Question

the sum of 4 times one number and 3 times a second number is 64. If the sum of the two numbers in 19, find the two new numbers

Answer 1

From the problem statement we derive a system of equations.

4a + 3b = 64
a + b = 19

There are a variety of ways to solve this, but I'll use substitution.

Isolate a variable.
a = 19 - b
... And substitute that into the other equation.

4(19 - b) + 3b = 64
76 - 4b + 3b = 64
-b = -12
b = 12

Plug that into the isolated equation.
a = 19 - 12
a = 7

So our answer is (a, b) = (7, 12).