We have two simultaneous equations:

(-1/2)x + (1/3)y = 0 ... (1)

x + 6y = 16 ... (2)

To solve theses equations for values of x and y, multiplying equation (1) by 2 we get:

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

Adding equation (2) and (3) we get:

x - x + 6y + (2/3)y = 16 - 0

(20/3)y = 16

Therefore:

y = 16/(20/3) = 12/5 = 2.4

Substituting the above value of y in equation (20 we get:

x + 6*2.4 = 16

x + 14.4 = 16

x = 16 - 14.4 = 1.6

Answer:

x = 1.6, y = 2.4

To add the quotients from the first equation, we have to have a common denominator, which in this case is 2*3=6.

So, we'll multiply (-x/2) by 3 and (y/3) by 2 and we'll get:

-3x + 2y = 0

x + 6y = 16

Now, we'll eliminate the unknown x in this way: we'll multiply the second equation by 3, all over, and we'll add it to the first one, like this:

3x+3*6y-3x+2y=3*16+0

We'll reduce the similar terms:

18y+2y = 48

20y=48

We'll divide by 4:

5y=12

We'll divide by 5:

**y=12/5**

Now, we'll substitute y in the second equation:x + 6y = 16.

x + 6*(12/5) = 16

x = 16 - 72/5

x = (5*16 - 72)/5

x = (80-72)/5

**x = 8/5**

**The solution of the system is (8/5 , 12/5).**

The given equations are :

-x/2+y/3 = 0 .....(1) and x+6y = 16......(2)

From the 2nd st equation, x+6y = 16, we get: x= 16-6y. substitute this 16-6y in place of x in the 1st equation and then solve the equation in the single variable y:

-(16-6y)/2 +y/3 = 0. Multiplying by 6, to get away with denominators:

-3(16-6y) +2y = 0.

-48+18y+2y = 0.

20y = 48

y = 48/20 = 2.4. Substitute this in 2nd eq : x+6(2.4) = 16. Or x = 16-14.4 = 1.6

So x=1.6 and y=2.4