Solve the following system in three variables using linear combinations. 3x−2y+z=0 4x+y−3z=−9 9x−2y+2z=20

The solution of the system of equations is

x=2,y=7,z=8

Step-by-step explanation:

we have

3x-2y+z=0

isolate the variable z

z=2y-3x > equation A

4x+y-3z=-9 > equation B

9x-2y+2z=20 > equation C                  

substitute equation A in equation B and equation C

4x+y-3(2y-3x)=-9

4x+y-6y+9x=-9

13x-5y=-9 > equation D

9x-2y+2(2y-3x)=20

9x-2y+4y-6x=20

3x+2y=20 > equation E

Solve the system

13x-5y=-9 > equation D

3x+2y=20 > equation E

Multiply equation E by 2.5 both sides

2.5*(3x+2y)=20*2.5

7.5x+5y=50 > equation F

Adds equation D and equation F

13x-5y=-9

7.5x+5y=50

13x+7.5x=-9+50

20.5x=41

x=41/20.5

x=2

Find the value of y

substitute the value of x in the equation E

3(2)+2y=20

6+2y=20

2y=20-6

2y=14

y=7

Find the value of z

Substitute the value of x and the value of y in equation A

z=2y-3x

z=2(7)-3(2)

z=14-6

z=8

therefore

The solution of the system of equations is

x=2,y=7,z=8

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

c. (3,2,6) is the answer.