Question

Algebra 2:
Which equation has the set of real numbers as its solution?

A) 2(-2r+4)= 4(-2r+6)
B) 3/5 (fraction) (z+4)= 4/5 (Fraction) z
C) 2x+5(x-3) -6=-8+7x-13
D) 0.7-(q+3) = 3.3-q
SHOW WORK TO ANSWER

Answer 1

the answer is all reals if the equation can be reduced to 0=0 somehow

Answer 2

not sure what you mean by
B) 3/5 (fraction) (z+4)= 4/5 (Fraction) z
what does (fraction) mean exactly ?

Answer 3

\[\frac{ 3 }{ 5}(z+4)=\frac{ 5 }{ 5 }z\]

Answer 4

ok, so for example lets do this one
multiple both sides by 5, what do you have?

Answer 5

25?
Im bad at math sooo sorry if i make u mad lol

Answer 6

\[\cancel5\cdot\frac3{\cancel5}=3\]

Answer 7

ahhh

Answer 8

so the left becomes\[3(x+4)\]what about the right side?

Answer 9

5z
or
1z

Answer 10

\[\cancel5\cdot\frac5{\cancel5}=5\]

Answer 11

so \[5z\]

Answer 12

yay

Answer 13

now we need to keep going\[3(z+4)=5z\]distribute to simplify the left...

Answer 14

3z+12=5z

Answer 15

but wait, your problem says it should be\[\frac35(z+4)=\frac45z\]is there a typo?

Answer 16

nop

Answer 17

well in your post you wrote
B) 3/5 (fraction) (z+4)= 4/5 (Fraction) z

and in the math feature you wrote\[\frac35(z+4)=\frac55z\]which is it?

Answer 18

4/5 or 5/5 ?

Answer 19

4/5
lol srry

Answer 20

okay, so what should the right be after multiplying by 5 then?

Answer 21

(everything else is still correct)

Answer 22

5z+20= 5z

Answer 23

no, you changed the left
nothing should change on the left; you did that correctly before\
the only change is on the right...

Answer 24

\[5\cdot\frac35(z+4)=5\cdot\frac45z\]simplifies to what?

Answer 25

5z+20=4z idk

Answer 26

you made a mistake on the left
you had it correctly before\[\cancel5\cdot\frac3{\cancel5}(z+4)=\cancel5\cdot\frac4{\cancel5}z\]

Answer 27

3z+12z=4z

Answer 28

yes
can you solve it for z?

Answer 29

oh wait, you should get\[3z+12=4z\](you have an extra z in there somehow)

Answer 30

ah and yea

3z+12=4z
-3z -3z
12=1z
z=12

Answer 31

correct
so there is only one answer:
z=12
that is not all real numbers, just one

Answer 32

so i got the answer?!?!?1
or do i do the others?

Answer 33

you now know that the answer to B) is *not* all real numbers since you cannot get it to be 0=0 through algebra

try A) and see what you get for that...

Answer 34

on a i can use any number then if it is a real numb it is the answer

Answer 35

just solve it and see what you get

Answer 36

2(-2r+4)= 4(-2r+6)

Answer 37

distribute the 2's on each side and you get...?

Answer 38

sorry I meant distribute the 2 on the left and the 4 on the right

Answer 39

-4r+8= -8r+24

Answer 40

right
try to solve it for r

Answer 41

4r=16

Answer 42

right, so r=?

Answer 43

0

Answer 44

4

Answer 45

right
again, that is just a single answer, so the answer to A) is also not ALL REALS because again we did not get 0=0

Answer 46

let's try C)

Answer 47

k

Answer 48

7x^2+21-6= -8+7x-13

Answer 49

not quite

2x+5(x-3) -6=-8+7x-13

distribute the 5 first

Answer 50

2x+5x-15-6= -8+7x-13

Answer 51

right :)
now simplify

Answer 52

7x-15-6= -8+7x-13

Answer 53

yep
now combine the constants and try to solve for x

Answer 54

soo to combine 7x+7x?

Answer 55

no
-15-6=?
-8-13=?

Answer 56

7x-9=-21+7x

Answer 57

-15-6 is not -9
try the left side again

Answer 58

-21

Answer 59

7x- -21=-21+7x
+

Answer 60

no you already have the - sign included...
7x+(-15-6)=( -8-13)+7x
7x+(-21)=-21+7x
7x-21=7x-21

Answer 61

its the same so were done?

Answer 62

right
or you can notice that we can subtract 7x-21 from both sides to get
0=0
which is always true, so the answer is all reals
you should still check D) to make sure that it also is not all reals, there could be more than one answer for all we know (at least from the information given it is not clear that there is only one answer)

Answer 63

yay tnx <3

Answer 64

welcome!