Solve the following system:
{x/3 + y/2 = 14 | (equation 1)
x/2 + y/5 = 10 | (equation 2)
Swap equation 1 with equation 2:
{x/2 + y/5 = 10 | (equation 1)
x/3 + y/2 = 14 | (equation 2)
Subtract 2/3 × (equation 1) from equation 2:
{x/2 + y/5 = 10 | (equation 1)
0 x+(11 y)/30 = 22/3 | (equation 2)
Multiply equation 1 by 10:
{5 x + 2 y = 100 | (equation 1)
0 x+(11 y)/30 = 22/3 | (equation 2)
Multiply equation 2 by 30/11:
{5 x + 2 y = 100 | (equation 1)
0 x+y = 20 | (equation 2)
Subtract 2 × (equation 2) from equation 1:
{5 x+0 y = 60 | (equation 1)
0 x+y = 20 | (equation 2)
Divide equation 1 by 5:
{x+0 y = 12 | (equation 1)
0 x+y = 20 | (equation 2)
Collect results:
Answer: |x = 12 and y = 20
