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It's just that some of the problems are different so it makes it harder to understand and to know you are doing the problems correctly because you can't go off of one problem if they are even the slightest bit different. Am I even making sense right now xD

Thank you for helping~

For (3). Just take the constant to the other side, you're left with x and y= constant.

Here, I'll show you!  

0=-36x + 14y + 8 

0 = 18y -9 -27x

Best thing to do first: 36x - 14y = 8 - 0, 36x - 14y = 8

And, for the second equation: 27x - 18y = 0+9, 27x - 18y = -9.

Now, divide the first equation by 2 and the second equation by 9, because the x will end up as a canceling factor.

So, 18x-7y=4, 3x-2y=-1, try to make both x's equal, subtract find x, then solve for y!

3)   0 =  -36x + 14y  + 8

      0 =  18y  - 9  - 27x

Rearrange these as

36x - 14y  = 8

27x  - 18 y   = -9

Divide the first equation through by 2  and the second one through  by 9

18x  -  7y   =  4

3x   -  2y   =  -1

I see that we  can  get  "-18x"  by multiplying the second equation by  -6  ...so we have

18x - 7y   =  4

-18x + 12y  = 6    add these

5y  = 10    divide by 5

y  = 2

Using   3x  - 2y  =  -1   to find x, we have

3x  - 2(2)  = -1

3x  - 4  =  -1

3x  = 3   divide by 3

x  = 1

So  (x,y)  = ( 1, 2)

You're not  "dumb"....it's just hard to  get the hang of this...

I'll do   2)  and  3).....maybe that will help some

2)  - 9x  - 14y   = -29

     10x + 7y  =   -2

Ok....do you see that we  can  eliminate y  by  multiplying the second equation by -2  ??...so we have

-9x - 14 y  =  -29

20x  + 14y  = -4     add these

11x  = -33   divide by 11

x  = -3

Using  10x + 7y   = -2 to find y, we have

10(-3) + 7y  = -2

-30  + 7y = -2    add 30  to both sides

7y  = 28   divide by 7

y  = 4

So...(x, y)  = ( -3, 4)

Thank you so much for your help, both of you. I have tried it and yet i'm still too dumb to solve these. >.<

That's what I think I'm just a little confused.

Wow, 1997-1998 and 1998-1999 are very close: 1997-1998: x+some amount, might be larger than the median income in 1998-1999.

Welcome, CPhill!

Thanks, Tertre  !!!!

Well, Guest.....maybe everyone isn't as brilliant as you are...besides....this elimnation stuff isn't that easy when you're new to  it....

(4) Hint: Try to rearrange the equation, the first and second: Take -10y to the other side, and see what you can do.

4)  -20  + 10 x  = -10y   

      45  = 6x  - 27 y

Rearrange each as

10x  + 10y   = 20

6x  - 27 y  = 45

I see that I can divide the top equation  by 10  and the bottom by 3  to  get :

x + y  = 2

2x - 9y  = 15

I see that I can  eliminate  'x"  by multiplying the top equation by  -2

-2x - 2y  = -4

2x - 9y  = 15     add these

-11y = 11     

y = -1

And using any equation to find x, we  have

x + y  = 2

x + (-1)  = 2

x - 1  = 2

x  =3

So  (x,y)  =  (3 , -1)

Do these steps help  ???

Sorry.... typo...should be 29....

@CPhill, 29 or -29? This was in step 2.

(2) Multiply the second equation by 2, so: -9x-14y=-29, 10x+7y=-2, -9x-14y=-29, 20x+14y=-4..

So, adding: y gets canceled, and 11x=-33, \(x=-3\), so y= -9(-3)-14y=-29, 27-14y=-29, 14y=27+29, 14y=56, \(y=4.\)

The keys to  solving  these

(1)  line up all variables  on  one side of the equation and constants on the other

(2)  pick a variable to eliminate....either will do....this is the REAL key

(3) not a necessity...but I always like to get rid of fractions, too, before  I start

7x + 2y  = -10

-14x  - 3y  = 29

I see that if I multiply the top equation by  2 and add it to the second equation,  the "x" term will disappear..so

14x  + 4 y  = -20

-14x - 3y  = 29    add these  and we get

y  = 9

To find x...we can  use any equation...so...

7x + 2(9)  = -10

7x + 18  = -10

7x = -28

x  = -4

So.....(x, y)  =(-4, 9)

Do you want to try a couple on your  own, RainbowPanda  ??

A vertical asymptote will  potentially  occur  where an x value makes the denominator of this rational function  = 0

To find these possibilities....set the denominator   =  0

x^2 + 6x  - 7   =  0

(x + 7) ( x - 1)    = 0

Setting both factors to 0 and solving for x produces   x  = -7  and  x  = 1

However.... since the  numerator  also  has the  factor (x -1), we will have a "hole"  at x  = 1

The only  true vertcal asymptote    will be at   x  =  -7

See the graph  here :  https://www.desmos.com/calculator/ohkiuoni6o

Tough to tell, NSS...but it looks like 1997 - 1998 

We are really  just looking for the countries where these two conditions exist :  the "orange" bar  is  "longer" than the " blue" bar  and the "gray" bar  is "longer" than the "orange" bar

Three  qualify  -  Australia, Canada and Egypt 

1. The first and thirteenth terms of an arithmetic sequence are  7/9  and 4/5 , respectively. What is the seventh term?

We have that

4/5  =  7/9 + d (12)

4/5 - 7/9 =  d (12)

1/45  = d(12)

1 (45 *12)  = d

1/540  = d

So...the seventh term is....    7/9  + (1/540)(6)    = 71 / 90

The common ratio, r, can be found  by   [ successive term  /  preceding term ]  = 16 / 12 =  4/3

Since...to get the third term...we had to multiply the first term  by  "r"  twice...se have this equation ;

(first term)  * r^2  = 12

(first term) (4/3)^2  = 12

(first term) ( 16/9)   = 12     multiply both sides by   9/16

first term   =  12 (9/16)   = (12/16) (9) = (3/4)(9) =  27/4

After the multiples of 2 are removed....there are 25  integers left

We are looking for the odd multiples of 3  [ the even ones have been accounted for ]  =   3 , 9 , 15, 21, 27, 33, 39, 45

Ther are 8  of these...so

25  - 8   = 17  integers   in set  S

Thanks!

We have two  equations in two unknowns...call the common  difference, d

(1/2) +  d  = x -1  ⇒    (1/2) + 1  = x - d      (1)

x -1 + d  = 3x    ⇒       -1 =    2x  - d    ⇒  1 = -2x + d      (2)

Add (1)  and (2)  and we get

(5/2)  = - x

-5/2  =  x

And  d  =  (3/2)  = (-5/2) - d ⇒   d  =  -5/2  - 3/2  = -8/2   = -4

Proof

1/2 , -5/2  -1 , 3 (-5/2)

1/2 , -7/2 , -15/2         with   d  =  -4