bwork math 2010002Spl lllnUTU EF part 2 Problem 4 Prev TUp T

bwork math -2010-002-Spl lllnUTU EF part 2: Problem 4 Prev TUp TNe Solve the system using elimination 6 12y +32 11 42 2y 2 14 Note: You can earn partial credit on this problem. Preview Answers l Submit Answers You have attempted this problem 0 times. You have unlimited attempts remaining

Solution

6x+2y+3z=11...(1)

5x-2y-4z=-3.....(2)

2x+3y+6z=14...(3)

(1)+(2)

(6x+2y+3z=11)+(5x-2y-4z=-3)=(11x-z=8)....(4)

3*(1)-2*(3)

2*(2x+3y+6z)=(4x+6y+12z=28)

3*(6x+2y+3z=11)=18x+6y+9z=33

3*(1)-2*(3)

(18x+6y+9z=33)-(4x+6y+12z=28)={12x-3z=5)...(5)

Now (4)*3-(5)

(33x-3z=24)-(12x-3z=5)={11x=19}

x=(19/11)

Now substitute x value in (4) then we get z

11*(19/11)-z=8

19-z=8

z=11

For y value keep x and z in any equation given

Let (1)

6*(19/11)+2y+3(11)=11

(114/11)+2y+33=11

114/11)+2y=-22

2y=-22-(114/11)

2y=((-22*11-114)/11)

2y=((-242-114)/11)

2y=(-356/11)

y=(-356/11)*(1/2)

y=(-118/11)

Therefore x=19/11,y=(-118/11)z=11

Proof:

6(19/11)+2(-118/11)+3(11)=11

114-356+(33*11)=121

114-356+363=121

477-356=121

121=121

Therefore

X=19/11

Y=-118/11

Z=11