Solve the system of linear equations using the elimination method.

Solve the following system:
{4 x+y/2 = 6 |     (equation 1)
2 y = 3 x+5 |     (equation 2)
Express the system in standard form:
{4 x+y/2 = 6 |     (equation 1)
-(3 x)+2 y = 5 |     (equation 2)
Add 3/4 × (equation 1) to equation 2:
{4 x+y/2 = 6 |     (equation 1)
0 x+(19 y)/8 = 19/2 |     (equation 2)
Multiply equation 1 by 2:
{8 x+y = 12 |     (equation 1)
0 x+(19 y)/8 = 19/2 |     (equation 2)
Multiply equation 2 by 8/19:
{8 x+y = 12 |     (equation 1)
0 x+y = 4 |     (equation 2)
Subtract equation 2 from equation 1:
{8 x+0 y = 8 |     (equation 1)
0 x+y = 4 |     (equation 2)
Divide equation 1 by 8:
{x+0 y = 1 |     (equation 1)
0 x+y = 4 |     (equation 2)
Collect results:
Answer: | 
| {x = 1
y = 4

1/2y+4x=6  

Multiply this equation by 4 on both sides....and we have

2y + 16x  = 24   (1)

2y = 3x + 5 

Rearrange this equation as 2y - 3x  = 5 

Now, multiply this equation by -1 on both sides.....and we have

-2y + 3x = -5    (2)

Add (1)  and (2), and we have

19x  = 19       divide both sides by 19

x = 1

And substituting this into (2), we have

-2y + 3(-)  = -5

-2y + 3  = - 5       subtract 3 from both sides

-2y = -8              divide both sides by -2

y = 4

So (x, y)   = (1, 4)