Quick Math Tricks

How to Quickly and Easily in Quick Math Tricks Math Grade 4, there are some requirements to master the grade 4 mathematics that should be considered include:
  1.  Mastery of multiplication and division is a prerequisite for students who sit in this class, why multiplication and division becomes absolutely necessary because if it is not mastered multiplication and division correctly that certainly will be left to follow the mathematics. Mathematics class 4 already includes many denominations variasianya in this matter, of course, using the name of multiplication and division so multiplication and division to be an absolute requirement.
  2. Flat wake mastery formula should have been mastered for students at this phase students must understand how to calculate the area and perimeter of a flat wake have learned in class 3 but remain flat given up on the class 4 as perfecting the material they have learned in class 3.
  3. The introduction of materials relating to the multiples fellowship and communion factors, it takes time to master the material Multiples fellowship and communion factors is so for students in required continuous practice on this matter, let tujuanya can later if there is a story about that are associated with Multiples fellowship and communion factors.
  4. Percent peel steeped material, roman numbers, the estimated value and estimated well below maupu but that needs to be sharpened is the deepening of material used per cent and the reverse fraction of a percent to percent.
  5. Requirements back to 5 is a repeat of conditions 1 to 4 conducted back so that later on the class 5 are no longer repeating the material. 4th grade math is a basic tumpuhan that will be used for math grades 5 and 6, so it should be built since the 4th grade.

Thus a fast and Quick Math Tricks Math Grade 4 who became mathematics exam initial success.

Easy Math Tips For Kids

The  Easy Math Tips For Kids Grade 3 math easy way to master it is a number of ways that should be done by the parents of the son of his daughter in this phase of parental involvement is crucial because the sons and daughters of this age still need proper guidance and direction.
What should be done? there are several steps, among others:

This stage is still something to do in the previous article about what should be done to master math grade two, there is an extra on the grade 3 mathematics mastery in addition to memorizing multiplication memorization added distribution such as memorizing multiplication is done in the article / tips before.
2.Tahapan second is to introduce fractions in fractions ranging from simple (add fractions with the same denominator) Example: 1/3 + 2/3 = 3/3, and so on, on the introduction of the same fractions denominator done a few times is enough .
A subsequent step (third) started to teach fractions whose denominators are not the same ordinary Example: 1/4 + 1/3 = 3/12 + 4/12 = 7/12, but are taught how to equalize the denominator of this can be seen or learned in grade 3 mathematics textbooks, please do as often as possible at this stage most children are still rather difficult way of summing fractions are not the same.
The fourth stage is taught up flat like Square, Rectangle, Triangle covers how to calculate the area and circumference up the flat.

So at a glance Tips
on How to Easy Math For Kids the Grade 3 mathematics although a bit of information that I can convey hopefully useful for the advancement of education our sons and daughters, especially in the field of mathematics.

Smart Math Simple Solution Methods

 How to get Smart Math Simple Solution Methods?
What must be prepared for children who step on to the next grade 2, at this age do not let the parents do not know what that should be required in 2nd grade math.
There are a few things to be prepared to be the best in math in grades 2, material that must be mastered in grade 2 about math is still about addition, subtraction, multiplication and division. So basically prioritizing children must master the system count the steps that must be taken are:

     If the child needs to enter in at the specialist tutoring math because math is the technique requires a relatively long time so it must immediately begin making decisions, it does not mean menomor two other subjects besides math lessons than can be read or recited (by mother ).
       Do not have an idea to incorporate the principles of tutoring to be that expensive does not mean it's cheap is not good, it's a lot of evidence that was not expensive to lift children's achievement to be the winner.
     In the second year, the parents immediately practice the multiplication of the species number in a week this has been done the way the tutor / teaching professionals in the fields of mathematics, as for the type of numbers that I mean for example, the number 9 is multiplied by 1 to 9 for one week is not replaced by other numbers. After the children have memorized one week then the next target is the number to another and apply the same way drill continues in one week.

So tips on how to master the math grade two, hopefully we can help to improve the intelligence of children, This Smart Math Simple Solution Methods.

Smart Math Tricks Easy way

How to smart math tricks Math Elementary First Class what to do or preparation?

Who would have been able to write the numbers 1 through 9 must be able to write numbers why, at the time was sitting in elementary school lessons are no longer teach start writing or reading other than that if you can not write numbers and letters would therefore prepare to miss your son or daughter at admission elementary was able to write or read.
Be sure to calculate a simple summation or subtraction without borrowing system units ranging from minus / plus the units, tens minus / plus dozens still at its initial commitment to make or give exercises in children figures not to meminjan up first because if the problem is too severe will cause fear in arithmetic (Numeric Phobia), it is good to avoid making the problem leads to a system borrowed depanya numbers.
Give about bagaima How to Quickly Mastering Math summation with storing system but remember do not directly made about tens plus dozens, but step by step as units plus units with saving techniques and this can be done several times to understand the essence of summer in math most large is 9 + 9.
Later in rather long teaching process, namely the reduction of the number of future borrowing system for this phase necessitates patience in conveying how because of the logic at this age is still unstable, so it should be given about the patient with a record of tens of reduced units such as 11-2 and 13-4 and onwards.
This way tips smart math tricks.


Smart Way Mathematics

The Smart Way to Early math is a step that must be taken by many older people especially the young mothers who always want their children so smart math since kindergarten age , a variety of ways such as looking for an agency pursued ranging from A to Z of financially affordable to the high level .Basically my goal is just a clever math or general subjects but math is a top priority because of what ? , Indeed mathematics requires fast and precise counting this is a long process because the child had often heard a number of points and a lot of practice , as for some Tips on how clever mathematics from an early age as follows :

In the golden age of children is very high level of intelligence seraph brain power , at this age an important role for parents to air an effort to provide guidance began writing numbers that can not be made better with the help of number of points , then the child is asked to be the point
thicken numbers - point proficiency level this is done with the sorts of numbers such as the number 2 performed many times and then figure be changed so that children write numbers memorized .
The second process is the task seoang why mothers should mother again because math is better when presented in the native language at this step mother prepare objects or toys that will be better shaped figure objects in the form of 3-dimensional shape earlier figures are colorful , in the form of 3 -dimensional will create a memory in a child's imagination .
After the third process can write numbers next step is to learn simple numbers do not add up over 10 because it is training or raising the left brain ( functions writes , menhitung ) is also useful in addition to enlarged right brain ( function for air imagination ) . since the right brain for imagination then air the parenting skills needed to make a story about dealing with a sum of not more than 10 .
Example : Anto has 2 marbles are then given 5 marbles father so how marbles Anto now ? . Note : This example is presented in the story just does not need to be written so kids just answered only


Formula Math Perimeter of a Circle

Perimeter is the distance around a closed figure and is typically measured in millimetres (mm), centimetres (cm), metres (m) and kilometres (km). These units are related as follows:
10 mm = 1 cm
100 cm = 1 m
1000 m = 1 km

The word 'perimeter' is also sometimes used instead of circumference.
If we know the radius
Given the radius of a circle, the circumference or perimeter can be calculated using the formula bwloe:

Perimeter (P) = 2 · π · R
where: R is the radius of the circle π is Pi, approximately 3.142
If we know the diameter
If we know the diameter of a circle, the circumference can be found using the formula
Perimeter (P) = π · D
where: D is the diameter of the circle π is Pi, approximately 3.142
If we know the area
If we know the area of a circle, the circumference can be found using the formula:
Perimeter (P) = √(4 · π · A )
where: A is the area of the circle π is Pi, approximately 3.142

Smart Math Example 1:
A circular flower-bed has a radius of 9 m. Find the perimeter/circumference of the flower-bed.
Smart Solution:

P = 2 · π · R
P = 2 · 3.1416 · 9
P = 56.5487 cm
So, the perimeter/circumference of the flower-bed is 56.5487 m.

Smath Math Example 2: Find the perimeter of the given circle whose diameter is 4.4 cm.
Smart Solution:
Given that:
Diameter of the circle (D) = 4.4 cm.

We know the formula to find the perimeter of the circle if the diameter is given, namely π·D.

Substitute the diameter 4.4 and Pi value as 3.14 in the above formula.
Perimeter = (3.14)(4.4) = 13.82
Therefore 13.82 cm is the perimeter of the given circle.

Smart Math Example 3: If the radius is 11.7 cm. Find perimeters (circumference) of the circle.
Smart Solution:
Given that:
Radius (r) = 11.7cm
Perimeter (circumference) of circle P = 2 π r
Substitute the r value in the formula, we have:

P = 2 x 3.14 x 11.7
P = 79.56 cm
Thus, the perimeter of the circle is 79.56cm

Smart Math Example 4: Find the perimeter and area of the circle, if the radius of the circle is 8cm.
Smart Solution: We have given the radius, which is 8cm. So, by using the formula of the perimeter of the circle, we have:

P = 2πr
P = 2×3.14×8
P = 50.24 cm
And for the area of the circle:-
A = π r2
A = 3.14×(8)2
A = 200.96cm2

Smart Math Example 5: The wheel of a bullock cart has a radius of 6 m. If the wheel rotates once how much distance does the cart move?
Smart Solution:
If the wheel rotates once, the cart will move by a distance equal to the perimeter of the wheel.
Step 1:
P = 2πr
P = 2× 3.14× 6 = 37.68 m
Thus, the bullock cart moves 37.68 m in one revolution of the wheel.

That is Formula Math Perimeter of a Circle

Formula Math Perimeter of a Rectangle

Formula Math Rectangle

Formula Math in rectangle, the distance around the outside of the rectangle is known as perimeter. A rectangle is 2-dimensional; however, perimeter is 1-dimensional and is measured in linear units such as feet or meter etc.The perimeter of a rectangle is the total length of all the four sides.
Perimeter of rectangle = 2L + 2W.
Math Example 1 : Rectangle has the length 13 cm and width 8 cm. solve for perimeter of rectangle.
Smart Solution:
Given that:
Length (l) = 13 cm
Width (w) = 8 cm

Perimeter of the rectangle = 2(l + w) units
P = 2(13 + 8)
P = 2 (21)
P = 42

Thus, the perimeter of the rectangle is 42 cm.

Math Example 2: If a rectangle's length is 2x + 1 and its width is 2x – 1. If its area is 15 cm2, what are the rectangle's dimensions and what is its perimeter?

Smart Solution:
We know that the dimensions of the rectangle in terms of x:
 l = 2x + 1
w = 2x – 1

Since the area of a rectangle is given by:
A = l * w

We can substitute the expressions for length and width into the equation for area in order to determine the value of x.

A = l * w
15 = (2x + 1) (2x -1)
15 = 4x2 – 1
16 = 4x2
x = ±2

 Note that the value of x must be positive and therefore in our case, the value of x is 2. And now we have:
l = 5 cm
w = 3 cm
Therefore, the dimensions are 5cm and 3cm.

Now, substituting these values in the formula for perimeter, we will get
P = 2l + 2w
P = 2(5)+2(3)
P = 10+6
P = 16 cm

Math Example 3: Find the area and the perimeter of a rectangle whose length is 24 m and width is 12m?
Smart Solution:
Given that:
length = L = 24m
width = W = 12m

Area of a rectangle:
A = L × W
A = 24 × 12
A = 188 m2

Perimeter of a rectangle:
P = 2L + 2W
P = 2(24) + 2(12)
P = 48 + 24
P = 72 m

Math Example 4: Find the area and perimeter of a rectangle whose breadth is 4 cm and the height 3 cm.
Smart Solution:
Area = b×h = 4×3 = 12 cm2.
Perimeter = 2(b) + 2(h) = 2(4) + 2(3) = 8 + 6 = 14.

Math Example 5: Calculate the perimeter of the rectangle whose length is 18cm and breadth 7cm
Smart Solution:
Given that:
L = 18 cm
B = 7 cm

Perimeter of rectangle = 2(length + breadth)
P = 2 (L + B)
P = 2 (18 + 7)
P = 50 cm

Math Example 6: Find the perimeter of rectangle whose length is 6 inches and width is 4 inches.
Smart Solution:
P = 2(L + B)
P = 2(6 + 4)
P = 20 in

Math Example 7: A boy walks 5 times around a park. If the size of the park is 100m by 50m, find the distance the boy has walked. If he walks 100m in 5 minutes, how long will it take for him in total?
Smart Solution:
Given that:
Length = L = 100m
Width = W = 50m
Rounds = 5
Time per 100m = 5minutes.

Perimeter of the park:
P = 2 L + 2 W.
P = 2 × 100 + 2 × 50
P = 200 + 100
P = 300 m

Total distance walked = 5 × Perimeter of the park.
= 5 × 300
= 1500 meters

Total time taken = Total distance walked × time taken to walk 1m.
= 1500 × 5/100
= 75 minutes or 1hr 15minutes

That is smart math formula of Perimeter of a Rectangle


Formula Math Surface Area of a Sphere

Difinition smart math formula of A sphere is a three-dimensional space, such as the shape of a football. A sphere is a body bounded by a surface whose every point is equidistant (i.e. the same distance) from a fixed point, called the centre or the origin of the sphere.
Like a circle in three dimensions, all points from the center are constant. The distance from the center to any points on boundary is known as the radius of the sphere. The maximum straight distance through the sphere is known as the diameter of the sphere. One-half of a sphere is called a hemisphere.

We can find the total surface area of a sphere by using the following formula:
SA = 4 π r2
where r is the radius.

NOTE: The value of π can never be calculated exactly, so the surface area of a sphere is only a approximation.

Surface area of sphere in terms of diameter = πd2
where d is the diameter of the sphere.

  •   What is the total surface area of a sphere whose radius is 5.5 meters?

Smart Solution: 
Given that:
r =5.5
Surface area of the sphere:
SA = 4 × π × r2
SA = 4 × π × (5.5)2
SA = 4 × 3.14 × 30.25
SA = 379.94
Thus the surface area of the sphere is 379.94 m2.  
A spherical ball has a surface area of 2464 cm2. Find the radius of the ball, correct to 2 decimal places, using π = 3.142.

Smart Solution:
SA = 4 × π × r2
In order to find r, we need to isolate it from the equation above:
r2 = SA / (4π)
r2 =2464 / (4 × π)
r2 =196.054
r = √(196.054)
r = 14.00 cm

  •   Find the surface area of the sphere whose radius is 18 cm. [π = 3.14]

Smart Solution:
r = 18 cm
The surface area of a sphere is given by:
SA = 4 × π × r2
SA = 4 × π × 182
SA = 4 × π × 342
SA = 4069.44 cm2

The surface area of the sphere is 4069.44 cm2.

  • Find the surface area of a sphere, whose radius is given as r = 11 cm.

Smart Solution:
The formula for calculating the surface area of sphere is given by:
SA = 4 × π × r2
SA = 4 × 3.14 × 112
SA = 1519.76
The surface area of sphere is 1519.76 cm2.

Example 5: A hemisphere has the radius measured to 8.3 cm. Find the surface area of it without the base.
r = 8.3 cm
The surface area of a hemisphere without the base is determined by using the following formula math:

SA = 2 × π × r2
SA = 2 × π × 8.32
SA = 432.62
The surface area of the hemisphere is therefore 432.62 cm2.

  •   Find the surface area of a sphere whose radius is 6cm?

Smart Solution:
SA = 4 × π × r2
SA = 4 × π × 62
SA = 4 × π × 36
SA = 452cm2

That is formula math in aplication mathematics of  Formula Math Surface Area of a Sphere.


Formula Math Volume of a Cuboid

Smart Math formula Volume of a Cuboid
formula math
In geometry, a cuboid is a solid shaped figure formed by six faces. There are two definitions for a cuboid. In the more general definition of a cuboid, the only supplementary condition is that each of these six faces is a quadrilateral. Otherwise, the word “cuboid” is sometimes used for referring a shape of this type in which each of the faces is a rectangle, and in which each pair of adjacent faces meets in a right angle.

 A cuboid with length l units, width w units and height h units has a volume of V cubic units given by: V = l × w × h
  •   A jewellery box that has the shape of a rectangular prism, has a height of 13 cm, a length of 35 cm and a width of 22cm. Find the volume of the jewellery box?
Math Solution:
V = l × w × h
V= 13 × 35 × 22 

V= 10010 cm3
  •   Find the volume of a brick whose size is 30 cm by 25 cm by 10 cm.
Math Solution:
The volume of the brick is given by:
V = l × w × h
V = 30 × 25 × 10
V = 7500 cm3

So, the volume of the brick is 7500cm3.

  •   Calculate the volume of a cuboid whose size is 8cm × 12cm × 6cm
Math Solution:
Volume of the cuboid is given by:
V = l × w × h
V = 8 × 12 × 6
V = 576 cm3
  • Given that the dimension of a cuboidal shape beam is 10m in length, 60 cm in width and 25 cm in thickness. How much does Nick have to pay for it, if it costs $250.00/ m3?
Math Solution:
Notice that all the measurements should be expressed in the same units (100 cm = 1 m). Volume of the beam = length x breadth x height (thickness here)
V = 10 x (60/100) x (25/100) = 1.5 m3
The final price of the beam will be: 1.5 x 250 = $375.00.
  •   A goods wagon shaped in the form of cuboid of measure 60m × 40m × 30m. How many cuboidal boxes can be stored in it if the volume of one box is 0.8 cubic meters.
Math Solution:
The volume of one box = 0.8 m3
The volume of the goods wagon is: 60 × 40 × 30 = 72000 m3
Number of boxes that can be stored in the goods wagon is therefore:
72000 / 0.8 = 90000
Hence the number of cuboidal boxes that can be stored in the goods wagon is 90,000.
  •   Find the height of a cuboid whose volume is 300 cubic cm and the base area is 30cm.
Math Solution:   
V = l × w × h
Since the base area is defined as: l × w, the height is therefore:
h = V / base area
h = 300 / 30
h = 10 cm
  •   A box of the size 60 cm × 40 cm × 30 cm. If the size of one chocolate bar is 24 cm × 12 cm × 4 cm each, how many bars can the box hold?
Math Solution:
The volume of the box = 60 × 40 × 30 = 72000 cm3
The volume of each bar = 24 × 12 × 4 = 1152 cm3

Therefore the number of chocolate bars that the box can hold is:
72000/1152 = 62
  •  The surface areas of the three coterminous faces of a cuboid are 6, 15 and 10 cm2 respectively. Find the volume of the cuboid.
Math Solution:
To begin with we need to determine the length, width and height of the cuboid.
l = √(6)
w = √(15)
h = √(10)

V = l × w × h
V = √(6) × √(15) × √(10)
V = 30 cm2

Surface Area of a Cuboid Formula Math

The total surface area (TSA) of a cuboid is the sum of the areas of its 6 faces,
which is given by:

TSA = 2 (lw + wh + hl)
Remember the surface area is the total area of all the faces of a 3D shape.
The lateral surface area of a cuboid is given by:
LSA = 2 (lh + wh) = 2 h (l + w)

Smart Math Formula in aplication :
Find the total surface area of a cuboid with dimensions 8 cm by 6 cm by 5 cm.

TSA = 2 (lw + wh + hl)
TSA = 2 (8*6 + 6*5 + 5*8)
TSA = 2 (48 + 30 + 40)
TSA = 236
So, the total surface area of this cuboid is 236 cm2.
Find the surface area of a cuboid of dimensions 4.8 cm, 3.4 cm and 7.2 cm.

Math Solution:
Area of Face 1: 4.8 × 7.2 = 34.56 cm²
Area of Face 2: 3.4 × 7.2 = 24.48 cm²
Area of Face 3: 4.8 × 3.4 = 16.32 cm²

Adding the area of these 3 faces gives 75.36 cm², since each face is duplicated on the opposite side, the total surface area of the cuboid will be:
TSA = 2 (75.36) = 150.72 cm²
The length, width and height of a cuboid are 10cm, 8cm and 7cm respectively. Find the lateral surface area of a cuboid.

Math Solution:
Lateral surface area of cuboid is given by:
LSA = 2h(l+w)
Note :
l = length = 10 cm
w = width = 8 cm
h = height = 7 cm

Insert these values into the formula we will get:
LSA = 2 ×7(10 + 8)
LSA = 14 × 18
LSA = 252 cm2

The length, breadth and height of a cuboid are 16cm, 14cm and 10cm respectively. Find the total surface area of the cuboid.

Math Solution:
The total surface area of a cuboid is given by:
TSA = 2 (l*b + b*h + h*l)
 Given that:
l = 16cm
b = 14cm
h = 10cm

Substituting the values in the equation we will get
TSA = 2 (16*4 + 14*10 + 10*16)
TSA = 2(224 + 140 + 160)
TSA = 2 * 524
TSA = 1048 cm2

Given a cereal box whose length is 20 cm, height is 30 cm and width is 8 cm. Find the surface area of the box.

Math Solution:
To find the surface are of the box we need to find the area of each rectangular face and add them all up.
The area of the front face is: 20 x 30 = 600 cm2.
The area of the top face is: 20 x 8 = 160 cm2.
The area of the side face is: 8 x 30 = 240 cm2.

Now add these values together we will get: 600 + 160 + 240 = 1000 cm2.

And the total surface area is therefore 1000 x 2 = 2000 cm2.

Find the surface area of a cuboid whose sides are 3cm by 6cm by 10cm.

Math Solution:
Surface area of the cuboid is given by:
TSA = 2 (16*4 + 14*10 + 10*16)
TSA = 2(3 x 6 + 6 x 10 + 3 x 10)
TSA = 2(18 + 60 + 30)
TSA = 216 cm2
That is smart math formula in aplication of Surface Area of a Cuboid. Formula Surface Area of a Cuboid Math

Formula Math Of Cylinder

Mathematics formula math of Cylinder :
rormula math

Definition of A cylinder as a solid figure that is bound by a curved surface and two flat surfaces. The surface area of a cylinder can be found by breaking it down into 2 parts:
1.  The two circles that make up the caps of the cylinder.
2.  The side of the cylinder, which when "unrolled" is a rectangle.

 Formula Math :
The area of each end cap can be found from the radius r of the circle, which is given by:
A = πr2

Thus the total area of the caps is 2πr2.

The area of a rectangle is given by:
A = height × width

The width is the height h of the cylinder, and the length is the distance around the end circles, or in other words the perimeter/circumference of the base/top circle and is given by:
P = 2πr

Thus the rectangle's area is rewritten as:
A = 2πr × h

Combining these parts together we will have the total surface area of a cylinder, and the final formula is given by:

A = 2πr2 + 2πrh

Smart Math Formula Note :
π  =  Pi, approximately 3.142,  r  = the radius of the cylinder, h  = height of the cylinder

By factoring 2πr from each term we can simplify the formula to:

A = 2πr(r + h)

The lateral surface area of a cylinder is simply given by: LSA = 2πr × h.
Smart Math Formula 
in the application :

  • Find the surface area of a cylinder with a radius of 4 cm, and a height of 3 cm.
Smart Solution:
SA = 2 × π × r2 + 2 × π × r × h
SA = 2 × 3.14 × 42 +  2 × 3.14 × 4 × 3
SA = 6.28 × 16 + 6.28 × 12
SA = 100.48 + 75.36
SA = 175.84
Surface Area = 175.84 cm2
  •  Find the surface area of the cylinder with a radius of 5.5cm and height of 10cm.
Smart Solution:
The radius of cylinder = 5.5 cm.
The height of cylinder = 10 cm.
The total surface area of the cylinder is therefore:
TSA = 2πr(r+h)
TSA = 11π (5.5+10)
TSA = 170.5 π
TSA = 535.6 cm2

  •   Find the total surface area of a cylindrical tin of radius 17 cm and height 3 cm.
Smart Solution:
Again as in the previous example:
TSA = 2πr(r+h)
TSA = 2π× 17(17+3)
TSA = 2π×17×20
TSA = 2136.56 cm2

  •  Find the surface area of the cylinder with radius of 6 cm and height of 9 cm.
Smart Solution:
The radius of cylinder: r = 6 cm
The height of cylinder: h = 9 cm
Total surface area of cylinder is therefore:
TSA = 2πr(r + h)
TSA = 12π (6+9)
TSA = 180 π
TSA = 565.56 cm2

  • Find the radius of cylinder whose lateral surface area is 150 cm2 and its height is 9 cm.
Smart Solution:
Lateral surface area of cylinder is given by:
LSA = 2πrh
Given that:
LSA = 150cm2
h = 9cm
π is the constant and its value = 3.14

Substitute the values in the formula and find the value of r by isolating it from the equation:
LSA = 2πrh
150 = 2× π × r × 9
r = 150 / (2×9× π)
r = 2.65cm
So the radius of the cylinder is 2.65 cm.

That is  Smart Math aplication from Cylinder Formula Math.


Easy Math of Linear (Equations and Inequalities) One Variable

Math of Linear (Equations and Inequalities) One Variable
Completion of the set of graphs of linear equations of the variables shown in a number line, in the form of dot (point).
Determine the completion of the set of equations
4 (2x + 3) = 10x + 8, if x
variables on the set of integers. Then, draw the number line
4 (2x + 3) = 10x + 8
8x + 12 = 10x + 8
8x + 12 - 12 = 10x + 8-12 (both sides minus 12)
8x = 10x - 4
8x - 10x = 10x - 4 - 10x (10x minus both sides)
-2x = -4
-2x: (-2) = -4 (-2) (both sides are divided by -2)
 x = 2
Thus, the solution set is {2}.

Easy Math of Linear Inequalities One Variable
In everyday life, surely you've come across or find sentences like the following.
a. Rebbecca weigh more than 52 kg.
b. Josh height 7 cm less than my height.
c. One of the requirements to be members of the Army is his height not less than 165 cm.
d. A bus can carry no more than 55 people.
How sentences are expressed in the form of mathematical sentence? To be able to learn to answer the following description.

1. understanding inequality
In order for you to understand the sense of inequality, try to remember back in elementary school about the matter in writing notation <,>,<,> , and not same.
a. 3 less than 5 written 3 <5.
b. 8 more than 4 written 8> 4.
c. x no more than 9 written x < 9.
d. Two times y is not less than 16 written 2y > 16.
Sentences 3 <5, 8> 4, x < 9, and  are called the inequality 2y >16 .
In general it can be written as follows.
An inequality is always marked by one of the following hyphen.
"<" For less than stated.
">" To declare over.
"<" to represent no more than or less than or equal to.
">" to represent not less than or more than or equal to.

2. Linear Inequalities One Variable
On the front you have learned that an equation is always marked with a hyphen "=". In this section you will learn the characteristics of an inequality.
Of the following forms, which specify a linear inequality with one variable.
a. x - 3 <5
b. A < 1 - 2b
c. x2 - 3x > 4
a. x - 3 <5
Inequality x - 3 <5 has one variable, namely x and rank 1, so x - 3 <5 is a variable linear inequality.
b. A < 1 - 2b
Inequality a < 1 - 2b has two variables, namely a and b, each of which rank 1.
Thus a < 1 - 2b is not a single variable linear inequalities.
c. x2 - 3x > 4
Due to the inequality x2 - 3x > 4 has variable x and x2, then x2 - 3x > 4 is not a single variable linear inequalities in Easy Math of Linear.