HELP

Solve the following system:
{4 q + 3 p = 8 | (equation 1)
3 q + 4 p = 13 | (equation 2)

 

Swap equation 1 with equation 2:
{4 p + 3 q = 13 | (equation 1)
3 p + 4 q = 8 | (equation 2)

 

Subtract 3/4 × (equation 1) from equation 2:
{4 p + 3 q = 13 | (equation 1)
0 p+(7 q)/4 = -7/4 | (equation 2)

 

Multiply equation 2 by 4/7:
{4 p + 3 q = 13 | (equation 1)
0 p+q = -1 | (equation 2)

 

Subtract 3 × (equation 2) from equation 1:
{4 p+0 q = 16 | (equation 1)
0 p+q = -1 | (equation 2)

 

Divide equation 1 by 4:
{p+0 q = 4 | (equation 1)
0 p+q = -1 | (equation 2 

 p = 4          and          q= -1
 

Multiply first equation by - 1 1/3 to get

-4p - 16/3 q = - 32/3    now add to second equation to get

-7/3 q = 7/3

q = -1

then  since   3 p + 4q = 8       3p -4 = 8       then p = 4