pls help me to solve them :(

Question

Solve : -4<2x+6≤3x-1

 

 

solve the equation 3y^2-9y+1=0 by using completing square

Answer 1

-4 < 2x+6 ≤ 3x-1

 

-4 < -x+6 ≤ -1

 

-10 < -x ≤ -7

 

10 > x ≥ 7    Solution:  [7,10)

 

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3y² - 9y + 1 = 0

 

  y² - 3y = -1/3     Add [½(-3)]² to both sides of the equation

 

  y² - 3y + 9/4 = -1/3 + 9/4

 

      (y-3/2)² = 23/12

 

       y - 3/2 = ±√(23/12)

 

              y = 3/2 ± √(23/12)  = 3/2 ± √23/(2√3)

 

                                           = 3/2 ± √69/6

 

                                           = (9 ± √69)/6

 

Answer 2

I would break up the conjunction inequality into two part.

 

-4 < 2x + 6               and                  2x + 6 ≤ 3x - 1

 

 

Now solve each part.  Then join the solutions.

 

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To complete the square, we group all the y terms to one side of the equation first.

 

3y² - 9y = -1

 

 

Next, divide both sides of the equation by 3.  We do this because we want the coefficient of y² to be 1.

 

y² - 3y = -1/3

 

 

Next, we need to add a term to both sides of the equation that will make the left side a perfect square trinomial.  If we divide the coefficient of y, then square the result, we get our term.  Add 9/4 to both sides of the equation.

 

y² - 3y + (9/4) = (9/4) - (1/3)

 

(y - 3/2)² = (27 - 4) / 12

 

(y - 3/2)² = 23/12

 

 

Now just square-root both sides of the equation.

 

y - 3/2 = ±√(23/12)

 

 

Can you solve for y from here?