algebra calculator online

Earlier we discussed multiplying polynomials worksheet and Angle Sum of a Triangle. In this post we will talk about Algebra calculator online. Algebra Calculator Online guides us to understand the concepts related to algebra in grade 4. It includes the introduction to the topic of algebra which makes the child aware of the use of algebra in day to day life.  It also helps us to learn the concepts of algebraic expressions and the meaning of the variables and the constants.
We are able to understand, how to express the statements in form of the mathematical algebraic expressions and which word will be represented by which expression. We also need to understand the use of different mathematical operators in the algebraic expressions. If we have a statement: five added to three times a number gives 10. Now here we have unknown number which is to be calculated using this expression. Let us say that the number is x. Now if we need to write three times of any number, it means 3 * x. Further, if we again look at the statement,  it says that 5 added to three times a number. So we are going to add ( + ) 5 to 3*x. Thus the expression will be written as follows:
 3x + 5 = 10,
In order to find the value of x, we will solve the given expression and we get:
3x = 10 – 5,
Or 3x = 5,
So we can get the value of x by dividing both sides of the equation by 3
So, x = 5 /3.
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multiplying polynomials worksheet

By the word polynomial, we mean an algebraic expression with two or more terms. We will not miss recalling here the definition of algebraic expression or in simpler words what an algebraic expression means. An algebraic expression is a relationship between the constants & variables bound together with the help of different mathematical operations. The expression is a group of terms related to each other by the operations of addition & subtraction, whereas the terms themselves are formed by the bond of multiplication & division of the constants & variables. By working out multiplying polynomial worksheet students can score high grades. Get more detail here.
To multiply the polynomials (also read Dividing Polynomials), there may be a binomial or a trinomial or even a polynomial on one side & one of these on the other side as well. You can also play multiplying polynomials worksheet online to improve your skills.
We must remember that while multiplying the polynomials, we keep one of the polynomials intact & of the other polynomial, each of the term; as many as may be ; is multiplied by the polynomial which is intact turn by turn. Then each multiplication relation is opened using the distributive property. Thereafter, the like terms are combined & finally the product is obtained.
Let us try to understand it with the help of an example here .
Let the two polynomials to be multiplied be ( 3x + 8y – 2 ) and ( x – 2y + 6 )
We can keep any of the two polynomials intact, let us take ( 3x + 8y – 2 )  multiply it by each of the terms of the other polynomial , i.e. , ( x – 2y + 6 ) . So , we get
                X * ( 3x + 8y – 2 ) +8y * ( 3x + 8y – 2 ) – 2 * ( 3x + 8y – 2 )
                =  3x>2 + 8xy -2x + 24xy + 64y>2 – 16y – 6x - 16y + 4
                = 3x>2 + 32xy - 8x + 64y>2 – 32y + 4

 In upcoming posts we will discuss about algebra calculator online and Area of a Triangle. Visit our website for information on chemistry syllabus for class 12 Maharashtra board

Solving by Substitution Method

Algebra 1 Help : In the algebra there are several system of equations that have the several variable so values of these variables are solved by Substitution method. When there are several equations in the system then these are solved using some predefined steps of substitution method to solve algebra linear equations that are as follows :
( a ) There are 1 variable for one equation as x ,y,z means if there are three variable given in an equation then there are also three equations to find the value of these variables .
( b ) For solving the missing variable substitute the expression into the other equation .
( c ) When find the value of a variable that find into the step (b) then put it into the first equation and find the answer .
( d ) Cross check the solution .
The above steps are using to find the values of the variable in the given example :
-p +q = 1
2 p+q = -2
step no ( a ) : Find one equation for one variable there are two variable p and q
-p+q = 1
-p+p+q = 1+p (add the +p for both side of equation to find the one variable )
q = 1+p or q = p+1 .
step no ( b ) : For solving the missing variable substitute the expression into the other equation as
2 p+q = -2
put the value of q by the step no ( a ) into this equation
2 p+(1+p) = -2
2 p+p+1 = -2
3 p+1 = -2
Solve the value of p by subtracting the 1 from both side of equation
3 p+1–1 = -2–1
3p = -3
Divide the both side of equation by 3
3p/3 = -3/3
p = -1
Step no ( c ) : Find the value of q by putting the value of p into one of the equation
q = p+1
q = -1+1
q = 0
step no ( d ) : Cross check the answer
put both the values into each equation
-p+q = 1 2p+q = -2
-(-1)+0 = 1 2(-1)+0 = -2
1 = 1 -2 = -2


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Solving by Elimination Method

What is a linear equation? Equation is a collection of terms, which contains the combination of variables and numbers to represent the certain values. Sometimes linear equation can be described as a statement of variables to represent the particular values in the mathematics. As we know that mathematics provides lots of tools like method, formulas, theorem and rules. We will discuss about Solving by Elimination Method for the equations which having more than two unknown variables. In the general aspect we include the problem, which is given below:
x + y = 13        (1)
x – y = -5         (2)
The above are the equation, in which we can see that there are two equations containing two unknown variables x and y. If we determine the values of above two unknown variables x and y, which are consider as a true value for equation (1) and (2) then we can say that equation as a simultaneous  equations. To solve the simultaneous equation, we generally makes the value of one unknown variable same in both equation by adding or subtracting the variable to form new equation, which contain only one variable. After that obtain the value of one variable is easy to find. This whole process is known as Elimination Method.
 Here we show you how to solve the equation by example:
Example: Solve the equation by elimination method of following equations?
                         4y + 3x = 9
                         3y - 3x = 5
Solution: first we give label to both given equations:
                         4y + 3x = 9         (1)
                         3y - 3x = 5       (2)
Now we add the equation (1) and (2) in the same manner:
            4y + 3x + 3y - 3x = 9 + 5         (here +3y and -3y are cancelled to each other)
                         7y  = 14
                            y = 14/ 7
                            y = 2

In upcoming posts we will discuss about Solving by Substitution Method and Right Triangle. Visit our website for information on higher secondary education Karnataka

Finding Solution of Simultaneous Equations by Graphing

Hello students in this session we are going to talk about Finding Solution Of Simultaneous Equations By Graphing. But before discussing it you should know about linear equation, it is an algebraic expression that contains either a static term or the outcome of a static and one variable. Meaning of Simultaneous is ‘occurring at the same time’ that means we have to find Solution of Equations when they are occurring at the same time. (get detail)
The meaning of this is a pair of linear equation in two variables is said to form a system of simultaneous linear equation like
x + 2y = 3, 2x – y = 5
We can find Solution of Simultaneous Equations by Graphing with the following methods and cases they are:-
Solving Linear Equations
Method :- let the given system of linear equation be
a₁ x + b₁ y + c₁ = 0
a₂ x + b₂ y + c₂ = 0 (or try linear equations calculator)
On the same graph paper, we draw the graph of each one of the given linear equation. Each such graph is always a straight line. Let’s suppose we have two lines L₁ and L₂ that can be represented in a graph. Then many cases arise, some of them are :-
Case 1 :- When the lines L₁ and L₂ intersect at a point.
Case 2 :- When the lines L₁ and L₂ are coincident. It means they have infinitely many common points.
Case 3 :- When the lines L₁ and L₂ are parallel. It means they do not have a common point and so the system has no solution that is non-consistent. If they have at least one solution then the system is consistent.
We can make a algorithm for the above mentioned method and cases by marking them step 1, 2 to 5 so that we can easily solve Simultaneous Equations by Graphing .


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Substitution Method to Solving Simultaneous Equations

If two linear equations are solved at the same time then these equations are known as simultaneous. We understand the simultaneous equations by the help of some examples as
p + q = 5 and p – q = 1 are described as the simultaneous equations (more detail here).
Simultaneous equations can be solved exactly with the help of either substitution method or elimination method. Here we will use Substitution Method to Solving Simultaneous Equations. (or try linear equation calculator)
We take an example of substitution Method to Solving Simultaneous Equations as follows:
p + q = 3
2 p + 3 q = 8 (you can also try linear equation solver)
Both the equations have the sane variables p and q and both have the same solutions, so these are simultaneous equations p = 1 , q = 2 .
Substituting p = 1 and q = 2 in both the equations:
1 + 2 = 3 and 2 * 1 + 3 * 2 = 8
3 = 3 and 2 + 6 = 8 that is 8 = 8
Thus the solutions of variables p = 1 and q = 2 is correct.
For solving simultaneous Equations by the substitution method we have to follow some steps as :
step 1 : From one side of equations pick the one variable ( p )
p + q = 3
Isolate p : p = 3 – q
Step 2 : In other equation isolate the other variable :
2 p + 3 q = 8
Substitute 3 – q in place of p
2 ( 3 – q ) + 3 q = 8
This above equation has only single variable so it can solve easily .
Step 3 : For another variable q solve this equation :
2 ( 3 – q ) + 3 q = 8
Brackets are expanded as :
6 – 2 q + 3 q = 8
6 + q = 8
q = 8 – 6 = 2
q = 2
Substitute the q = 2 in equation for getting p
p = 3 – q
p = 3 – 2
Then p = 1


In upcoming posts we will discuss about Finding Solution of Simultaneous Equations by Graphing and Equilateral Triangle. Visit our website for information on CBSE board home science syllabus for class 11

Solving Linear Equations Examples

Hi friends, I am going to discuss about how to graph linear equations?
Linear equation is a pair of equation in two variables. There are two types of linear equations. First one is simultaneous linear equation of two variables and second one is graphical representation of linear equation. In this session, we will be discussing solving linear equations with two variables and solving simple problems from different areas.
In general form a linear equation in two variables x and y is
a₁x+b₁y+c₁=0
a₂x+b₂y+c₂=0
where a₁, b₁, c₁  and a₂, b₂, c₂ are all real numbers and x, y are variables. This is known as the algebraic representation of linear equation in two variables.


We take some examples of linear equation.
Example 1:- Show that x=2, y=1 is a solution of the simultaneous linear equations.
3x-2y=4
2x+y=5
solution :- the given system of equation is
3x-2y=4
2x+y=5
putting the x=2 and y=1 in equation (1) we have
L.H.S= 3*2-2*1=4= R.H.S
putting the x=2 and y=1 in equation (11) we have
L.H.S=2*2+1*1=5=R.H.S
thus, x=2 and y=1 and this satisfies both the equations of the given system.
Hence, x=2 and y=1 is a solution of the given system.
Now, we discuss about linear equation with fraction such that express by x/y.
We take some example to solving linear equations with fractions
Example 1:- solution equation (x+1)/3 =(2x+1)/5 solving by linear equation with fraction.
Solution :- to solve the equation step1:- (x+1)/3 =(2x+1)/5
step 2:- (x+1)/3 =(2x+1)/5 (we can start with left side and we can use cross multiply it means to one multiply numerator of one fraction by denominator of another fraction).
Step 3 :- 3(2x+1) = 5(x+1)
step 4:- 6x+3 =5x+5
step 5:- 6x-5x = 5-3
step 6:- x = 2 (this is answer for x)
now, we can check this solution putting the x = 2 in equation (1) we have
step 1:- L.H.S = (2+1)/3 =3/3 = 1
now, again putting the x = 2 in equation (2) we have
step :- R.H.S = (2*2+1)/5 =5/5 =1
so, L.H.S =R.H.S

In upcoming posts we will discuss about Substitution Method to Solving Simultaneous Equations and Congruent Triangles. Visit our website for information on CBSE board fashion studies syllabus for class 11