Help for 3rd Grade Math

Introduction to help for 3rd grade math:
In this article  we are going to help for 3rd grade  math student ,those 3rd grade students need help in basic operation of math like place value, compare numbers, addition, subtraction, multiplication, division, conversion of units like kilometer, meter, centimeter, millimeter, how to draw the graph  and also they need help in geometry.

I like to share this Tangent Line Equation with you all through my article.

Example - help for 3rd grade math


Which symbols are used for compare the numbers?
Solution:
Generally in comparison of number we use the symbol as < (less than),> (greater than), <=(less than or equal to),>= (greater than or equal to)
suppose take the  number as 24 and 12 ,here the number 24 is less than the 12 ,so it will be used in the form of 24>12

Example - help for 3rd grade math
8 hundreds+4 tens. How those expression will be written in number format
Solution:
Here 8 hundreds will be written as 800 then 4 tens will be written as 40 after that we have to note one thing in-between the plus sign will be there so after converging in the number format we have to add 800+40=840
So the answer is 840.

Example - help for 3rd grade math
What is the value of the underlined digit 789?
Solution:
Here place value of 7 will be written as 70, place value of 8 will be written as 80 and place value of 9 will be written as 9, so the value of the underlined digit is 80.

Understanding hard math word problems is always challenging for me but thanks to all math help websites to help me out.

Example - help for 3rd grade math


Solve 32-11
Solution:
Here we are doing the operation as subtraction it means removing some of the objects from the group here 11 will be removed from 22
First we subtract ones digit place 2-1=1 then do the operation is tens digit place 3-1=2
Combine these two digits we get the answer 21, it will be shown below.

Math Notes for Grade 8

Math notes for grade 8:

In this article we are going to discuss about find the math notes for grade 8.Math notes for grade 8 are easy to understand and solve. Math notes for grade 8 problems involve basic addition, subtraction, multiplication and division problems. Those arithmetic problems involve biggest numbers. The following topics are studied in the 8 grade,

Numbers
Measures
Algebra
Geometry
Handling data.
The math notes for grade 8 solving problems are given below.

Is this topic How to Find the Surface Area of a Cube hard for you? Watch out for my coming posts.

Example problems of math notes for grade 8:


Example 1:

Sum of 3 consecutive odd numbers is 51. Find the numbers.

Solution:

Framing the equation:

Let the first odd numbers be x.

Then the second and third odd numbers are (x+2) and (x+4)

There sum is 51.

x+(x+2)+(x+4) = 51

x+x+2+x+4 = 51

3x + 6 = 51(adding the like terms)

3x+6 = 51 is the required equation.

Solving the equation:

3x+6 = 51

3x = 51 – 6 (by rule 1)

= 45

x = 45 * `1/3 ` (by rule 3)

x = 15

The consecutive odd numbers are x, (x=2), (x=4)

15, (15+2) and (15+4)

The required numbers are 15, 17 and 19.

Example 2:

If office managers fixed in the pay scale 3200 – 85 – 4900, when will he reach his maximum?

Solution:

Pay scale: 3200 – 85 – 4900

Starting salary = $ 3200= a, Annual increment = $85 = d,

Maximum salary = $4900 = tn

tn = a + (n – 1)d = 4900 = 3200+(n-1) 85

n – 1 = 1700/85 = 20

n =20+1 = 21

The manager will reach his maximum in his 21st year of service.Having problem with math problems for 3rd grade keep reading my upcoming posts, i will try to help you.


Example 3:


The base and height of a right triangular ground are 60 m and 45 m. Find the cost of leveling the ground at Rs 150 per are ( 1 are = 100 sq.m)

Solution:

Given base b = 60m; height, h = 45m

Cost of leveling 1 are = Rs.150

Area of the triangular, A = `1/2` * 60 * 45

= 1350m2

100m2 = 1are

1350m2 = `1350/100` = 13.5are

Cost of leveling 1 are = Rs.150

Cost of leveling 13.5 Ares =Rs. 13.5 * 150 = Rs. 2025

Cost of leveling the right triangular ground = Rs. 2025

Example 4:

Find the circumference of a circle whose radius is 7cm.

Solution:

Radius r = 7cm

Circumference, C = 2`pi` r

= 2 * 22/7 * 7

= 44

Circumference of the circle = 44cm.

Example 5:

Solve 6x + 12 = 4x – 2

Solution:

Given expression 6x + 12 = 4x – 2

Subtract 12 on both sides of the equation

6x +12 – 12 = 4x – 2 -12

6x = 4x -14

Subtract 4x on both sides of the equation

6x – 4x = 4x – 4x -14

2x = -14

Divide 2 on both sides of the equation

`(2x)/2` =` -14/2`

x = -7

Solution is x = -7

Writing Equation Solver

Introduction to writing equation solver:

A linear equation solver is an arithmetical equation in which every word is either a stable or the multiplication of a stable and (the first power of) a distinct variable.

Linear equations solver can have single or many variables. Linear equations take place with great regularity in practical arithmetic. whereas they happen fairly logically when replica many phenomenon, they are mainly helpful because a lot of non-linear equations can be summary to linear equations by imagining that extent of interest vary to just a small extent from some "conditions" state

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Different types of linear equation solver:


They are three different types for writing linear equation solver:

Writing linear equation solver type: 1

Solve the given linear equation:  x + 7 = 5

Solution:

x + 7 = 5

x + 7 – 7 = 5 – 7          (Subtract 7 on both sides)

x = -2

So, the answer is x = -2

Writing linear equation solver type: 2

Solve the given linear equation: 3x – 8 = 7

Solution:

3x – 8 = 7

3x – 8 + 8 = 7 + 8       (Add 8 on both sides)

3x = 15

3x / 3 = 15 / 3              (Divide 3 on both sides)

x = 3.

So, the answer is x = 3.

Writing linear equation solver type: 3

Solve the given linear equation: 15x + 3 = 10x + 13

Solution:

15x + 3 = 10x + 13

15x + 3 - 3 = 10x + 13 – 3      (Subtract 3 on both sides)

15x = 10x + 10

15x – 10x = 10x – 10x + 10    (subtract by 10x on both sides)

5x = 10

5x / 5 = 10/ 5               (Divide 5 on both sides)

x = 2.

therefore, the answer is x = 2.

Is this topic Random Variables hard for you? Watch out for my coming posts.

Writing Linear equation in two variables:


A general structure of writing linear equation in the two variables x and y is

Y = mx + b.

Where m and b are select constants. The sources of the first name “linear” arrive from the information that the place of explanation of such an equation appearance a straight line in the plane. In this fastidious equation, the invariable m concludes the slope or gradient of that line, and the constant term b conclude the point at which the line traverse the y-axis, or else known as the y-intercept.

Algebra Solver Program

Introduction to algebra solver program:
Algebra solver program is a program which solves algebra problems with detailed solution in the stepwise manner. Solver program solves algebra problems. Algebra is the main branch in mathematics which deals the process of determining the unknown variable values . In algebra program the numbers are considered as constants. The following program shows the solved algebra problems.

Problems in algebra include simplifying expressions, finding the value of unknown variable, solve one step linear equations , functions , word problems of day to day life and so on. Algebra deals with the solutions of the problems in a structured , simple step by step procedure. Each step is small and the solution is thus obtained by a set of such steps.Here is a sample of the solver program that solves linear inequalities:

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Examples solved using the solver program of Algebra :


Ex 1:Solve the expression using algebra solver.

6(x -3) + 5y - 2(x -y -2) + 1

Sol:Step 1:Given the algebraic expression

6(x -3) + 5y - 2(x -y -2) + 1

Step 2:Multiplying the integer with above terms

= 6x - 18 + 5y -2x + 2y + 4 + 1

Step 3: Grouping the above terms

= 4x + 7y – 13 is the solution.

Ex 2:Solve the expression using algebra solver.

5(-2y - 2) - (-2y - 4) = -4(4y + 4) + 15

Sol:

Step 1:Given equation is

5(-2y - 2) - (-2y - 4) = -4(4y + 4) + 15

Step 2:Multiplying the integer with above terms
-10y - 10 + 2y + 4 = -16y - 16 +15

Step 3:Grouping the above terms

-8y - 6 = -16y - 1

Step 4:Add 8y and 1  on both sides, the above equation becomes

-5 =-8y

Y= `5/8`  is the solution

Ex 3:Solve the expression using algebra solver.

-2(y - 3) - 6y - 1 = 8(y +2) - 2y

Sol:

Step 1:Given
-2(y - 3) - 6y - 1 = 8(y + 2) - 2y

Step 2:Multiplying the factors
-2y + 6 - 6y - 1 = 8y + 16 - 2y

Step 3:Grouping the above terms
-8y + 5 = 6y + 16

Step 4:Subtract 6y + 5 on both sides
-8y + 5 – 6y -5 = 6y +16 -6y-5

Step 5:Grouping the above terms
-14y = 11

Y = -`11/14` is the solution.

Understanding algebra 2 help free is always challenging for me but thanks to all math help websites to help me out.

Practicing to solve algebra problems using the solver program:


1) Solve the expression using algebra solver.

6(-8y - 3) - (-5y - 5) = -8(2y + 4) + 9

Ans: y = `10/9` is the solution.

2) Solve the expression using algebra solver.

5(x -8) + 11y - 5(x -y +6) + 4

Ans: 16y – 66 is the solution.

Math Multi Steps Equations

Introduction to math multi steps equations:

An equation is a mathematical statement that asserts the equality of two expressions. Equations consist of the expressions that to be equal on opposite sides of an equal sign. (Source: Wikipedia).

The following rules are used to simplify the equation and the equation does not change.

1) Add or subtract any variable or number to the both sides of the equation

2) Multiply or divide any variable or number to the both sides of the equation.

3) Distributive law also used to eliminate the parentheses in the given equation. Distributive law is a(b + c) = a b + a c.

More than two steps are used to solve the equation is called as multistep equation. Now, we are going to see some of the problems on math multistep equations.

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Problems on math multistep equations:


Example problem 1:

Solve for the variable x: -6 = -4 (9 x + 6)

Solution:

-6 = -4 (9 x + 6)

Use the distributive law, to eliminate the parentheses

-6 = -4 (9 x) - 4 (6)

-6 = -36 x – 24

Add 24 on both sides of the equation

-6 + 24 = -36 x – 24 + 24

18 = - 36 x

Divide by -36 on both sides of the equation

(18 / -36) = -36 x / -36

-1 / 2 = x

So, the answer is x = -1 / 2.

Example problem 2:

Solve for the variable m: m + 10 = -5m + 34

Solution:

m + 10 = -5m + 34

Subtract 10 on both sides of the equation

m + 10 – 10 = -5m + 34 – 10

m = -5m + 24

Add 5m on both sides of the equation

m + 5m = -5m + 5m + 24

6m = 24

Divide by 6 on both sides of the equation

6m / 6 = 24 / 6

m = 4

So, the answer is m = 4.


Few more math multi step equations problems:


Example problem 3:

Solve for the variable x: 7(4t -3) – 1 = 13 + 21t

Solution:

7(4t -3) – 1 = 13 + 21t

To eliminate the parentheses, use the distributive law a (b + c) = a b + a c.

28t – 21 – 1 = 13 + 21t

28t – 22 = 13 + 21t

Add 22 on both sides of the equation

28t – 22 + 22 = 13 + 21t + 22

28t = 21t + 35

Subtract 21t on both sides of the equation

28t – 21t = 21t + 35 – 21t

7t = 35

Divide by 7 on both sides of the equation

7t / 7 = 35 / 7

t = 5

So, the answer is t = 5.

Example problem 4:

Solve the multi step equation for x: - (17 + x) – 6(-2 – x) = 35

Solution:

- (17 + x) – 6(-2 – x) = 35

To eliminate the parentheses, use the distributive law a(b + c) = a b + a c.

-17 – x + 12 + 6x = 35

-1x + 6x – 17 + 12 = 35

5x - 5 =35

Add 5 on both sides of the equation

5x - 5 =35

5x – 5 + 5 = 35 + 5

5x = 40

Divide by 5 on both sides of the equation

5x / 5 = 40 / 5

x = 8

So, the solution is x = 8.

Practice problems on multistep equations:


1)    Solve for the variable t: -1t + 25 = t + 45

(Answer: t = -10)

2)    Solve for the variable y: 3(y + 1) = 7 + y - 13

(Answer: y = -4.5)

Definition of Base in Math

Introduction to definition of base in math:

In arithmetic, the radix or base refers the number b in an expression of form bn. The number n is called the exponent and the expression is known formally as exponentiation of b by n or the exponential of n with base b. It is more commonly expressed as "the nth power of b", "b to the nth power". (Source: Wikipedia).

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Examples for definition of base in math:


Example 1 for definition of base in math:

Find the base 2 for base 10 of 28.

Solution:

The given base 2 number is (28)10.

We have to convert the base 10 number to base 2 numbers that is in the binary format.

The binary representation for the number 2 is 0010 and the binary representation for the number 8 is 1000. For 28 the binary value is 0010 1000.

So the value of base 10 for the number (28)10 is (0010 1000)2.

Example 2 for definition of base in math:

Find the base 10 for (1111)2.

Solution:

The given base 2 value is (1111)2.

The binary value for 15 is 1111 using the 8 4 2 1 method.

So the decimal value for the (1111)2 is 15.

Example 3 for definition of base in math:

Convert the base 2 value 11010101 into base 6 of the hexadecimal.

Solution:

The given base 2 value is 11010101.

Separate the binary values as 4 digits.

11010101 = 1101 0101

The hexadecimal values are 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D. . . . . . So,

11010101 = D 5

The hexadecimal value for 11010101 is D5.

Example 4 for definition of base in math:

Convert the base 2 value 10011011 into base 6 of the hexadecimal.

Solution:

The given base 2 value is 10011011.

Separate the binary values as 4 digits.

10011011 = 1001 1011

The hexadecimal values are 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D. . . . . . So,

11010101 = 9 B

The hexadecimal value for 11010101 is 9B.

Practice problem for definition of base in math:


Convert the (1110)2 into the decimal number.
Answer: 14

Convert 43 into the binary number.
Answer: 01000011

Convert the 10111010 into hexadecimal number.
Answer: B A.

Free Statistics Course Online

Introduction to free statistics course online:

Statistics is done normally with the data collected for specific purposes.  The method used in statistics for finding a representative value of the given data is called as the measure of central tendency. The three measures of central tendency of statistics were Mean, also median and mode. Statistics are all largely used in banking sectors, educationist, industrialist, economist, agriculturalists etc. Free statistics course online is nothing but free course material available on internet for the sake of helping students. Statistics is one of the important topics in mathematics. Let us see free statistics course online.

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Free statistics course online:


Example 1:

Find the mean deviation of the median for the following data: 6,12,7,5,14,12,20,6,8,21,23.

Solution :

Here the total number of observations is 11 which is odd. Arranging the data into ascending order,

5 , 6 , 6 , 7 , 8 , 12 , 12 , 14 , 20 , 21 , 23

Now Median =(11+1/2) or 6th observation = 11

The absolute values of the respective deviations from the median, i.e.,|xi − M| are

6 , 5 , 5 , 4 , 3 , 0 , 0 , 3 , 9 , 10 , 12
Therefore

∑ 11i-1 |xi - M| = 57

M.D.(M) =(1/11) ∑ 11i-1 |xi - M| = (1/11)*57  =5.181.

Example 2:

Find the mean deviation of  the median for the following data: 11,6, 6, 3, 13, 11, 18, 4, 7, 18, 22.
Solution:

Here the total number of observations is 11 which is odd. Arranging the data into

ascending order, we have 3,4,6,6,7,11,11,13,18,18,22

Now Median =(11+1/2) or 6th observation = 11

The absolute values of the respective deviations from the median, i.e.,|xi − M| are

8 , 7 , 5 , 5 ,4, 0 , 0, 2 , 7 , 7 , 11
Therefore
`sum` 11i-1 | - M| = `56`

and  M.D.(M) = `1/11` `sum` 11i-1 |xi - M| =`56/11` =5.6.

Is this topic solving proportions examples hard for you? Watch out for my coming posts.

Free statistics course online-examples:


Example 3:

Find the mean deviation of the mean for given data:

6,5,15,9,13,2,7,15

Solution:

Step 1 Mean of the given data are   `barx`


`barx` = `(6+5+15+9+13+2+7+15)/8`   =`72/8` =8

Step 2 The deviations of the respective observations from the mean x, i.e., xi– x are
6– 8,5–8,15–8,9–8,13–8,2–8,7–8,15–8      ( or )   –2,–3,7,1,5,–6,–1,7

Step 3 The absolute values of the deviations, i.e.,|xi − x |are  2,3,7,1,5,6,1,7


Step 4 The required mean deviation about the mean is

M.D. ( `barx` ) = `sum` 8 i-1 |xi-x| / 8

= `(2+3+7+1+5+6+1+7)/8` = `32/8` = 4

Example 4:

Find the mean deviation of the mean for given data :

12, 5, 19, 18, 4, 11, 18, 21, 20, 8, 15, 18, 2, 4, 15, 11, 3, 1, 10, 5

Solution:

find the mean ( `barx` ) of the given data

`barx`   =1/20`sum` 20i-1  xi = `220/20` = 11


The respective absolute values of the deviations from mean, i.e.,|x- `barx` | are

1,6,8,7,7,0,7,10,9,3,4,7,9,3,4,0,8,10,1,6

Therefore
20i-1 |xi - `barx` | = 118

and M.D. ( `barx` ) =`118/20` = 5.9

4th Grade Math Area

Introduction to 4th grade math:

In 4th grade mathematics, we learn about basic numbers, numerals, letters, integers and area of the different shapes. It provides basic idea to the students for solving the simple math problems. Topics under 4th grade mathematics is given below,

Numbers and its operations
Fractions and mixed numbers
Addition and subtraction
Division and multiplication
Geometrical shapes and areas
Algebraic expressions and equations
Logical reasoning
Charts and graphs

Example problems for 4th grade math area


4th grade math problem 1:

Find the area of the square with the side length of 19 cm.

Solution:

Given, side length (a) = 19 cm

Formula:

Area of the square = a2

Substitute the given side length value in the above formula, we get

Area of the square = 192 cm^2

= 361 cm^2

Answer:

Area of the square is 361 cm^2



4th grade math problem 2:

Find the area of the rectangle with the side length of 34 cm and width is 18 cm.

Solution:

Given, side length (l) = 34 cm and width (w) = 18 cm

Formula:

Area of the rectangle = length * width

Substitute the given side length and width values in the above formula, we get

Area of the rectangle = 34 * 18 cm^2

= 612 cm^2

Answer:

Area of the rectangle is 612 cm^2



4th grade math problem 3:

Find the area of the circle with the radius 42 cm.

Solution:

Given, radius of the circle (r) = 42 cm

Formula:

Area of the circle = pr2

Substitute the given side length value in the above formula, we get

Area of the circle = (p * 422) cm^2

= 1764p cm^2

Answer:

Area of the circle is 1764p cm^2

I have recently faced lot of problem while learning Mixed Numbers to Improper Fractions, But thank to online resources of math which helped me to learn myself easily on net.

Practice problems for 4th grade math area


4th grade math problem 1:

Find the area of the square with the side length of 81 cm.

Answer:

Area of the square is 6561cm^2

4th grade math problem 2:

Find the area of the circle with the radius 27.3 cm.

Answer:

Area of the circle is 745.29 cm^2

4th grade math problem 3:

Find the area of the rectangle with the side length of 24.67 cm and width is 29.5 cm.

Answer:

Area of the rectangle is 727.765 cm^2

Variable Solver

Introduction to variable solver:

The variable solver is the mathematical function to solve the equation of the one variable, two variables and also three variables. The variable solver is used to get the values for the given algebraic equation. This variable solver is also used in the electronic circuits to find the voltage across the circuits, current and also the capacity of the circuit.


Examples for variable solver:


Example 1 for variable solver:

Find the value of x for the algebraic equation 5x + 4=94.

Solution:

The given equation is 5x + 4=94.

Subtract 4 on both sides of the algebraic equation.

5x + 4- 4= 94- 4

5x =90

Divide by 5 on both sides of the above algebraic equation

5x/5 = 90/5

x= 18

The value of x for the algebraic equation 5x + 4=94 is 18.

Example 2 for variable solver:

Find the value of x and y for the algebraic equations    2x+ 3y =4 and 2x- 3y =4.

Solution:

The given algebraic equations are 2x+ 3y =4 and             2x- 3y =4.

Add the two equations to get first the value of x.

2x+ 3y =4

2x- 3y = 4

______________

4x = 8

Divide 4 on both sides of the above equation.

4x/ 4 = 8/ 4

x= 2

To get the value of y substitute the value of x in the given equation. Take any one of the given equation to get the value of y.

Substitute the value of x in this algebraic equation    2x+ 3y =4.

2 (2) + 3y=4

4+ 3y =4

Subtract 4 from the above equation.

4-4 + 3y = 4-4

3y= 0

Divide 3 on both sides of the above quadratic equation to get the value of y.

3y/3 = 0/3

y= 0

The values of x and y for the given algebraic equations are x=2 and y=0.

Exercise problem for variable solver:


Find the value of x for the algebraic equation 2x+ 3=13.
Justification: x=5.

Find the values of x and y for the algebraic equations 4x- 2y =14 and 2x + 2y =10
Justification: x=4 and y= 1.

Equation as Relations Solver

Introduction to equation as relations solver:

In this article equation as relations solver, we will solve the equation which is formed by some relation. Let us see how to make the equation with given statement and how to solve the equation. Generally the algebraic equation constitutes  of variables and expressions and we have to solve the equation to find the solution. Let us solve some example problems for equation as relations solver.


Example problems for equation as relations solver:


Example problem 1 - Equation as relations solver

A number is eight more than another, sum of the smallest integer and five times the greatest is 70.What is the greatest integer?

Solution:

Consider Smallest number x

Greatest number x+8

x+5(x+8) =70

x+5x+40=70

6x=30

x=5

Smallest number x=5

Greatest number x+8=13

Example problem 2 - Equation as relations solver

The perimeter of a rectangle is given by 58 m and the length is nine more than three times the breadth find out the breadth?

Given:

Perimeter=58 m

l=3b+9

Solution:

Perimeter of a rectangle=2(l+b)

2(3b+9+b)=58

2(4b+9)=58

8b+18 = 58

8b=40

b=5 m

Example problem 3 - Equation as relations solver


The price of one pencil and note book is $15 then what is the price of 5 pencil and 5 note book?

Solution:

Consider pencil as x

Note book as y

So from the given statement x+y=15-------1

The price of 5 pencil and 5 note book=5x+5y=?

We can write the above equation as

The price of 5 pencil and 5 note book =5(x+y)

From the first equation we can substitute the x+y value

The price of 5 pencil and 5 note book =5(x+y)

=5(15)=75

The price of 5 pencil and 5 note book =$75

Example problem 4 - Equation as relations solver

The ages of John and Krithick differ by 12 years when comparing. If 3 years ago, the elder one be 5 times as old as the younger boy, find their present ages.

Solution:

Consider the age of the John be x years.
The age of the Krithick = (x + 12) years.
:. 5 (x - 3) = (x + 12 - 3)

5x-15=x+12-3

4x=15+12-3

4x=27-3

4x=24

x=6

John present age is 6 years

Krithick present age is 18 years

Easy Algebra Solver

Introduction to easy algebra solver:

Algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. Together with geometry, analysis, topology, combinatorial, and number theory, algebra is one of the main branches of pure mathematics.

Source: From Wikipedia.

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Example problems for easy algebra solver:


Some example problems are solved with solution for easy algebra solver.

Example 1:

Determine the x intercept of the equation.

4x - 7y = 12

Solution:

Given the equation

4x - 7y = 12

To find the x intercept we set y = 0 and solve for x.

4x - 0 = 12

Solve for x.

x = 12 / 4

x = 3

The x intercept is at the point (3, 0).

Example 2:

Evaluate f (3) - f (2)

f(x) = 6x + 3

Solution:

Given the function

f(x) = 6x + 3

f (3) - f (2) is given by.

f(3) = (6(3) + 3) = 18 + 3 = 21

f(2) = (6(2) + 3) = 12 + 3 = 15

f (3) - f(2) = (21) - (15) = 6

Solution to the given function is f(3) - f(2) = 6

Example 3:

Find the slope of the line or gradient passing through the points

(-5, -5) and (7, 7).

Solution:

Given the points (-5, -5) and (7, 7), the slope m is given by

m = (y2 - y1) / (x2 - x1)

= (7 – (-5)) / (7 – (-5))

= 12 / 12

= 1

m = 1

Solution to the given slope of the line m = 1.

Example 4:

Find the slope of the line

12x - 8y = 18

Solution:

Given slope of the line

12x - 8y = 18

Write the above equation in the form of slope intercept y = mx + b and identify the value of m the slope.

-8y = -12x + 18        both side can be divided by -8

-8y/-8 = -12x/-8 + 18/-8

y = 3/4x - 9/4

The slope is given by the coefficient of x which is 3/ 4.

Slope = 3/ 4

Example 5:

Solve x for the following linear equation 7x - 3 = 144
Solution:
Add by 3 on both sides, we get
7x - 3 + 3 = 144 + 3
Simplify both sides:
7x = 147
Divided by both sides on 7:
7x/7 = 147/7
Simplify both sides:

x = 21.

Solution to the linear equation is x = 21.

Easy algebra solver practice problems:


Practice problems for easy algebra solver are given below.

1) Determine the x intercept of the equation.

8x - 12y = 24

Answer: The x intercept is at the point (3, 0)

2) Evaluate f (4) - f (3)

f(x) = 8x + 7

Answer: f(4) – f(3) = 8

3) Find the slope of the line or gradient passing through the points

(-6, -8) and (10, 12).

Answer: m = 4/ 5

4) Find the slope of the line

15x - 4y = 25

Answer: slope = 15/4

5). Solve x for the following linear equation 8x – 9 = 87

Answer: Solution to the linear equation is x = 12

Problem Solving Unit

Problem solving unit:

A unit of measurement is a definite magnitude of a physical quantity, defined and adopted by convention and/or by law, that is used as a standard for measurement of the same physical quantity. Any other value of the physical quantity can be expressed as a simple multiple of the unit of measurement.

For example, length is a physical quantity. The meter is a unit of length that represents a definite predetermined length. When we say 10 meters (or 10 m), we actually mean 10 times the definite predetermined length called "meter". (Source-Wikipedia)

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Problem solving unit-Example 1:


Convert 748.6 cm into meter.

Solution:

Here the conversion is from centimeter to meter. i.e. from lower unit to higher unit.

100cm = 1m.

Hence 748.6cm = (748.6/100) m

= 7.486 m (shifting the decimal two digits to the left)

The solution is 7.486 m.

Problem solving unit-Example 2:

Convert 50.1735 km into meter.

Solution:

Here the conversion is from higher to lower. Hence we have to shift the decimal point to the right.

1km = 1000 m

50.1735 km = 50.1735 × 1000 m

= 50173.5 m

The solution is 50173.5m

Problem solving unit-Example 3:

Convert 6m into millimeter.

Solution:

Note: Here we are converting the unit meter into millimeter. i.e. from higher unit to lower unit. So the operation should be multiplication.

1 m = 1000 mm

Hence 81 m = 81 × 1000 mm

= 81000 mm

The solution is 81000 mm

Problem solving unit-Example 4:


Express 8 kg 3 dag in grams

Solution:

1 kg = 1000 g

1 dag = 10 g

Hence 6 kg 5 dag = 6 × 1000 g + 5 × 10 g

= 6000 g + 50 g

= 6050 g

The solution is 6050g

Problem solving unit-Example 5:

Express 2769 g in kilograms.

Solution:

1000 g = 1 kg

Hence 3766 g = (3766/1000) kg

= 3.766 kg

The solution is 3.766 kg

Problem solving unit-Example 6:

Express 7402 kg into quintals.

Solution:

100 kg =1q

8402 kg =8402/100 q

= 84.02 q

The solution is 84.02 q

Problem solving unit-Example 7:

Convert 10.1835 km into meter.

Solution:

Here the conversion is from higher to lower. Hence we have to shift the decimal point to the right.

1km = 1000 m

10.1835 km = 10.1835 × 1000 m

= 10183.5 m

The solution is 10183.5 m

Free Online Tutors

Introduction to free online tutors:

Online tutors are those tutors who can teach from the distance part through computer and internet.  Online tutors can be in any part of the world and the student can also be any part of the world.  They both interact through an interface.  Computer and internet is the prime requirement of the online tutoring.

Online tutors are able to each form grade I or even kindergarden students using voice transfer technology.

Free online tutors-tutorvista as online tutoring company

Tutorvista is one of the important online tutoring company which provides quality services to all the parts of the world catering to the need of variety of need.  Tutorvista is primary based in Bangalore in India.  But it hires tutors all over the globe. It is established 5 years back and is one of the prime operator in the online tutoring.  Tutorvista has proved how successful the online tutoring and how useful it can be.  It makes a difference in the grade  which students gets.  Tutorvista primarily deals with science, Math and English for all grade of students.

It Now focus on the US student, but it had catered to the needs of the Korean, British and Australian students also.

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free online tutors-tutorvista-online tutoring company


Tutorvista uses one of the sophisticated technology for tutoring in the world.  Students can log in at any time and can get access to tutor immediately.  Tutorvista has a tutor base of more than 2000 and is expanding.  The number of students who had benefited from the company is enormous.

Tutorvista uses the white board which can be seen by both the student and tutor and on which both can write.  Which is visible in both the side.  It is a personalised one to one teaching with voice also available.  The session are normally of 45 minutes duration and the session can be replayed to clarify the doubt.

There are chat box also which can be used for chatting is very power ful tool.  Files can be shared in the board and can be seen in both the ends.

Tutorvista gives free online demo session which can be used by the students to get a first hand experience of the site and the method of teaching

Find Percentage Difference

Introduction to find percentage difference:

Percentage is expressing any number in a fraction with the denominator as 100. The symbol used for the percentage is ‘%’. Percentage difference is defined as the comparing a previous value to the new value.

Formula for percentage difference = [Previous value – new value] / 100

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Word problem on find percentage difference:


David lived in a United States of America. When David left the country the population was 920 million. David recently heard that the population has decreased by 6%. What is the present population?

Solution:

Find out the decrease percentage 920*6/100 = 5520/100 = 55.2 million

Then the new population is old population minus the decreased population

920-55.2 = 864.8 million

The population is now 864.8 million



Word problem 2:

Shan diets and decrease his weight 67kg to 65kg. Find the percentage decrease?

Solution:

Initially find the weight loss = 67-65 = 2kg

This 2kg is decrease the original weight. So the percentage will modify the original,

Let us take x as the decrease percentage

2=x * 67

Divide by 67 on both sides to make the variable alone

2/67 = 67x/67

0.0298 = x

Convert the decimal to percentage we get,

X=2.98%

The loss of weight in percentage is 2.98%

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Word problem on find percentage difference


If 58 students in a class are boys and 42 students are in girls. Find the percentage difference between boys and girls?

Solution:

First find out the total number of students in a class

Boys + girls = 58+42 = 100

After that find out the percentage of boys in a class

Boys percentage = total number of boys / total number of students

= 58/100 = 58 %

After that find out the percentage of girls in a class

Girls percentage = total number of girls / total number of students

= 42/100 =42 %

Percentage difference = 58-42 =16

The girls are 16 % percentage lesser than boys

Free Number Look Up

Introduction for free number look up:

In mathematics, number system is called as system of numeration. Number systems are used to express the quantities for counting, defining order, comparing the quantities, calculating numbers and denoting values. Number system includes natural numbers, integers, rational numbers, algebraic numbers, real numbers, complex numbers, p-adic numbers, surreal numbers and hyper real numbers.

The free number look up is give with examples and practice problems very interactively So students are getting free number look up to their studies.


Examples for free number look up:


Example 1:

Add the following decimal numbers 105.45 +10.892

Solution:

Addition operation for decimal numbers is just like the integers addition. In this problem

105. 45 has two decimal place but 10.892 has three decimal place. Hence, we have to add 0 with the number 105.45, so we get 105.450

That is, 105.45+10.892 = 105.450+10.892

11
105.450
+ 10.892
116.342

Example 2:

Find the square root of the following numbers `sqrt(625)` .

Solution:

`sqrt(625)` = `sqrt(25 xx 25)` = 25

Example 3:

Write the standard for the following number 109.089 `xx` `10^3`

Solution:

109.089 `xx` `10^3 ` which can be written as

109.089 `xx ` 10 `xx` 10 `xx` 10

109.089 `xx` 1000 now we have to shift the decimal point to three decimal points. So that we will get

109089

Example 4:

Add the following mixed numbers 11 2/3 and 11 2/4.

Solution:

We have to convert the following mixed numbers in to improper fraction. For this, denominators are multiplied with the whole number and then add the result of the product with numerator. So that, we will get the improper fraction.

`11 2/3` => `((11 xx 3) + 2)/3` => `(33 + 2)/3` => `35/3`

`11 2/4` => `((11 xx 4) + 2)/4` => `(44 + 2)/4` => `46/4`

Now we can add both improper fraction

`35/3` + `46/4` here, denominators are not same. So that, we have to find LCM. The LCM is 12

The denominator 3 from the fraction `35/3` is 4 times in the LCM. So we have to multiply the numerator 35 by 4. So we get `140/12`

The denominator 4 from the fraction 46/4 is 3 times in the LCM. So we have to multiply the numerator 46 by 3. So we get `138/12`

`(140 + 138)/12`

`278/12`


Practice Problems for free number look up:

Problem 1:

Add the following decimal numbers 10.45 + 10.82

The answer is 21.27

Problem 2:

Find the square root of the following numbers `sqrt(676)` .

The answer is 26

Problem 3:

Write the standard for the following number 1075.89 xx 10^3

The answer is 1075890

Problem 4:

Add the following mixed numbers `5 2/3` and `5 2/4` .

The answer is 13

Solving Conditional Statements

Introduction solving conditional statements:

Let P and Q be two logical conditional statements while solving. While solving conditional statement, the statement P `|->` Q is called a conditional statement. In logic, the relationship“If …then” between two conditional statements, the first statement is the hypothesis and the second statement is the conclusion in solving conditional statement. If P is, the second conditional statement is the conclusion. If P is, the implication if P, then Q   written     as `P ``|->` Q and read as p implies q.let us see problems for solving conditional statements. I like to share this absolute and conditional convergence with you all through my article.


Solving conditional statements:


Let us see the truth table for conditional statement

          P              Q              P `|->` Q

1.       T              T                 T

2.       T              F                 F

3.       F              T                 T

4.       F              F                 T

Let us see how to solve the conditional statements and prove the above statements are true

Example 1 on conditional statements:

Let us take an example and prove the condition if p is true, q is also true then p `|->` q proves true.

P:  If a figure given is a square

Q: then it has four side

Here in P: it is clearly given the figure is a square then Q: definitely, it has four sides

Both P and Q is true then the result, P `|->` Q  is true

Example 2 on conditional statements :

Let us take an example and prove the condition if p is true, q is not true then p `|->` q proves false.

P:  If a figure has four side,

Q: then it is square

Here in P: Given the figure, have four sides then Q: given it is square though it may be rectangle, which is of four sides.

P is true, Q is proved false then the result, P`->` Q  is false


Solving some more examples on conditional statements:


Example 3 on conditional statements :

Let us take an example and prove the condition if p is false, q is true then p`|->` q proves true.

P:  If it is not a weekday,

Q: It is Friday

Here in P: if we consider it is not a weekday as false statement, Q: definitely, it is weekday given Friday is weekday

P is false; Q proved true then the result, P `|->` Q is true

Example 4 on conditional statements:

Let us take an example and prove the condition if p is false, q is false then p `|->` q proves true.

P:  If you are not paid,

Q: then you do not work

Here consider if both the given statement is falsely given, the real statement is

“If you do not work, you do not get paid”

P is false; Q is also false then the result, P `->` Q is true

Practice Free Algebra Answers

Introduction of “Practice free algebra answers”:

For practice free algebra answers, we have to do the following basic operations such as addition, subtraction, multiplication and division. For practice free algebra answers, we have to know variables, constant, coefficients, exponents, terms and expressions. Practice free algebra answers is balancing the algebraic equations on both sides. For practice free algebra answers, also we have to know the basic properties such as commutative, associative, identities and inverse. I like to share this Equations with Decimals with you all through my article.


Order of the operation for practice free algebra answers:


1. First, Reduce what ever in the parentheses.

2. Next, Reduce the exponents.

3. Next, Reduce multiplication or division operations.

4. Finally, Reduce the addition or subtraction operation.

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Examples For practice free algebra answers:


Example 1:

3(a-2)+4b-3(a-b-3)+10

Solution:

= 3(a-2)+4b-3(a-b-3)+10

= 3a–6+4b–3a–3b–9+10

= 3a–3a+4b–3b-6–9+10

= b–5

Example 2:

3x - 2 = 2x - 3

Solution:

3x - 2 = 2x – 3

3x – 2 + 2 =2x -3 + 2 (Add 2 on both sides)

3x =2x -1

3x – 2x =2x -2x - 1 (Add -2x on both sides)

x = -1

Example 3:

Solve the equation 4x + 8 = -50

Solution

4x + 8 = -50

4x + 8 - 8 = -50 - 8(Add -8 on both sides)

4x = -58

4x / 4 = - 58 / 4 (Divided both sides by 4, so we get)

x = - 14.5

Example 4:

Solve the equation |-4x + 4| -6 = -6

Solution:

|-4x + 4| -6 = -6

|-4x+4| -6 + 6 = -6 +6 (Add 6 on both sides)

|-4x + 4| = 0

|-4x + 4| is same as -4x + 4, now solve for x

-4x + 4 = 0

-4x + 4 - 4= 0 (Now add -4 on both sides)

-4x=-4

-4x / 4 = -4 / 4 (Now divide both sides by -4)

-x = - 1 are equal to x = 1

Practice Problems:

1.  Solve for a, 2(a-5)=0

2. Solve for x, 7x - 15 = 2x – 5

3. Solve for x, 2x + 5 = 85

4. Solve for x, |-6x + 4| -3 = -6

Answer Key:

a=5
x=2
x=40
x=7/6

Free Algebra Homework

Introduction to online free algebra homework:

In mathematics an online free algebra is one of the main part . Online pre algebra explains the main properties of algebraically expressions and relations. This online free algebra explains the quantity, letters and other symbols. It indicates  variables, equations, objects, polynomials, and expressions etc. An online free algebra, math problems are solved subject of an online pre-algebra, Online free algebra one, algebra two, geometry of algebra. Fundamentally, an online free algebra involved  from the properties and processes of arithmetic which begins with the 4 processes: addition, subtraction, multiplication and division of numbers.

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online free algebra homework - Some Definition


Numerals, some literal numbers, and other algebraic signs of operations are made by algebraical symbol. Which expression denotes one number or one quantity. That is, just like the sum of 4 and 2 is one quantity, which result is 6, The sum of C and D is one quantity, that is, C + D. Similarly divide b,vb , a*b, a – b are algebraic expressions each of these represents one quantity or one number. Some signs and mathematical symbols are used in the algebraic expression.

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online free algebra homework - Examples:


There are two different ways you can solve by  algebraic and graphic.

Example 1:

The first home work example is finding  the x value in given equation.

Solve  x in the following equation.

X - 4 = 10;       x = 14

The answer is  x = 14

To determine the result by substitute 14 in the original equation for x. If the left side of the equation equals the accurate to right side of the equation after the substitution, you can find out the correct answer.

Example 2:

The second home work example is graphical algebraic model.

2 y – 6 = 10

Add 6 in both sides on the above equation.Take y is the common for both sides.

2 y = 16

Divide both sides by 2:

The answer is y = 8.

Check the solution by substituting 8 in the original equation for x. If the left side of the equation equals to right side of the equation its following by the substitution, then you can find out the correct answer.

2(8) - 6 = 16 - 6 = 10.

Distance Rate Problem

Introduction:

Distance word problems, repeatedly entitled as uniform rate problems, engage something travelling at set and stable speed or as well moving at an average speed. The formulas that used in distance and rate problems are two basic formulas that relate distance, rate and time.

Distance =speed * time
Speed = `(distance) / (time)`

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Distance Rate - Example problems:


Distance Rate - Example: 1

If a bus travelled 540 miles in 4 hours find the speed of the bus

Solution:

Bus travelled distance =540 miles

Total time =4 hours

Already we know the formula

Rate =`(Distance) / (time)`

=`540/4`

=135

Therefore the rate of the bus is 135 miles per hour

Distance Rate - Example 2:


At 11:00am, a bus (1) leaves city "1" at a constant rate of 60 mi/hr toward city "2". At the same time a second bus (2) leaves city "2" toward city "1" at the constant rate of 50 miles/hour. The distance between cities1 and 2 is 220 miles and these cities are connected by a highway used by the two bus. At what time will the two buses cross each other?

Solution:

Distance d1, between city "1" and bus (1), changes with the time t as

d1 = 60t, t = 0 corresponds to 11:00am.

Distance d2, between city "2" and bus (2),

Changes with the time t as

d2 = 220 - 50t

When the bus cross each other d1 = d2

60t = 220 - 50t

110t = 220

t = 2 hours

The two buses cross each other at

11:00am + 2hours = 13:00pm

Distance Rate - Example: 3

Two buses started from the same point, at 5 am, traveling in opposite directions at 40 and 50 mph respectively. At what time will they be 450 miles apart?

Solution:

After t hours the distances D2 and D1, in miles per hour, traveled by the two bus are given by

D1 = 40 t and D2 = 50 t

After t hours the distance D separating the two bus is given by

D = D1 + D2 = 40 t + 50 t = 90 t

Distance D will be equal to 450 miles when

D = 90 t = 450 miles

To find the time t for D to be 450 miles, solve the above equation for t to obtain

t = 5 hours.

5 am + 5 hours = 10 am

Free Algebra Problems

Introduction of Free Algebra 1 Problems:

Algebra is the division of mathematics in relation to the study of the rules of operations and relations, and the constructions and concepts arising from them, together with terms, polynomials, equations and algebraic structures. In algebraic expression represents a scale, what is done on one side of the scale with a number is also done to the other side of the scale. The numbers are the constants. Algebra consist of real numbers, complex numbers, matrices, vectors etc Understanding Associative Property of Addition Example is always challenging for me but thanks to all math help websites to help me out.


Free algebra 1 problems examples:


Ex  1 :  A number increased by nine

Sol :     x + 9

Ex  2 :  The difference between a number and ten

Sol :     x – 10

Ex 3 : Three fourths of some number

Sol :   `3/4` x

Ex 4  :  The sum of a number and five

Sol:    x + 5

Ex  5 : Eighteen subtracted from some number

Sol :    x – 18

Ex  6 :  Five times the sum of a twice a number and three

Sol :     5 ( 2x + 3 ); 10x + 15

Ex 7 :  Eight times the difference between four and some number

Sol :     8 ( 4 – x ); 32 – 8x

Ex  8 :  Twelve more than the sum of a number and negative twenty

Sol  :     ( x + ( −20 ) ) + 12; x – 8

Ex 9 : Four times the difference between a number and six, decreased by fourteen

Sol :   4 ( x – 6 ) – 14; 4x – 38

Ex 10 : The sum of a number and eight is negative ten.

Sol :  => x + 8 = −10; −18

Ex  11 : The difference between seventeen and a number is twenty-five.

Sol :   =>17 – x = 25; −8

Ex  12 :  Eight less than four times a number is twelve.

Sol :  =>4x – 8 = 12; 5

Ex  -13 : Ten added to the product of two and a number is forty-two.

Sol :   =>2x + 10 = 42; 16

Ex 14 : Twenty subtracted from some number

Sol :    x - 20

Ex  15 : nine more than the sum of a number and negative ten

Sol :     ( x + ( −10 ) ) + 9; x - 1

Please express your views of this topic College Algebra Problem Solver by commenting on blog.

Free algebra 1 problems:

Q 1:  How many pounds of coffee worth 120 a pound must be mixed with 10 pounds of coffee worth 90 cents a pound to produce a mixture worth 1.00 a pound?

Q 2:  James has 20 ounces of a 20% of salt solution. How much water should he evaporate to make it a 30% solution?