Solve the following system of equations using matrices​ (row operations). x-y=3
6x-5z=36
6y+2z=18

(6,3,0)

Step-by-step explanation:

Given,

x -  y + 0z  =  3

6x -  0y - 2z = 36

0x + 6y + 2z = 18

You have the following augmented matrix:

(1) 1     -1     0      3

(2) 6    0   -2      36

(3) 0    6    2       18

Let's (3) add to (2) and divide by 6

(1)  1    -1    0       3

(2) 1     1    0       9

(3) 0    6   2       18

Now, let's (2) add  to (1) and divide by 2

(1)  1   0    0     6

(2) 1    1    0     9

(3) 0   6   2     18

Let's (3) ÷ 2

(1)  1    0   0     6

(2) 1    1   0      9

(3) 0   3   1      9  

Let's (1) subtract from (2)

(1)  1   0   0     6

(2) 0   1   0     3

(3) 0   3   1     9

Finally, let's (2) multiply by 3 and subtract from (3)

(1)  1   0   0     6

(2) 0   1   0     3

(3) 0   0   1     0

Thus,

x = 6, y = 3 and z = 0

The answer is (6, 3, 0)

We have to verify the answer

6 - 3 = 3  ⇒⇒ 3 = 3

6*6 - 5*0 = 36   ⇒⇒ 36 = 36

6*3 + 2*0 = 18    ⇒⇒ 18 = 18  

sinn is the total amount of students now and m were the students before, you subtract them both (n-m) and divide that by m. then, since to get a percentage from a fraction you have to multiply by 100, your answer should look like this: n-m/m*100

step-by-step explanation:

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.