Find the exact solution to the system of equations Check you

Find the exact solution to the system of equations. Check your answer algebraically. y=x^2-6x +1 y+2x=6 16.12x^2-4x=y+1 x+y=1

Solution

15) Given that y = x²-6x+1 ------- Eq (1)

                     y + 2x= 6 ---------- Eq (2)

From eq (2) ,

                     y = 6-2x    -------------- Eq(3)

From Eqs (1) and (3),

x²-6x+1 = 6-2x

  x²-4x-5 = 0

x² - 5x +x -5 = 0

x (x-5) +1 (x-5) = 0

(x-5) (x+1)= 0

x = 5 ,-1

Substitute values of x in Eq (3) to get values of y,

y = 6-2x    -------------- Eq(3)

x = 5 then y = 6- 2.5 = - 4

x = -1 then y = 6 -2 . -1 = 8

Therefore,

solutions are

(x,y) = (5,-4) , (-1,8)

16) 2x²-4x = y+ 1 -------- Eq (1)

      x+y = 1 -------   Eq (2)

    From Eq (2),

          y = 1-x ------ Eq(3)

     Substitute value of y in eq (1),

Eq (1) ----->   2x²-4x = y+ 1

                    2x²-4x = 1-x+ 1                [ y = 1-x ]

                   2x²-3x -2 = 0

                2x2 -4x+x -2 =0

                2x(x-2) + (x-2) = 0

                  (x-2) (2x+1) = 0

             x = 2 , x = -1/2

Substitute values of x in Eq (3) to get values of y,

            Eq (3) is   y = 1-x

x =2 then y = 1-2 = -1

x = -1/2 then y = 1- (-1/2) = 3/2

Therefore,

solutions are

(x,y) = (2,-1) , (-1/2, 3/2)