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
Algebra ->
Customizable Word Problem Solvers
-> Numbers
-> 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
Log On
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):
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
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