SOLUTION: The sum of two numbers is 56. The larger exceeds twice the smaller by 2. Find the numbers. I already tried x+2+2y = 56 with the substitution of x= 56-y

Question 220990: The sum of two numbers is 56. The larger exceeds twice the smaller by 2. Find the numbers.
I already tried x+2+2y = 56 with the substitution of x= 56-y
Found 2 solutions by drj, checkley77:
Answer by drj(1380) (Show Source): You can put this solution on YOUR website!
The sum of two numbers is 56. The larger exceeds twice the smaller by 2. Find the numbers.
I already tried x+2+2y = 56 with the substitution of x= 56-y

Good effort...I think you made the wrong substitution or equation

Step 1. Let x be the smaller integer and y be the larger integer.

Step 2. Then y=2x+2 since the larger exceeds twice the smaller by 2.

Step 3. x+y=56 or x=56-y

Step 4. Then substituting into yields the following

Add 2y to both sides

Divide by 3 to both sides of the equation

and

Check equation if true in Step 2... 2*x+2=2*18+2=38...which is a true statement.

Step 5. ANSWER: The numbers are 18 and 38.

I hope the above steps were helpful.

For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

Good luck in your studies!

Respectfully,
Dr J

Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
x+y=56 or x=56-y
y=2x+2
y=2(56-y)+2
y=112-2y+2
y+2y=112+2
3y=114
y=114/3
y=38 ans.
x+38=56
x=56-38
x=18 ans.
Proof:
38+18=56
56=56

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