anonymous
  • anonymous
How many solutions does the equation have? 2(2x+5)=4(x+3) A. one solution B. infinite solutions C. no solution
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
SolomonZelman
  • SolomonZelman
2(2x+5)=4(x+3) 4x+10=4x+12
SolomonZelman
  • SolomonZelman
Is that true for any values of x?
anonymous
  • anonymous
no solution?

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SolomonZelman
  • SolomonZelman
Yes
anonymous
  • anonymous
Thank You!
SolomonZelman
  • SolomonZelman
Yw !
anonymous
  • anonymous
How many solutions does the equation have? a+5=1/5(5a+25) A. one solution B. infinite solutions C. no solution What about this one i think infinite solutions
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ a+5=\frac{1}{5}(5a+25)}\) like this?
anonymous
  • anonymous
yes
SolomonZelman
  • SolomonZelman
then you are correct - "infinte solutions" is the right answer.
SolomonZelman
  • SolomonZelman
Because when you expand the right side, you get: \(\large\color{black}{ a+5=\frac{1}{5}(5a+25)}\) \(\large\color{black}{ a+5=a+5}\) and that is true for any value of a.
anonymous
  • anonymous
Thank you again!
SolomonZelman
  • SolomonZelman
You welcome, again! :)
anonymous
  • anonymous
I have 3 more
anonymous
  • anonymous
if you could check them
SolomonZelman
  • SolomonZelman
Alright...
anonymous
  • anonymous
How many solutions does the equation have? 4x + 2(x-3)=8x + 12 A. one solution B. infinite solutions C. no solution
SolomonZelman
  • SolomonZelman
what did you choose?
anonymous
  • anonymous
A?
SolomonZelman
  • SolomonZelman
Yes, correct!
anonymous
  • anonymous
Thank You!!
SolomonZelman
  • SolomonZelman
yw
anonymous
  • anonymous
Which expressions would complete this equation so that it has infinitely many solutions? 8 + 2(8x – 6) = Choose exactly two answers that are correct. A. 2(4x + 7) B. 9x – 10 C. 16x – 4 D. 4(4x – 1)
SolomonZelman
  • SolomonZelman
and your choice was?
anonymous
  • anonymous
C and D?
anonymous
  • anonymous
Or A and D?
SolomonZelman
  • SolomonZelman
8 + 2(8x – 6) = 8 + 16x – 12 = 16x – 4. ` same = same ` ----> ` 8 + 2(8x – 6) = 16x - 4 ` C is right
anonymous
  • anonymous
So C and D?
SolomonZelman
  • SolomonZelman
8 + 2(8x – 6) = 8 + 4(4x – 3) = 8 + 4(4x – 1 - 2) = 8 + 4(4x – 1) + (-2)(4) = 8 + 4(4x – 1) -8 = 4(4x – 1) =
SolomonZelman
  • SolomonZelman
yes, C and D
anonymous
  • anonymous
Thanks Again!!! I have one more
SolomonZelman
  • SolomonZelman
k
anonymous
  • anonymous
What is the best way to classify each equation? Column AColumn B 1. 8x + 24=2(4x+12) 2. 5x + 18-x=2(2x+8) 3. 7(3x-2)=20x-13 4. 3x + 2(x-10)=5(x-4) A. identity B. contradiction C. neither @SolomonZelman
SolomonZelman
  • SolomonZelman
Can you write the Columns more clearly please?
anonymous
  • anonymous
dont worry about the columns... Column A is with the long equations and Column B is A. identity B. contradiction C. neither
SolomonZelman
  • SolomonZelman
Column A Column B 1. 8x + 24=2(4x+12) 2. 5x + 18-x=2(2x+8) 3. 7(3x-2)=20x-13 4. 3x + 2(x-10)=5(x-4) like this?
anonymous
  • anonymous
what you have underneath column B is suppose to be under Column A and under Column B is suppose to be this A. identity B. contradiction C. neither
SolomonZelman
  • SolomonZelman
Column A Column B 8x+24=2(4x+12) Identity 5x+18-x=2(2x+8) Contradiction 7(3x-2)=20x-13 Niether 3x+2(x-10)=5(x-4) like this?
anonymous
  • anonymous
yes
SolomonZelman
  • SolomonZelman
`(When you expand 1st row, 1st column)` 8x+24=2(4x+12) \(\longrightarrow\) 8x+24=8x+24
SolomonZelman
  • SolomonZelman
`(Expand 2nd row, 1st column)` 5x+18-x=2(2x+8) \(\longrightarrow\) 5x+18-x=4x+16 `(Subtract x from 5x: 5x-x=4x)` 5x+18-x=4x+16 \(\longrightarrow\) 4x+18=4x+16
SolomonZelman
  • SolomonZelman
So at this point you should be able to tell me whether A is an *identity* (true for all values of x), a *contradiction* (true for NO values of x - i.e. always false) OR *Niether* (not contradition or identity - i.e. has 1 solution)
SolomonZelman
  • SolomonZelman
`(Expand 3rd row, 1st column)` 7(3x-2)=20x-13 \(\longrightarrow\) 21x-14=20x-13 `(Subtract 20x from both sides)` 21x-14=20x-13 \(\longrightarrow\) x-14=13 `(Add 14 to both sides)` x-14=13 \(\longrightarrow\) x=27 (and this is Niether)
SolomonZelman
  • SolomonZelman
`(Expand 4th row, 1st column)` 3x+2(x-10)=5(x-4) \(\longrightarrow\) 3x+2x-20=5x-20 `(Add like terms on the left side)` 3x+2x-20=5x-20 \(\longrightarrow\) 5x-20=5x-20 Can you identify this one as either: *identity*, *contradiction*, or *Neither*
anonymous
  • anonymous
sorry I had to leave but im back @SolomonZelman
SolomonZelman
  • SolomonZelman
Ok, read over my posts, please. This should clarify every thing....
anonymous
  • anonymous
ok
anonymous
  • anonymous
so 1.A 2.B. 3.C 4.A?
SolomonZelman
  • SolomonZelman
yes, Identity Contradition Neither Identity *CORRECT!*
anonymous
  • anonymous
THANK YOU!!!!!!!
SolomonZelman
  • SolomonZelman
Anytime:)
anonymous
  • anonymous
I got 100%
SolomonZelman
  • SolomonZelman
Nice to hear that:)
SolomonZelman
  • SolomonZelman
have a good night!

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