text: 1. ## Is my answers page completely broken? or am I?
  
  $y=3x+11$
  $y=x^2+4x+5$
  
  • $3x+11=x^2+4x+5$
  • $11=x^2+x+5$
  • $6=x^2+x$
  • $x^2+x-6=0$
  • $(x-3)(x+2)$
  • So! I think $x = 2 or -3$
  
  so
  When $x=2$
  $y=6+11$
  $y=17$
  
  $x=-3$
  $y=-9+11$
  $y=2
  $
  
  My answers page says the co-ords are $(2,-3)$ as in $x =2$ and $y = -3$
  
  Putting x as -3 in $y=x^2+4x+5$ Gives $y=9-12+5 = 2$
  
  SO to me x = -3 and y = 2. But the answers page says other wise, who's right?
  
  2. $3x + 11 = x^2 + x + 5\;\; \to \;\;x^2 - 2x - 6 = 0$
  
  3. Sorry I wrote it down wrong, try again
  
  4. Originally Posted by sxsniper
  $y=3x+11$
  $y=x^2+4x+5$
  The way you solved above tired ME out!
  Try keeping it simpler (so your teacher will not give up!), like:
  
  x^2 + 4x + 5 = 3x + 11
  x^2 + 4x - 3x + 5 - 11 = 0
  x^2 + x - 6 = 0
  (x + 3)(x - 2) = 0
  x = -3 or x = 2
  
  Now, using y = 3x + 11 :
  when x = -3, y = 3(-3) + 11 = 2
  when x = 2, y = 3(2) + 11 = 17
  
  Now you and/or your book can rant and rave, but the (x,y) solutions are:
  (-3,2) and (2,17) ; over and out
  
  5. Thanks, I don't have a teacher, I'm learning for the fun of it.
  It tires me out too! But at least I can see exactly what I've done.
  
  This book sucks, I think someone just put random numbers in the answers page.
  
  6. Originally Posted by sxsniper
  $y=3x+11$
  $y=x^2+4x+5$
  
  • $3x+11=x^2+4x+5$
  • $11=x^2+x+5$
  • $6=x^2+x$
  • $x^2+x-6=0$
  • $(x-3)(x+2)$
  • So! I think $x = 2 or -3$
  This is incorrect. "If ab= 0 then either a= 0 or b= 0". You have (x- 3)(x+ 2)= 0 so either x- 3= 0 or x+ 2= 0. Then x= 3 or x= -2, not "x= 2 or -3"
  
  so
  When $x=2$
  $y=6+11$
  $y=17$
  
  $x=-3$
  $y=-9+11$
  $y=2
  $
  
  My answers page says the co-ords are $(2,-3)$ as in $x =2$ and $y = -3$
  As I said before, x= -2 or x= 3. If x= -2, then y= 3(-2)+ 11= 5. If x= 3, y= 3(3)+ 11= 20. the correct solutions are (-2, 5) and (3, 20).
  
  Putting x as -3 in $y=x^2+4x+5$ Gives $y=9-12+5 = 2$
  
  SO to me x = -3 and y = 2. But the answers page says other wise, who's right?
  Did you check your answers by putting them back into the equations?
  
  7. You've just wrecked every hope I had, I understood it until you posted that. It's the format mess the post is in that's throwing me off.
  
  So X could equal either 3 or -2. I actually wrote that down wrong and continued from my error (I even wrote it down wrong when typing this up!). I am prone to writing things down wrong ! And failing the maths due to negligence.
  
  After de-cyphering the format of your post it all makes sense.
  
  8. $x^2+x-6$
  is $(x+3)(x-2)$
explanation: The text extract is a discussion about solving a system of equations involving a linear equation and a quadratic equation. It includes a question about the correctness of the solutions and a detailed step-by-step solution provided by the user. Another user points out an error in the factorization and provides the correct solutions. The text also includes a response from the original poster acknowledging the mistake and expressing confusion.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: 1. ## Is my answers page completely broken? or am I?
  
  $y=3x+11$
  $y=x^2+4x+5$
  
  • $3x+11=x^2+4x+5$
  • $11=x^2+x+5$
  • $6=x^2+x$
  • $x^2+x-6=0$
  • $(x-3)(x+2)$
  • So! I think $x = 2 or -3$
  
  so
  When $x=2$
  $y=6+11$
  $y=17$
  
  $x=-3$
  $y=-9+11$
  $y=2
  $
  
  My answers page says the co-ords are $(2,-3)$ as in $x =2$ and $y = -3$
  
  Putting x as -3 in $y=x^2+4x+5$ Gives $y=9-12+5 = 2$
  
  SO to me x = -3 and y = 2. But the answers page says other wise, who's right?
  
  2. $3x + 11 = x^2 + x + 5\;\; \to \;\;x^2 - 2x - 6 = 0$
  
  3. Sorry I wrote it down wrong, try again
  
  4. Originally Posted by sxsniper
  $y=3x+11$
  $y=x^2+4x+5$
  The way you solved above tired ME out!
  Try keeping it simpler (so your teacher will not give up!), like:
  
  x^2 + 4x + 5 = 3x + 11
  x^2 + 4x - 3x + 5 - 11 = 0
  x^2 + x - 6 = 0
  (x + 3)(x - 2) = 0
  x = -3 or x = 2
  
  Now, using y = 3x + 11 :
  when x = -3, y = 3(-3) + 11 = 2
  when x = 2, y = 3(2) + 11 = 17
  
  Now you and/or your book can rant and rave, but the (x,y) solutions are:
  (-3,2) and (2,17) ; over and out
  
  5. Thanks, I don't have a teacher, I'm learning for the fun of it.
  It tires me out too! But at least I can see exactly what I've done.
  
  This book sucks, I think someone just put random numbers in the answers page.
  
  6. Originally Posted by sxsniper
  $y=3x+11$
  $y=x^2+4x+5$
  
  • $3x+11=x^2+4x+5$
  • $11=x^2+x+5$
  • $6=x^2+x$
  • $x^2+x-6=0$
  • $(x-3)(x+2)$
  • So! I think $x = 2 or -3$
  This is incorrect. "If ab= 0 then either a= 0 or b= 0". You have (x- 3)(x+ 2)= 0 so either x- 3= 0 or x+ 2= 0. Then x= 3 or x= -2, not "x= 2 or -3"
  
  so
  When $x=2$
  $y=6+11$
  $y=17$
  
  $x=-3$
  $y=-9+11$
  $y=2
  $
  
  My answers page says the co-ords are $(2,-3)$ as in $x =2$ and $y = -3$
  As I said before, x= -2 or x= 3. If x= -2, then y= 3(-2)+ 11= 5. If x= 3, y= 3(3)+ 11= 20. the correct solutions are (-2, 5) and (3, 20).
  
  Putting x as -3 in $y=x^2+4x+5$ Gives $y=9-12+5 = 2$
  
  SO to me x = -3 and y = 2. But the answers page says other wise, who's right?
  Did you check your answers by putting them back into the equations?
  
  7. You've just wrecked every hope I had, I understood it until you posted that. It's the format mess the post is in that's throwing me off.
  
  So X could equal either 3 or -2. I actually wrote that down wrong and continued from my error (I even wrote it down wrong when typing this up!). I am prone to writing things down wrong ! And failing the maths due to negligence.
  
  After de-cyphering the format of your post it all makes sense.
  
  8. $x^2+x-6$
  is $(x+3)(x-2)$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The text extract is a discussion about solving a system of equations involving a linear equation and a quadratic equation. It includes a question about the correctness of the solutions and a detailed step-by-step solution provided by the user. Another user points out an error in the factorization and provides the correct solutions. The text also includes a response from the original poster acknowledging the mistake and expressing confusion.
  
  Label: question-and-answer