Solve the system of equations. 4x+2y+5z=8 5x+3y+4z=9 6x+3y+3z=3

a) (x=-3,y=6,z=1)
b) (x=-5, y=8,z=3)

x = -4, y = 7,  z = 2.

Step-by-step explanation:

4x + 2y + 5z = 8(1)

5x + 3y + 4z = 9(2)

6x + 3y + 3z = 3(3)

Subtract:  equation (2) - equation (3):

-x + z = 6 (4)

Multiply   (1) by 3 and (2) by 2:

12x + 6y + 15z = 24 (5)

10x + 6y + 8z = 18 (6)

Subtract: (5) - (6):

2x + 7z = 6(7)

Multiply  equation (4) by 2:

- 2x + 2z = 12(8)

Add (7) + (8):

9z = 18 , so  z = 2.

Substitute for z in equation (4)

-x + 2 = 6

-x =  4 so x = -4.

Now find y by substituting in equation (1):

4(-4) + 2y + 5(2) = 8

2y =  8 + 16 - 10 = 14

y = 7.

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solution:

the given function is ,

      =

the three roots of this quadratic function is , 0 ,8 and 10.all the roots are real and rational.

if a quadratic equation is of the form ,

sum of the roots = a + b +c= 0+8+10=18=  

product of roots taken two at a time = a b + b c+ c a = 0+80+0=80 =  

product of roots = a b c = 0=  

there are 11 doors in a hall starting from 0 to 11.the entry in the hall is permitted through gate number 0 and exit is allowed through only gate number 8 and 10.so, three roots(0,8,10) of quadratic function is given by entry and exit route of this hall.