need help with 2 3unit questions from my topic test (1 Viewer)

[Riot09]

alright the first one is:

1.find the values of x or which

3/x>-5 i got x>-3/5 but whats the other solution?.

and 2.solve simultaeously

3x/4-2y/3=4

x/2-y/6=6

thnx in advance.

 

[XTsquared]

1.
3/x > -5
(therefore x cannot = 0)
3x > -5x^2 (multiply by x^2 as for x could be positive or negative, therefore we multiply by a positive number so we don't have to change to sign around).
5x^2 - 3x > 0 (bring to one side)
x(5x + 3) >0 (factorise)
x > 0 and x < -3/5

And as for 2:
Sorry I can't be bothered to type up what I did (in the middle of my assignments) but:
Make eqt 1 and 2 have common denominators, multiply or whatever you need to do so that you have the same number of x's or y's (whatever you want to cancel). Solve for x or y, then sub in and find the other.
I got x = 16, y = 12

 

[muzeikchun852]

1.

3/x > -5

(therefore x cannot = 0)

3x > -5x^2 (multiply by x^2 as for x could be positive or negative, therefore we multiply by a positive number so we don't have to change to sign around).

5x^2 - 3x > 0 (bring to one side)

x(5x + 3) >0 (factorise)

x > 0 and x > -3/5

thats wrongggg!



3/x > -5
3/x +5 > 0
(3 + 5x) / x > 0
therefore x = 0, x = -3/5

do it on a number plane.

_______-3/5________0_____

sub one value.

x = 1,
3+5 / 1 = 8 (positive)

_______-3/5________0__+__

therefore:

___+___-3/5___-____0___+_


therefore:
x < -3/5, x > 0


question two:
x12
9x - 8y = 48 ...1

x6
3x - y = 36 ...2

...2x3
9x - 3y = 108 ...3

...1 - ...3
-5y = -60
y = 12

sub y value to ...2
3x - 12 = 36
3x = 48
x = 16

 

[XTsquared]

woops yeah, stuffed up the sign on x < -3/5
sorry. Otherwise I was right.