4x 1 2x 1

4x 1 2x 1 DEFAULT

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\mathrm{substitution} \mathrm{elimination} \mathrm{cramer} \mathrm{gauss\:jordan} \mathrm{simplify}
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Examples

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y=2x+1,y=4x-1

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Sours: https://www.symbolab.com/solver/system-of-equations-calculator/y=2x+1,y=4x-1

Linear equations with one unknown

Step by step solution :

Step  1  :

Equation at the end of step  1  :

(4x - 1) • (2x + 1) = 0

Step  2  :

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  :    4x-1 = 0 

 Add  1  to both sides of the equation : 
                      4x = 1
Divide both sides of the equation by 4:
                     x = 1/4 =

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 =

Two solutions were found :

  1.  x = -1/2 =
  2.  x = 1/4 =
Sours: https://www.tiger-algebra.com/drill/(4x-1)(2x_1)=0/
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DERIVADA 16 GRANVILLE Y= 2X^3/4+4X-1/4 DERIVADA DE UNA SUMA CAP 4 PAG 44

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2x 4x 1 1

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Solve 4x + 1 = 2x + 12

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Well, 1. 65 is the maximum adjusted for the circumstances. Oh, damn it.



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