Anonymous A. answered 07/27/17
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The question isn't clear but it will be the same process if I've interpreted it incorrectly.
Let's say the smaller number is x1 and the larger number is x2.
The smaller number is 3 times larger and nine more than 2 times the larger number. We can write this as an equation by saying,
9+3x1=2x2 ...(1)
We have two unknowns and one equation so we need an additional equation to solve for both unknowns.
The bigger number is 26 more than 2 times the smaller number,
x2+26=2x1 ...(2)
We have a system of equations with two equations and two unknowns, we can use substitution to solve for one unknown.
x2+26 = 2x1-->x1= (x2+26)/2
Now plug this into equation (1) for x1
9+3[(x2+26)/2] = 2x2 ---> 9+(3/2)x2+39=2x2
48 = (1/2)x2 ---> x2 = 96
Now plug 96 into either equation (1) or (2) for x2 to solve for x1
Using equation (1),
9+3x1=2*96 ---> x1 = (192-9)/3 = 61
So the smaller number is 61 and the larger number is 96