of the variables, even and operators, which represent a straight line. For example x+y = three Here x and y are variables. Three is the constant +, = are operators. For study the system of equations, Substitution method is used. There are three methods to study substitution a system of linear equations Substitution method, Elimination Method, Graphical method. In this article let us see study substitution method. Please express your views of this topic solving systems of equations by substitution answers by commenting on blog.


Steps involved in study Substitution:-


For study the system of linear equations using the method of substitution, the subsequent steps are to be followed:

Step 1: study anyone of the equation to write one variable in terms of other variable.

Step 2: Then Substitute this in the next equation to get a single variable equation.

Step 3: The after that step is to solve the single variable equation to find the value of that variable.

Step 4: Once we get the rate of one variable, substitute the rate in any of the equation to get the rate of the subsequent variable.

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Example problems for study substitution:-


Example 1:-

Study the following system of linear equations using the method of substitution.

x - y = -5

3x+8y = -48

Solution:-

Rearrange the first equation,

x - y = -5

y = x + 5

Substitute this value for y into the second equation;

3x + 8(x + 5) = -48

Expand and simplify the equation:

3x + 8x + 40 = -48

11x = -88

x = -8

Substitute x back into one of the original equations;

-8 - y = -5

y = -3

Solution:-

x = -8, y = -3

Example 2:-

Study the following system of linear equations using Substitution method:.

x + y = 25

-4x + y = 10.

Solution:-

Rearrange the first equation,

x + y = 25

y = 25 - x

Substitute this value for y into the second equation;

- 4x + (25 - x) = 10

Expand and simplify the equation:

-4x + 25 - x = 10

-5x = 10 - 25

-5x = -15

x = 3

Substitute x back into one of the original equations;

3 + y = 25

y = 22

Solution:-

x = 3, y = 22

These are the examples for solving substitution method.

How To Reduce Math

Introduction to reduce in math

In math, reduction or reduce refers to the process of rewriting an expression into a simpler form. For example, the process of rewriting a fraction into one with the smallest whole-number denominator possible (while keeping the numerator an integer) is called "reduce a fraction". Rewriting a radical (or "root") expression with the smallest possible whole number under the radical symbol is called "reduce a radical". (Source: From Wikipedia). Here we will see some example problems to how to reduce a fraction or a radical expression in math.


Example problems to reduce fractions in math


Here we will see some example problems to learn how to reduce a fraction in math.

Example 1

Reduce the fraction `25/365` to simplest form.

Solution

The given fraction `25/365` is not the simplest form, because the numerator and denominator of the fraction has some common factors. By finding the common factors between the numerator and denominator, we can reduce the fraction further into simplest form.

To find the common factors of the numerator and denominator, the prime factorization is given as,

25 = 5 * 5

365 = 5 * 73

So the fraction `25/365` can be written as, `((5)(5))/((5)(73))`

So the simplest form or reduced form of the fraction `25/365` is `5/73`

Example 2

Reduce the fraction `28/118`

Solution

The prime factorization of 28 = 2 * 2 * 7

The prime factorization of 118 = 2 * 59

So, the fraction `28/118` can be written as `((2)(2)(7))/((2)(59))`

So the reduced form of the fraction `28/118` is `14/59`


Example problems to reduce radical expressions in math


Here we will see some example problems to learn how to reduce a radical expression in math.

Example 1

Reduce the radical expression, `sqrt(856)`

Solution

The prime factorization of 856 = 2 * 2 * 2 * 107

So the expression `sqrt856` can be written as, `sqrt((2)(2)(2)(107))`

= 2`sqrt214`

2`sqrt214` is the reduced form of `sqrt856`

Example 2

Reduce the radical expression `sqrt250` into simplest form

Solution

The prime factorization of 250 = 2 * 5 * 5 * 5

So, `sqrt250` = `sqrt((2)(5)(5)(5))`

= `5sqrt((2)(5))`

= `5sqrt10`

Probability Math 7

Introduction to probability math for grade 7:

Probability is the method of expressing knowledge or belief that an event will occur or has occurred. In mathematics the concept has been given an exact meaning in probability theory,  that is used extensively in  areas of study as mathematics, statistics, finance, gambling, science, and philosophy to draw conclusions about the likelihood of potential events and the underlying mechanics of complex systems.

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Example problems for probability math 7:


Ex: 1    For two events A and B, P(A) = 0.5, P(B) = 0.6 and P(A `U ` B) = 0.8. Find the

(i) P(`A/B` )

(ii) P(`B/A` ).

Sol:   P(A) = 0.5,

P(B) = 0.6 and

P(A`uu` B) = 0.8

Now,

P(A`uu` B) = P(A) + P(B) –P(A`nn` B)

P(A`nn` B) = P(A) +P(B) –P(A `U` B)

= (0.5 + 0.6 – 0.8)

= 0.3

Thus, P(A`nn` B) = 0.3

Therefore (i) P(`A/B` ) =`(P(AnnB))/(P(B))` =`0.3/0.6` = `3/6` = `1/2 ` = 0.5

(ii) P(`B/A` ) =` (P(AnnB))/(P(A))` = `0.3/0.5` = `3/5` = 0.6

Hence, P(`A/B` ) = 0.5 and P(`B/A` ) = 0.6

Ex: 2    A die is rolled. If the outcome is an odd number, What is probability that it is prime?

Sol:   When a die is rolled, sample space is S = {1, 2, 3, 4, 5, 6}

Let A = Event of getting an odd number, and

B = Event of getting a prime number.

Then, A = {1, 3, 5}, B = {2, 3, 5} and A B ={3,5}.

Therefore  P(A) = `(n(A))/(n(S))` = `3/6` = `1/2`

P(B) = `(n(A))/(n(S))` = `3/6` = `1/2` and

P(A`nn` B) =` (n(AnnB))/(n(S))` = `2/6` = `1/3.`

Suppose A has already occurred and then B occurs.

Now, `P(B/A)` = `(P(AnnB))/(P(A))` = `(1/3) / (1/2)` = `(1/3 xx 2/1)` = `2/3.`

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Practice problems for probability math 7:


1. A die is rolled. If the outcome is an even number, What is the probability that it is a number greater than 2?

[Ans: `2/3` ]

2. A pair of fair dice is thrown. Find the probability that the sum is 10 or greater if 5 appears on the first die.

[Ans: `1/3` ]

Angles to Learn in Math

Introduction to angles to learn in math:

In geometry, the angles are formed by the two line segments arising from a particular point. Thus, it forms the angles in the vertices and it can be also called as the vertex angles. There are many types of angles to learn in math. Now we are going to see about the angles to learn in math.

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Angles to learn in math:


Now we are going to see the angles to learn in math and the types of angles in math as explained one by one below.

Acute angle:

The acute angle is the angle where the measurements will be with in 90 degrees. Some of the example degrees are 28, 37, 46 etc.

Obtuse angle:

The obtuse angle is the angle where the measurements will be higher than 90 and lesser than 180 degrees only. Some example degrees are 91, 100, 179 etc.

Straight angle:

The straight angle is the angle nothing but the line having 180 degrees.

Reflex angle:

The reflex angle is the angle where the measurements will be higher than 180 degrees and lesser than 360 degrees. Some example degrees for reflex angles are 190, 250 etc.

Complementary angle:

The complementary angles are the angles whose measurements can be calculate as the sum of the two angles equals 90 degrees. Some example degrees are 45 and 45

Supplementary angle:

The supplementary angles are the angles whose measurements can be calculated as the sum of the two angles equals to 180 degrees. Some example degrees are 50 and 130.

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Problems for angles to learn in math:


Example 1:

Find the angles of supplementary in the ratio 6: 12.

Solution:

The angles are given in the ratio 6: 12

Let us assume the two angles be 6y and 12y

The angles are supplementary and so it can be given as

6y + 12y =180

18y =180, divide by 18 on both the sides,

y = 10

The supplementary angles are 60° and 120°

Example 2:

Find the acute angle of a triangle when one angle is 30 degree.

Solution:

Now we calculate the acute angle from the data as follows,

The two acute angles measures must be less than 90°.

If one of the acute angles a triangle is 30°, then its measure of the acute angle is 89° - 30° = 59°. The angles should be less than 90 degrees.

Math Problem Solving

Introduction to math problem solving:

Mathematics is the study of quantity, structure, space, and change. Mathematicians seek out patterns, formulate new conjectures, and establish truth by rigorous deduction from appropriately chosen axioms and definitions. Mathematics is used throughout the world as an essential tool in many fields, including natural science, engineering, medicine, and the social sciences. (Source: From Wikipedia). Now, we are going to see some of the solving problems in math.

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Solving math problems:


Example problem 1:

Solve the equation for x: 5x + 20 = 8x + 50

Solution:

5x + 20 = 8x + 50

Subtract 20 on both sides of the equation

5x + 20 - 20 = 8x + 50 – 20

5x = 8x + 30

Subtract 8x on both sides of the equation

5x – 8x = 8x + 30 – 8x

-3x = 30

Divide by -3 on both sides of the equation

-3x / -3 = 30 / -3

x = -10

So, the answer is x = -10.

Example problem 2:

Simplify the expression: 12x – 12 + 22x - x + 22

Solution:

Add the like terms in the given expression

12x – 12 + 22x - x + 22 = 12x + 22x - x – 12 + 22

= (12 + 22 -1) x + (-12 + 22)

= 33 x + 10

So, the answer is 33x + 10.

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Few more solving math problems:


Example problem 3:

Length and breadth of a rectangle are 20 cm and 13 cm respectively.

(i) Find its area.

(ii) Find its perimeter.

Solution:

(i)    Area of the rectangle = Length × Breadth

= l × b

= 20 * 13

= 260 Sq. cm.

(ii)Perimeter of rectangle = 2 (l + B)

= 2 (l + b)

= 2 (20 + 13)

= 2 * 33 = 66 cm.

So, the answer is

(i) Area = 260 sq. cm.

(ii) Perimeter = 66 cm.

Example problem 4:

Two angles of a triangle are of measures 80 and 42. Find the measure of the third angle.

Solution:

Let us take the third angle be x.

Sum of three interior angles of the triangle is 180 degrees.

80 + 42 +x = 180 degree

By solving this, we get

122 + x = 180 degree

Subtract 122 on both sides, we get

122 + x - 122 = 180 – 122

x = 58 degrees.

Practice math problems with answers:

1)  Two angles of a triangle are of measures 70 and 42. Find the measure of the third angle.(answer: 68 degrees)

2)  Solve the equation for x: 10x + 20 = 8x + 50 (Answer: x=15)

Making Rubrics Math

Introduction to making rubric math:
A rubrics math is a plan, a graph or chart that defines accurately what the expectations are for an assignment. The rubrics help to change more subjective prospect into very define and detailed expectations. Students can use rubrics when doing assignments to help conduct their content and presentation and teachers can use a rubric to help make grading easier and less subjective.

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More about rubrics math:


Rubrics are fairly easy to making and use and it can really have a constructive impact on student’s act and teacher grading.
In fact, the time taken to making rubrics can be a fraction of the time the rubrics really saves in the long run.
The rubrics math gives extra information to students and parents than; a math problem is correct or incorrect.
It tells accurately what was good or bad.
If rubrics were used, students should not have to ask "why" they established a particular grade and teachers will have an easy break of the score if that query is asked.

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Making Rubrics Math:

Step 1:

To making rubrics, first, define all goals for a project.
For example, a math assignment can have goals such as set up terms problems correctly, computation, checking of work, etc.
Step 2:

Arrange the list of goals in sort of result.
The objectives will be written in a column beside the left side of a paper.
Step 3:

Under each objective, listing each one of the specific criteria for that objective.
For example, one objective could be viewing good punctuation in a written assignment.
In a math class, the calculation could be busted down into borrow, carrying, etc.
Step 4:

Assign a precise percentage of the rating or number of point to each one objective.
For example, In a math class, a teacher might grade evenly on each of several criteria.
Step 5:

To make grading easier, use a scale for rating each goal.
List the scale along the top of the paper along with the criteria for each point on the scale.
This could be a Like art style scale where each norm is rated from 1 to 5.
A rating could also be "poor", adequate", "good", and "excellent."
Step 6:

Save copies of each rubric after making.
After you have made one, you can change it but you should never have to make one from scratch again.
Making a file for the rubrics so they can easily be pulling when needed.

Learn Online Factoring Radicals

Introduction  to learn online factoring radicals:

The factorization is the  process of factoring the given polynomial equation . basically  factorization  is used to find the common factors  of polynomial equation .. we factorize the radicals  equation  .

These are the steps to  solve  factoring radical

Step 1: to remove the  radical symbol for the given equation

Step 2: factorize   the equation

Step 3:  Solve the equation


learn online factoring radicals problem explanation:


we learn how to solve radical problem:

Solve for x if √2x+3=x  Squaring both sides of the equation gives us  2x+3= x2

Setting terms equal to zero gives  0x2-2x-3

The expression factors  0=(x-3)(x+1)  Setting each factor equal to 0 gives two possible answers: x = 3 or x = – 1.

We check each answer in the original equation:  If x = 3 we have

√2(3)+3=3

√9=3

If x = – 1 we have  √2(-1)+3=-1

is impossible since the square root cannot be negative.

Therefore the only answer is x = 3.

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Examples for learn online factoring radicals


some problems explain for  online  factoring radicals:

1. To solve radical problem for online learning  √x2-2=9 ?

Solution :

1. To solve √x2-2=9 we first square both sides of the equation. The result is x - 2 = 81. This equation is simple to solve. We have x = 83

2. A more complicated situation is√x+2=x In this case we still begin by squaring both sides of the equation. The result is x+2=X2

To finish solving this needs us to set all terms equal to zero and either factor or use the quadratic formula. We get x2-x-2=0

This factors (x-2)(x-1)=0  and the solutions are x = 2 or x = - 1.

We must check each of these solution in the original equation to see if the value of x gives a solution x = 2 gives

√2+2=2   or √4=2 is correct

x = - 1 gives √-1+2=-1  and √1= -1 is impossible

2.   Solve for x if√x+2=√2-x

Solution :

We square both sides. This gives x + 2 = 2 – x.

Solving for x gives 2x = 0. The only solution is x = 0.

Checking this in the original equation gives√0+2=√2-0  or √2=√2

Therefore the solution is x = 0.