Solve the system by finding the reduced row-echelon form of the augmented matrix.
x-4y-2z=2
2x-11y-10z=13
-x+6y+6z=-8

If there is one solution, give its coordinates in the answer spaces below. If there are infinitely many solutions, enter t in the answer blank for z, enter a formula for yy in terms of tt in the answer blank for y and enter a formula for x in terms of tt in the answer blank for x. If there are no solutions, leave the answer blanks for x, y and z empty.

Therefore the solution are

x=-6t-10

y=-3-2t

z=t

Step-by-step explanation:

Given system of the equation is

x-4y -2z=2

2x-11y-10z=13

-x+6y+6z=-8

Transforming matrix of the reduce row echelon form

add -2 times the first row to the second row

add 1 times the row to the third row

multiply the second row by

add -2 times the second row to the third row

add 4 times the second row to the first row

The reduce form of the system of equation is

x+0.y+6.z=-10......(1)

0.x+y+2z=-3.........(2)

There are infinity of many solution

z= t (say)

Putting the value of z in the equations of (1) and (2)

x +6t=-10

⇒x=-6t-10

and

y+2t=-3

⇒y= -3-2t

Therefore the solution are

x=-6t-10

y=-3-2t

z=t

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?

a is 8.b is 10.c is 10.5. d is -.5

step-by-step explanation:

just minus the smallest number from the biggest.