2 3x 2 answer

Learn how to use the Algebra Calculator to check your answers to algebra problems.

Example Problem

Solve 2x+3=

Check Answer

x=6

How to Check Your Answer with Algebra Calculator

First go to the Algebra Calculator main page.

Type the following:

1. First type the equation 2x+3=15.
2. Then type the @ symbol.
3. Then type x=6.

Try it now: 2x+3=15 @ x=6

Clickable Demo

Try entering 2x+3=15 @ x=6into the text box.

After you enter the expression, Algebra Calculator will plug x=6 in for the equation 2x+3= 2(6)+3 =

The calculator prints "True" to let you know that the answer is right.

More Examples

Here are more examples of how to check your answers with Algebra Calculator. Feel free to try them now.

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Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                     2*x^*x-(2)=0 

Step by step solution :

Step  1  :

Equation at the end of step  1  :

(2x² - 3x) - 2 = 0

Step  2  :

Trying to factor by splitting the middle term

      Factoring  2x²-3x-2 

The first term is,  2x²  its coefficient is  2 .
The middle term is,  -3x  its coefficient is  -3 .
The last term, "the constant", is  -2 

Step-1 : Multiply the coefficient of the first term by the constant   2 • -2 = -4 

Step-2 : Find two factors of  -4  whose sum equals the coefficient of the middle term, which is   -3 .

-4

+

1

=

-3

That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -4  and  1 
                     2x² - 4x + 1x - 2

Step-4 : Add up the first 2 terms, pulling out like factors :
                    2x • (x-2)
              Add up the last 2 terms, pulling out common factors :
                     1 • (x-2)
Step-5 : Add up the four terms of step 4 :
                    (2x+1)  •  (x-2)
             Which is the desired factorization

Equation at the end of step  2  :

(x - 2) • (2x + 1) = 0

Step  3  :

Theory - Roots of a product :

     A product of several terms equals zero. 

 When a product of two or more terms equals zero, then at least one of the terms must be zero. 

 We shall now solve each term = 0 separately 

 In other words, we are going to solve as many equations as there are terms in the product 

 Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation :

       Solve  :    x-2 = 0 

 Add  2  to both sides of the equation : 
                      x = 2

Solving a Single Variable Equation :

       Solve  :    2x+1 = 0 

 Subtract  1  from both sides of the equation : 
                      2x = -1
Divide both sides of the equation by 2:
                     x = -1/2 =

Supplement : Solving Quadratic Equation Directly

Solving  2x²-3x-2  = 0 directly

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Parabola, Finding the Vertex :

       Find the Vertex of   y = 2x²-3x-2

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  "y"  because the coefficient of the first term, 2 , is positive (greater than zero). 

 Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions. 

 Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex. 

 For any parabola,Ax²+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is     

 Plugging into the parabola formula     for  x  we can calculate the  y -coordinate : 
  y = * * - * -
or   y =

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = 2x²-3x-2
Axis of Symmetry (dashed)  {x}={ } 
Vertex at  {x,y} = { ,} 
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {, } 
Root 2 at  {x,y} = { , } 

Solve Quadratic Equation by Completing The Square

      Solving   2x²-3x-2 = 0 by Completing The Square .

 Divide both sides of the equation by  2  to have 1 as the coefficient of the first term :
   x²-(3/2)x-1 = 0

Add  1  to both side of the equation :
   x²-(3/2)x = 1

Now the clever bit: Take the coefficient of  x , which is  3/2 , divide by two, giving  3/4 , and finally square it giving  9/16 

Add  9/16  to both sides of the equation :
  On the right hand side we have :
   1  +  9/16    or,  (1/1)+(9/16) 
  The common denominator of the two fractions is  16   Adding  (16/16)+(9/16)  gives  25/16 
  So adding to both sides we finally get :
   x²-(3/2)x+(9/16) = 25/16

Adding  9/16  has completed the left hand side into a perfect square :
   x²-(3/2)x+(9/16)  =
   (x-(3/4)) • (x-(3/4))  =
  (x-(3/4))²
Things which are equal to the same thing are also equal to one another. Since
   x²-(3/2)x+(9/16) = 25/16 and
   x²-(3/2)x+(9/16) = (x-(3/4))²
then, according to the law of transitivity,
   (x-(3/4))² = 25/16

We'll refer to this Equation as  Eq. #  

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of
   (x-(3/4))²  is
   (x-(3/4))^2/2 =
  (x-(3/4))¹ =
   x-(3/4)

Now, applying the Square Root Principle to  Eq. #  we get:
   x-(3/4) = √ 25/16

Add  3/4  to both sides to obtain:
   x = 3/4 + √ 25/16

Since a square root has two values, one positive and the other negative
   x² - (3/2)x - 1 = 0
   has two solutions:
  x = 3/4 + √ 25/16
   or
  x = 3/4 - √ 25/16

Note that  √ 25/16 can be written as
  √ 25  / √ 16   which is 5 / 4

Solve Quadratic Equation using the Quadratic Formula

      Solving    2x²-3x-2 = 0 by the Quadratic Formula .

 According to the Quadratic Formula,  x  , the solution for   Ax²+Bx+C  = 0  , where  A, B  and  C  are numbers, often called coefficients, is given by :
                                     
            - B  ±  √ B²-4AC
  x =   ————————
                      2A

  In our case,  A   =     2
                      B   =    -3
                      C   =   -2

Accordingly,  B²  -  4AC   =
                     9 - () =
                     25

Applying the quadratic formula :

               3 ± √ 25
   x  =    —————
                    4

Can  √ 25 be simplified ?

Yes!   The prime factorization of  25   is
   5•5 
To be able to remove something from under the radical, there have to be  2  instances of it (because we are taking a square i.e. second root).

√ 25   =  √ 5•5  =
                ±  5 • √ 1   =
                ±  5

So now we are looking at:
           x  =  ( 3 ± 5) / 4

Two real solutions:

x =(3+√25)/4=(3+5)/4=

or:

x =(3-√25)/4=()/4=

Two solutions were found :

1.  x = -1/2 =
   
2.  x = 2

This is a tutorial on how to use the Algebra Calculator, a step-by-step calculator for algebra.

Solving Equations

First go to the Algebra Calculator main page. In the Calculator's text box, you can enter a math problem that you want to calculate.

For example, try entering the equation 3x+2=14 into the text box.

After you enter the expression, Algebra Calculator will print a step-by-step explanation of how to solve 3x+2=

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Examples

To see more examples of problems that Algebra Calculator understands, visit the Examplespage. Feel free to try them now.

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Math Symbols

If you would like to create your own math expressions, here are some symbols that Algebra Calculator understands:

+ (Addition)
- (Subtraction)
* (Multiplication)
/ (Division)
^ (Exponent: "raised to the power")

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Graphing

To graph an equation, enter an equation that starts with "y=" or "x=". Here are some examples: y=2x^2+1, y=3x-1, x=5, x=y^2.

To graph a point, enter an ordered pair with the x-coordinate and y-coordinate separated by a comma, e.g., (3,4).

To graph two objects, simply place a semicolon between the two commands, e.g., y=2x^2+1; y=3x

Polynomials

Algebra Calculator can simplify polynomials, but it only supports polynomials containing the variable x.

Here are some examples: x^2 + x + 2 + (2x^2 - 2x), (x+3)^2.

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Evaluating Expressions

Algebra Calculator can evaluate expressions that contain the variable x.

To evaluate an expression containing x, enter the expression you want to evaluate, followed by the @ sign and the value you want to plug in for x. For example the command 2x @ 3 evaluates the expression 2x for x=3, which is equal to 2*3 or 6.

Algebra Calculator can also evaluate expressions that contain variables x and y. To evaluate an expression containing x and y, enter the expression you want to evaluate, followed by the @ sign and an ordered pair containing your x-value and y-value. Here is an example evaluating the expression xy at the point (3,4): xy @ (3,4).

Checking Answers For Solving Equations

Just as Algebra Calculator can be used to evaluate expressions, Algebra Calculator can also be used to check answers for solving equations containing x.

As an example, suppose we solved 2x+3=7 and got x=2. If we want to plug 2 back into the original equation to check our work, we can do so: 2x+3=7 @ 2. Since the answer is right, Algebra Calculator shows a green equals sign.

If we instead try a value that doesn't work, say x=3 (try 2x+3=7 @ 3), Algebra Calculator shows a red "not equals" sign instead.

To check an answer to a system of equations containing x and y, enter the two equations separated by a semicolon, followed by the @ sign and an ordered pair containing your x-value and y-value. Example: x+y=7; x+2y=11 @ (3,4).

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Tablet Mode

If you are using a tablet such as the iPad, enter Tablet Mode to display a touch keypad.

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