1. ## Is my answers page completely broken? or am I?

$\displaystyle y=3x+11$
$\displaystyle y=x^2+4x+5$

• $\displaystyle 3x+11=x^2+4x+5$
• $\displaystyle 11=x^2+x+5$
• $\displaystyle 6=x^2+x$
• $\displaystyle x^2+x-6=0$
• $\displaystyle (x-3)(x+2)$
• So! I think $\displaystyle x = 2 or -3$

so
When $\displaystyle x=2$
$\displaystyle y=6+11$
$\displaystyle y=17$

$\displaystyle x=-3$
$\displaystyle y=-9+11$
$\displaystyle y=2$
My answers page says the co-ords are $\displaystyle (2,-3)$ as in $\displaystyle x =2$ and $\displaystyle y = -3$

Putting x as -3 in $\displaystyle y=x^2+4x+5$ Gives $\displaystyle y=9-12+5 = 2$

SO to me x = -3 and y = 2. But the answers page says other wise, who's right?

2. $\displaystyle 3x + 11 = x^2 + x + 5\;\; \to \;\;x^2 - 2x - 6 = 0$

3. Sorry I wrote it down wrong, try again

4. Originally Posted by sxsniper
$\displaystyle y=3x+11$
$\displaystyle y=x^2+4x+5$
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
$\displaystyle y=3x+11$
$\displaystyle y=x^2+4x+5$

• $\displaystyle 3x+11=x^2+4x+5$
• $\displaystyle 11=x^2+x+5$
• $\displaystyle 6=x^2+x$
• $\displaystyle x^2+x-6=0$
• $\displaystyle (x-3)(x+2)$
• So! I think $\displaystyle x = 2 or -3$
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 $\displaystyle x=2$
$\displaystyle y=6+11$
$\displaystyle y=17$

$\displaystyle x=-3$
$\displaystyle y=-9+11$
$\displaystyle y=2$
My answers page says the co-ords are $\displaystyle (2,-3)$ as in $\displaystyle x =2$ and $\displaystyle y = -3$
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 $\displaystyle y=x^2+4x+5$ Gives $\displaystyle y=9-12+5 = 2$

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. $\displaystyle x^2+x-6$
is $\displaystyle (x+3)(x-2)$