SOLUTION: Hi Checkley, Are you still there? I have one more problem involving integers. This particular problem: The sume of the two integers and then the second is 8 more than 3 times

Question 111747: Hi Checkley,
Are you still there? I have one more problem involving integers. This particular problem: The sume of the two integers and then the second is 8 more than 3 times the first.
So, I think that I should also set up this problem similar.
I worked it out and I know my answer was wrong.
Here is what I did.
The sume of two numbers is 76. The second is 8 more than 3 times the first.
What are the two numbers?
I put
x+y = 76
y= 3x+8
Is that correct? I am not certain how to deal with this problem correctly.
is there any way you can help me with this one.
Thank you Checkley.
Tracy
Found 3 solutions by jim_thompson5910, checkley71, kalyanam:
Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
You set it up correctly

Start with the given system

Plug in into the first equation. In other words, replace each with . Notice we've eliminated the variables. So we now have a simple equation with one unknown.

Combine like terms on the left side

Subtract 8 from both sides

Combine like terms on the right side

Divide both sides by 4 to isolate x

Divide

Now that we know that , we can plug this into to find

Substitute for each

Simplify

So our answer is and

So our two numbers are 17 and 59

Check:

works
Answer by checkley71(8403)   (Show Source): You can put this solution on YOUR website!
y=3x+8 substitute (3x+8) for y in the other equation.
x+(3x+8)=76
x+3x+8=76
4x=76-8
4x=68
x=68/4
x=17 answer.
y=3*17+8
y=51+8
y=59 answer.
proof
17+59=76
76=76
Answer by kalyanam(1)   (Show Source): You can put this solution on YOUR website!
x+y=76
y=3x+8
y-3x = 8 subtract y+x = 76
-4x= -68
x= 17
y=76-17
y= 59

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