# Thread: A lot of algebra problems, a little time to do them!

1. ## A lot of algebra problems, a little time to do them!

Hi everybody!

I have few problems that needs to be solve by to tomorrow!!! Some of the problems are quite easy, but I just don't seem to get the right answers.

Here are the problems:

1) (x-y)(x+y)-(x-y)^2-(x+y)^2

2) 2/5x^2 - 5/2x(x-2)-3x

3) (x+4)^2 / x^2-16

4) 3a/4 + 1/4b(b-a/b) + 1/2b^2

5) x^2+2x-35 < 0

6) x-2y+3z=4
3x+2y-4z=10
x+y+z=9

7) x+y= -1
x*y= -6

8) Let Y be the number of units of a product and let p be the price per unit of the same product. One of the equations below describes the correspondence between the demand for the product and the price per unit. The other equation describes the correspondence between the supply and the price per unit.

Y=900-100p
Y=300+50p

a) Which of the equations describes the demand situation and which describes the supply situation? Explain why.

b) Draw the graph of the two equations in the same coordinate system.

c) Solve the equation system both graphically and algebraically.

I know theres a lot of question, but I'll appreciate if you could help!

Thank you!!

2. Originally Posted by Ana-Maria
Hi everybody!

1) (x-y)(x+y)-(x-y)^2-(x+y)^2
Multiply it out, but keep everything in parantheses!
(x^2-xy+xy-y^2)-(x^2-xy-xy+y^2)-(x^2+xy+xy+y^2)
Combine what you can still in parantheses,
(x^2-y^2)-(x^2-2xy+y^2)-(x^2+2xy+y^2)
Open parantheses, careful with signs
x^2-y^2-x^2+2xy-y^2-x^2-2xy-y^2
Combine (some will die),
-x^2-y^2=-(x^2+y^2)
2) 2/5x^2 - 5/2x(x-2)-3x
Common denominator is: 10x^2(x-2)
The first fraction becomes,
4(x-2)/10x^2(x-2)
The second fraction becomes,
25x/10x^2(x-2)
The third fraction becomes,
3x(10x^2)(x-2)/10x^2(x-2)
Find the denominators are all the same combine the numerator:
4(x-2)-25x-3x(10x^2)(x-2)
Work with third term,
4(x-2)-25x-30x^3(x-2)
Open parantheses,
4x-8-25x-30x^4+60x^3
Combine,
-30x^4+60x^3-21x-8
-30x^4+60x^3-21x-8/10x^2(x-2)

3. Originally Posted by Ana-Maria

3) (x+4)^2 / x^2-16
Factor the numerator,
(x+4)(x+4)
Factor the denominator,
(x-4)(x+4)
Kill what is the same,
(x+4)/(x-4)

5) x^2+2x-35 < 0
Factor,
(x+7)(x-5)<0
Thus,
Case 1 (x+7)>0 and (x-5)<0

OR

Case 2 (x+7)<0 and (x-5)>0

Work with the frist case,
x>-7 and x<5 thus, -7<x<5

Work with the second case,
x>-7 and x<5 impossible thus no such numbers.

The only thing that we are left with is,
x=(-7,5)

6) x-2y+3z=4
3x+2y-4z=10
x+y+z=9
4x-z=14

2)Mutliply third equation by 2:
2x+2y+2z=18

3)Add first equation with one in 2):
3x+5z=22

4)Thus,
4x-z=14
3x+5z=22

5)Multiply first equation in 4) by 5 and add,
23x=92
Thus,
x=4

6)Substitute x=4 into first equation in 4) and solve for z:
16-z=14 thus, z=2

7)Substite x=4 and z=2 into third equation:
x+y+z=9 thus, 4+y+2=9 thus, y=3

8)Solution (if exists) must be,
(x,y,z)=(4,3,2)

9)Check to confirm it is a solution.

This is my 29th post!!!

4. Thank you so much!!! I just have few questions.

Originally Posted by Ana-Maria

2) 2/5x^2 - 5/2x(x-2)-3x
The second question is that x(x-2)-3x in in the middle of the 5/2.
so the x(x-2)-3x in no denumerator, what should I do then?

Originally Posted by Ana-Maria

5) x^2+2x-35 < 0
The answer that you gave me in the question 5, does it mean that you have two ways to do it ? or is there just one answer?

There is just two more questions

Thanks again!

5. Originally Posted by Ana-Maria
The answer that you gave me in the question 5, does it mean that you have two ways to do it ? or is there just one answer?
Thanks again!

(x+7)(x-5)<0 can be negative (less then 0) only if one multiple is negative and other positive.

So you have two cases as ThePerfectHacker told you when result can be negative.

Case 1 (x+7)>0 and (x-5)<0

OR

Case 2 (x+7)<0 and (x-5)>0

Simplest way to explain is:
a*b < 0 if and only if
a>0 and b<0
or
a<0 and b>0

Substitute a=(x+7) and b=(x-5).

6. Originally Posted by Ana-Maria
7) x+y= -1
x*y= -6

From first equation we have x=-1-y

Substite x in second equation:
(-1-y)*y=-6
-y^2-y=-6
-y^2-y+6=0 {*-1}
y^2+y-6=0

(y-2)(y+3)=0

y=2 and y=-3

Substitue both y values in first equation to get x.
x+2=-1 --> x=-3

x-3=-1 --> x=2

So you have solutions
x=-3,y=2
and
x=2, y=-3

7. Thank you for your help!!

8. Originally Posted by Ana-Maria

8) Let Y be the number of units of a product and let p be the price per unit of the same product. One of the equations below describes the correspondence between the demand for the product and the price per unit. The other equation describes the correspondence between the supply and the price per unit.

Y=900-100p
Y=300+50p

a) Which of the equations describes the demand situation and which describes the supply situation? Explain why.

b) Draw the graph of the two equations in the same coordinate system.

c) Solve the equation system both graphically and algebraically.

b) and c)
Algebraically, system solution is:
900-100p=300+50p
600=150p
p=4

Then Y = 500.

Graphically system is shown in attached picture.

Blue line represents Y=900-100p and red line represents Y=300+50p.

a) Y=900-100p (blue line) represents demand because when you increase price p you see that Y is getting less which means that higher price means lower selling (lower demand). Y=300+50p (red line) represents supply.

9. Originally Posted by OReilly
b) and c)
Algebraically, system solution is:
900-100p=300+50p
600=150p
p=4

Then Y = 500.
I understand the first part, but how did you get Y=500?

10. Originally Posted by Ana-Maria
I understand the first part, but how did you get Y=500?
Substite p in one of equations with 4 and you will get Y=500.

11. I got it !

Do you know answer to the question 2 & 4 ?

Thank you so much !

12. Originally Posted by Ana-Maria
The second question is that x(x-2)-3x in in the middle of the 5/2.
so the x(x-2)-3x in no denumerator, what should I do then?
I am not sure what is the second question.

Since LaTex is working now can you type it again?

Is maybe this question 2)?

$\displaystyle \frac{2}{5}x^2 - \frac{5}{2}x(x - 2) - 3x$

13. Originally Posted by Ana-Maria
I got it !

Do you know answer to the question 2 & 4 ?

Thank you so much !
Also, for question 4) please use parenthesses or even better use LaTex to show it.
I don't see exactly what is what.

14. Originally Posted by Ana-Maria
I understand the first part, but how did you get Y=500?
Hi,

I've modified OReilly's diagram.

You can get the values of p and Y directly from the graph: