Question

What values of x make the two expressions below equal?
(2x+1)(x-7)/11(x-7)= 2x+1/11
A. All real numbers
B. All real numbers except -1/2
C. All real numbers except 7
D. All real numbers except -1/2 and 7

Answer 1

missing parenthesis

Answer 2

maybe

Answer 3

The best way to go about this problem is for you to plug in the numbers where stated "except"" to see if they work.

Answer 4

\[\frac{ (2x+1)(x-7) }{ 11(x-7) }=\frac{ 2x+1 }{ 11 }\]

is that it

Answer 5

yes that's what I wrote

Answer 6

If not A, but if they do you'd be picking one of the B to D. Simple right?

Answer 7

group the entire numerator in a bra kets

Answer 8

simplify, (x-7) can cancel in the first, but you must keep in mind that x cant be 7, it will make the original expression undefined

Answer 9

then both are over 11, and you can multiply both sides by 11 to cancel those...

left with

2x+1 = 2x+1 , true, all reals

Answer 10

so if you cancel x-7 out, top and bottom you're left with (2x+1)-11 right?

Answer 11

/11

Answer 12

yes,
\[\frac{ 2x+1 }{ 11 }= \frac{ 2x+1 }{ 11 }\]

Answer 13

It is like saying, 1 = 1, True statement, any value for x is possible there

Answer 14

remember you canceled out (x-7) in the denominator though, so x-7 can not be zero

Answer 15

same exact thing on both sides except the left one has parentheses

Answer 16

right, if you have the same things on both sides, then the variable can be any real number. Except here it can not be 7 since you cancelled out x-7 in the denominator, you cant have x-7=0 or x=7

Answer 17

zero in the bottom is bad

Answer 18

so it would be all real numbers except 7?

Answer 19

yes

Answer 20

thank you ;)-

Answer 21

:)

Answer 22

i forget what they call those solutions x=7, some term

Answer 23

did not know that myself haha

Answer 24

you can check, put x=7 and see if both sides are the same...

Answer 25

i did

Answer 26

extraneous solutions, just popped into my head, check that definition out if it sounds familiar.