prealgebra problem, really confused, please help!!!

lillybeth

lillybeth

Junior Member
Joined
Nov 1, 2012
Messages
211
My math homeworkstated this problem-

 -14x + 14= -8 (solve for x)

 I already know the answer to the problem: x= 11/17

My question is: How do you find the answer? :confused:

Thanks for the help guys! (and/or girls)
 
lillybeth said:
My math homeworkstated this problem-

-14x + 14= -8 (solve for x)

I already know the answer to the problem: x= 11/17 ====> Are you sure about this? Is there a typo here?

My question is: How do you find the answer? :confused:

Thanks for the help guys! (and/or girls)
Click to expand...

See my comment above in green. Did you calculate 11/17 or is that what the answer key says? Because the answer is not 11/17.
 
answer

srmichael said:
See my comment above in green. Did you calculate 11/17 or is that what the answer key says? Because the answer is not 11/17.
Click to expand...

On mathway.com, you type in your algebra problems and it answers them for you. 11/17 is what it gave me.
 
lillybeth said:
On mathway.com, you type in your algebra problems and it answers them for you. 11/17 is what it gave me.
Click to expand...

One way to check the answer (for these problems) is to replace "x" in the original equation and check if it makes sense.

Then the original equation is

-14 *x + 14 = -8

The as given by computer is x = 11/17

Then we check

- 14 * (11/17) + 14 = 14 * 6/17 = 4.941176 ← surely not equal '8'

hence the answer ( x = 11/17) cannot be correct.

So

may be you are posting the wrong problem here

may be you "input" wrong problem at mathway.com

may be mathway gone crazy and gave wrong answer

may be you are posting wrong answer here (most probably)

may be combination of all above.......

The correct answer to the posted problem is x = 11/7 (the denominator is 7 not 17)
 
Last edited by a moderator:
Subhotosh Khan said:
One way to check the answer (for these problems) is to replace "x" in the original equation and check if it makes sense.

Then the original equation is

-14 *x + 14 = -8

The as given by computer is x = 11/17

Then we check

- 14 * (11/17) + 14 = 14 * 6/17 = 4.941176 ← surely not equal '8'

hence the answer ( x = 11/17) cannot be correct.

So

may be you are posting the wrong problem here

may be you "input" wrong problem at mathway.com

may be mathway gone crazy and gave wrong answer

may be you are posting wrong answer here (most probably)

may be combination of all above.......

The correct answer to the posted problem is x = 11/7 (the denominator is 7 not 17)
Click to expand...

Mathway gone crazy. Ok. I am still trying to figure out the real answer, can you to?

 
lillybeth said:
My math homeworkstated this problem-

 -14x + 14= -8 (solve for x)

 I already know the answer to the problem: x= 11/17

My question is: How do you find the answer? :confused:

Thanks for the help guys! (and/or girls)
Click to expand...

-14x + 14 = -8

Minus 14 on both sides:
-14x = -22

Both sides *-1 for ease:
14x = 22

Both sides divided by 14:
x = 22/14

Which is the same as:
x = 11/7
 
Oh. Thanks!

phoenix9124 said:
-14x + 14 = -8

Minus 14 on both sides:
-14x = -22

Both sides *-1 for ease:
14x = 22

Both sides divided by 14:
x = 22/14

Which is the same as:
x = 11/7
Click to expand...

Thanks Pheonix! This has been the most helpful post to me in this thread. :) I see now.
 
Oh no!

Sorry guys, mathway did it right, I accidently typed the problem wrong here. Thanks anyway.:D
 
Denis said:
Originally Posted by phoenix9124
Want to try a new equation?
How about this one:
12 - 7x = 4
What is x ?

Yes...why are you unsure?

Here's how YOU can make sure; substitute your answer in equation:
12 - 7(8/7) = 4 ?
12 - 8 = 4 ?
4 = 4 : SUCCESS!!!!!
Click to expand...
Sweet!
 
This method of substituting your solution(s) into the original equation is very useful when extraneous solutions may have been introduced in the solving process. Consider the equation:

\(\displaystyle x=\sqrt{2-x}\)

Squaring both sides gives:

\(\displaystyle x^2=2-x\)

\(\displaystyle x^2+x-2=0\)

\(\displaystyle (x-1)(x+2)=0\)

So, we have the two roots \(\displaystyle x=-2,\,1\)

Since we squared the equation in the process of solving, we need to verify by substitution that neither of the roots is extraneous. In doing so we find:

i) \(\displaystyle x=-2\)

\(\displaystyle -2=\sqrt{2-(-2)}\)

\(\displaystyle -2=2\)

This root is extraneous.

ii) \(\displaystyle x=1\)

\(\displaystyle 1=\sqrt{2-1}\)

\(\displaystyle 1=1\)

This root is valid.
 
Cool.

MarkFL said:
This method of substituting your solution(s) into the original equation is very useful when extraneous solutions may have been introduced in the solving process. Consider the equation:

\(\displaystyle x=\sqrt{2-x}\)

Squaring both sides gives:

\(\displaystyle x^2=2-x\)

\(\displaystyle x^2+x-2=0\)

\(\displaystyle (x-1)(x+2)=0\)

So, we have the two roots \(\displaystyle x=-2,\,1\)

Since we squared the equation in the process of solving, we need to verify by substitution that neither of the roots is extraneous. In doing so we find:

i) \(\displaystyle x=-2\)

\(\displaystyle -2=\sqrt{2-(-2)}\)

\(\displaystyle -2=2\)

This root is extraneous.

ii) \(\displaystyle x=1\)

\(\displaystyle 1=\sqrt{2-1}\)

\(\displaystyle 1=1\)

This root is valid.
Click to expand...
cooleo. thanks! :)
 
You must log in or register to reply here.
Top