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it's obvious

2 + 2 = 22

HINT:    factorise  5xy-2y

well here you have   5x lots of y minus  2 lots of y  so you must end up with (5x-2) lots of y

5xy-2y = (5x-2)y   or    y(5x-2)

consider 2x+3x = 2lots of x + 3 lots of x = (2+3)lots of x = 5x   [you knew the answer was 5x in the first place]

so

6(x+3)+2(x+3) = (6+2)(x+3)=8(x+3)

7(y-4)+y(y-4)=(7+y)(y-4)

These facctorizations questions are very easy.  Do you understand what I am trying to show you?

The hyperlink is a lot more elegant than having to have the whole web address in your post, glad it worked for you.

1) 5x(x+3)-2(x+3)

= (x+3)(5x-2)   I guess this is my answer. Not that familiar with this, seemed a bit too simple so worked it the other way to check as usual. Expanded the brackets;

= 5x²-2x+15x-6   Then combined like terms;

= 5x²+13x-6      Then tried to factorise this, looking for something that would multiply to make -30 and add to make 13;

= 5x²-2x+15x-6   Continued factorising and got back to your original question so should be good.

2) -7x(2x-1)-(2x-1)

= (2x-1)(-7x-1)   I think this is my answer. Threw me a bit only having the - in front of the second bracket with no number, but went through the same process and it seems to work ok. Expanded;

= -14x²-2x+7x+1   Combined like terms;

= -14x²+5x+1   Factorise, look for something that multiplies to -14 and adds to 5.

= -14x²+7x-2x+1   Then back to where we started again.

I am quite slow on these. I think I used to just struggle playing around with numbers in brackets until it worked but this is a more deliberate method. 

I am a bit thrown by your hint "HINT:    factorise  5xy-2y     Can you see the similarity?" Was this to relate to question 1?

It is not $$3x^2$$

it is     $$(3x)^2=3x*3x=9x^2$$

Difference of 2 squares            $$\boxed{a^2-b^2=(a-b)(a+b)}$$

It is difference because it is subtraction.

Here are some for you

$$\\simplify\\
1) (a+3)(a-3)\\
2) (2n-4)(2n+4)\\
3) (5-2t)(5+2t)\\\\\\
factorize\\
4)\;\; x^2-16\\
5) \;\;9-4m^2\\
6)\;\;16x^2-a^2b^2$$

$$\\(x+1)^2=x^2(x+7)^2\\
x^2+2x+1=x^2(x^2+14x+49)\\
x^2+2x+1=x^4+14x^3+49x^2\\
x^4+14x^3+49x^2-x^2-2x-1=0\\
x^4+14x^3+48x^2-2x-1=0\\$$

It is not easy to find the roots of this.  so I consulted Wolfram|Alpha

$$\small{\text{The best solutions: $< 0.001$ \$ difference }}\\\\
\small{\text{
$
\begin{array}{rcccr}
\# & & & & \small{\text{difference in \$ }} \\
1. & \$ 4.22 & \$ 2.08 & \$ 0.81 & 0.000144 \\
2. & \$ 3.21 & \$ 3.21 & \$ 0.69 & 0.000171 \\
3. & \$ 4.62 & \$ 1.35 & \$ 1.14 & -0.000180 \\
4. & \$ 3.22 & \$ 3.20 & \$ 0.69 & 0.000240 \\
5. & \$ 4.37 & \$ 1.87 & \$ 0.87 & 0.000447 \\
6. & \$ 3.23 & \$ 3.19 & \$ 0.69 & 0.000447 \\
7. & \$ 3.85 & \$ 2.53 & \$ 0.73 & -0.000565 \\
8. & \$ 4.25 & \$ 2.04 & \$ 0.82 & 0.000600 \\
9. & \$ 3.24 & \$ 3.18 & \$ 0.69 & 0.000792
\end{array}
$
}}$$

(by  Computing)

3.141592653589793... it goes on forever. If you need more digits than that, then google it.

I am going to test that hyperlink here

Yes, unsolvable for x as it is, I was hoping for more information if they replied.

(3x+5)(3x-5)   If I expand this I get;

3x²-15x+15x-25

3x²-25      How you have got to your answer, I have no idea.

I have heard the term "difference of two squares" but that's my only exposure to it.

Thanks Melody, that's quite useful.

"Brute Force" and "Ignorance"  are two of my favorite math techniques....in fact, I would feel lost without them....!!!

Thanks, Alan...for those answers....

Here are a couple of other solutions (though all are approximate, including heureka's, as the products are only accurate to just under 2*10^-2 of a cent!):

$0.69  $3.21  $3.21

$0.81  $2.08  $4.22

Solutions obtained by brute force and ignorance!  (i.e. running through all possibilities to the nearest cent, from 1cent to $7.11, for two of the items and calculating the third by subtracting the sum from $7.11, then checking to see if the product differs from $7.11 by less than 0.02cents - I didn't do this by hand of course!

I found no solutions when looking for a product that differs from $7.11 by less than 0.01 cents).

.

This one looks interesting Chris  

That is very cute Chris.   

Very nice, heureka......can you show us how you arrived at your solution???

......what are the possible prices of the three items : $$\\ \$\ 4.62
\qquad \$\ 1.35
\qquad \$\ 1.14
\\
\dots$$

The general equation for a quadratic is:  y = a*x² + b*x + c  where a, b and c are constants.  To find what they are here, substitute your three sets of (x, y) values into this general equation to give you three simultaneous equations for a, b and c:

 -11 = a*(-2)² + b*(-2) + c

 13 = a*4² + b*4 + c

 29 = a*6² + b*6 + c

or:

-11 = 4a - 2b + c

 13 = 16a +4b + c

 29 = 36a + 6b + c

Can you take it from here?

(You should find a = 1/2, b = 3, c = -7)

.

Pete Zarilla sells 10 inch pizzas for $4.30, 12 inch pizzas for $5.88, and 14 inch pizzas for $7.78. Assume that the price is a quadratic function of the diameter.

We have this set of equations

a(10)^2 + b(10) + c = 4.30   → 100a + 10b + c = 4.30      (1)

a(12)^2 + b(12) + c = 5.88   → 144a + 12b + c = 5.88      (2)

a(14)^2 + b(14) +c  = 7.78   → 196a + 14b + c = 7.78      (3)

Subtract  (1) from (2)   and (2) from (3)    so we have

44a + 2b = 1.58     (4)

52a + 2b = 1.90     (5)

Subtract  (4) from (5)   and we have that 8a = .32 → a = .04

And using (5), we have 52(.04) +2b = 1.90 → b = -.09

And using (1) we have

100(.04) + 10(-.09) + c = 4.30  → c = 1.2

So, our function is

P(d) = (.04)d^2 - .09d + 1.2

I assume this is

[(1/3)-(2/3)]  / [(1/6)-(5/6)] ?? ....if so, we have....

[-1/3] / [ -4 /6 ]  =

[-1/3] / [-2/3] =     { a negative divided by a negative is positive}

[1/3] / [2/3] =

[1/3] x [ 3/2] =

1/2

5 + (7-1)2/6

Using the order of operations......work inside the parentheses, first

5 + (6)*2/6

Next, multiply

5 + 12/6  

Next, divide

5 + 2

And we get.....   7....!!!

I'll second Melody's appreciation for heureka....he's one of the best mathematicians on here, and I usually always learn a new "trick"  - or two - from him.....

Same as the old one.......[ it's a Beatles' reference.....in case you don't know]

Number 9 .... Number 9.....Number 9......Number 9......Number 9.......Number 9

@@ End of Day Wrap   Fri 23/1/15     Sydney, Australia      Time 4:25 pm

Hi everyone,

Today our intrepid answerers were CPhill, den-one, alan, Rosala, Heureka, Geno3141, Ang2061, Tenacious, will85237, Tetration, Saseflower, TakahiroMaeda and TeamCapt22.  Thank you all - your blood is worth bottling 

CONGRATULATIONS TO HEUREKA

Heureka is now on the Top Answerers Board !!!   I am so pleased.  

Heureka you have always been an extremely valued member of our answer team.    

Interest posts:

As the number of posts increase, the number of interest post I collect decreases.  This is because I actually spend less time looking at the questions and mor time just ticking them off.  I need people to send me the addresses of posts that they find interesting and that they would like to share with others.  If you want to stay anonymous you can do that.  I will know who made the suggestion but no one else needs to if you want it that way.

1) A puzzle   (I haven't looked at it yet. Sorry.   )   Thanks Chris

Sat 24/1/15

1) Quadratic simultaneous equation pizza problem