Question

3x-3y+7z=17
3x+y-3z=-5
27x-y-7z=0

Answer 1

solve usinh elimination!

Answer 2

Right away, i see the first two equations both have 3x, so if you subtract them both, you will eliminate the x's. In the second and third equation, they both have 1y, so subtract them together (in this case, actually just add them since one in positive and one is negative) and you will eliminate the y's.

Answer 3

3x-3y+7z=17 -----(1
3x+y-3z=-5 -----(2
27x-y-7z=0 -----(3

Multiplying eq (2 by 3 and adding it with eq (1 we find
3x-3y+7z=17 -----(1
9x+3y-9z=-15 -----(2
------------------------
12x -2z=2
ie. 6x-z = 1_-----(4

Now add equations (2 and (3 in order to get another eq in vaiariable in x and z

Answer 4

Now add equations (2 and (3
3x+y-3z=-5 -----(2
27x-y-7z=0 -----(3
-------------------
30x-10z =-5
i.e. 3x-z=-5/10

i.e. 3x-z=-0.5
i.e. 3x+0.5=z ---(5
substituting z= 3x+0.5 in eq (4 we have ....
6x-( 3x+0.5)= 1
i.e. 6x- 3x-0.5= 1
i.e. 3x= 1.5
i.e. x= 1.5/3
i.e. x= 0.5= 1/2

substituting x= 0.5 in eq (5 we have ..
z= 3x+0.5 = 3.(0.5) +0.5 = 1.5+0.5 = 2.0=2

Now substituting x= 0.5& z=2 in eq (2 we have ..

3x+y-3z=-5 -----(2

i.e. 3(0.5)+y-3(2)=-5
i.e. 1.5+y-6=-5
i.e. y=6-5 -1.5
i.e. y=6-6.5
i.e. y=-0.5
Thus
x=0.5, y= -0.5 & z=2 are the required solution of the given system of linear equations.

@KenzieNoel

Answer 5

thanks hang on let me get it down

Answer 6

that explanation is very confusing @dpasingh

Answer 7

@KenzieNoel do you have any other idea to solve it