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)

I like to share this Significant Figures Calculator with you all through my article.

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