Question

Use mateix method.

x + 2y + z = 79
x + 3z = 11
2x-3y =1​

Answer 1

Answer:

We have,

x+2y+z=7 .(i)

x+3z=11 .(ii)

2x−3y=1 ..(iii)

From equation (i), we get

z=7−x−2y

Substituting z=7−x−2y in equation (ii), we get

x+3(7−x−2y)=11

⇒x+21−3x−6y=11

⇒−2x−6y=−10  ..(iv)

Adding equations (iii) and (iv), we get

−9y=−9⇒y=1

Putting y=1 in equation (iii), we get x=2.

Putting x=2,y=1 in equation (i), we get

2+2+z+7⇒z=3

Hence, x=2,y=1,z=3.

Answer 2

Answer:

X+ 2Y + Z =79

then X+ z +2y=79

X+z+y=79/2=39.5

...

X+3Z=11

X=11/3z

=3.666667

2x -3y=1

2x-y= 4y

2x=4y