Question

Solve the following: Three times the 1st number plus the 2nd number plus twice the 3rd is 5. If three times the 2nd number is subtracted from the sum of the 1st and three times the 3rd number, the result is 2. If the 3rd number is subtracted from two times the 1st number and three times the 2nd, giving a result of 1. Find the three numbers.

Answer 1

Three times the 1st number plus the 2nd number plus twice the 3rd is 5 is the same as 3x+y+2z=5. If three times the 2nd number is subtracted from the sum of the 1st and three times the 3rd number, the result is 2 is just x+3z-3y=2. And if the 3rd number is subtracted from two times the 1st number and three times the 2nd, giving a result of 1 means 2x+3y-z=1. Then you use substition on these equations to get a equation where one variable equals 2 others, like using the first to get y=5-2z-3x and then this can be substituted into the other two to get x+3z-3(5-2z-3x)=2 and 2x+3(5-2z-3x)-z=1 we can then simplify and subtract the equations. After simplification we have 10x+9z=17 and 7z+7x=16 which can be turned into 70x+63z=119 and 70x+70z=160 which can be then subtracted to get that 7z=41 and z=41/7. Now we backtrack to a two variable equation like 7z+7x=16 and plug in to find x. So after plugging in we get 41+7x=16 and 7x=-25 so x=-25/7. Now we choose a 3 variable equation and plug in. So taking y=5-2z-3x we plug in 41/7 for z and -25/7 for x to get y=5-82/7+75/7 and y=5-7/7 and y=4. Therefore x = -25/7 y = 4 and z = 41/7.