Algebra 1/Fractions - Checking the solution of an equation

[Michaela03]

Need help with these two math problems please:

1. 2p+ 1/3= 7/3

2p + 1/3-1/3=7/3-1/3=6/3=2 This is the best answer I could come up with. How would I do this problem and check my math?


2. 5n+ 1/2 =10 1/2

5n+1/2 - 1/2=10 1/2 - 1/2 =10 5n/5=10/5=2 This is the best answer I could come up with. How I would I do this problem and check my math?

 

[Dr.Peterson]

Need help with these two math problems please:

1. 2p+ 1/3= 7/3


2. 5n+ 1/2 =10 1/2

If you want us to check your solutions, you'll have to show us your solutions!

That would include both the answers you get, and the work you did, so we can point out any errors you made.

 

[Michaela03]

I apologize for any confusion; it's my first time, I got the math problems now. Until next time, thank you!

 

[Dr.Peterson]

Need help with these two math problems please:

1. 2p+ 1/3= 7/3

2p + 1/3-1/3=7/3-1/3=6/3=2 This is the best answer I could come up with. How would I do this problem and check my math?


2. 5n+ 1/2 =10 1/2

5n+1/2 - 1/2=10 1/2 - 1/2 =10 5n/5=10/5=2 This is the best answer I could come up with. How I would I do this problem and check my math?

The first is correct as far as you went, but you didn't solve for p!

If 2p = 2, then what is p?

The second is correct; you first found that 5p = 10, and then that p = 2.

To check a solution, you just think about what it means for this to be a solution. When you say that p = 2, you are claiming that this is the value of p for which the equation you started with is true. So replace p with 2; the left-hand side becomes 5(2) + 1/2 = 10 + 1/2 = 10 1/2, which is just what it has to be. So you got it.

For the first, if you claimed that p = 2, you would be wrong, because the left-hand side becomes 2(2) + 1/3 = 4 1/3, while the right-hand side is 7/3 = 2 1/3. That tells you to go back and find the error, which is just that you didn't finish.