Algebra help !!

Question

The sum of three number is 12. the sum of twice the first number , 5 times the second number , & 6 times the third number is 64. The difference between 4 times the first number and the second number is 0. Find the three numbers

Answer 1

1. A + B + C = 12
2. 2A + 5B + 6C = 64
3. 4A - B = 0

4A - B = 0

4A - B+ B = 0+B

4A=B

A + 4A + C= 12

5A + C = 12

5A-5A + C =12-5A

C=12-5A

2A + 5(4A) + 6(12-5A) = 64

2A + 20A + 72 -30A = 64

72-8A=64

-8A+72-72=64-72

-8A=-8

-8  -8

A=1

B=4(1)=4

C=12-5(1)=7

A + B + C = 12

1 + 4 + 7 = 12

12=12

2A + 5B + 6C = 64

2(1) + 5(4) + 6(7) = 64

2 + 20 + 42 =64

64=64

4A - B = 0

4(1) - 4 = 0

4 - 4 = 0

0=0

Numbers are 1,4,&7