Answer this plz math?

Question

6x - 10y =12

-15x + 25y =-30

what are x and y?

Answer 1

method of elimination:

FIRST YOU MULTIPLY the first equation by 5 and the secondequation by 2. therefore u get the equations:

30x-30y=360

-30x+50y=-60 now add the equations

___________

0x+20y=300

y=300/20

y=15

now substitute:

6x-10(15)=12

6x-150=12

6x=162

x=162/6=27

hope this helps:):):)

Answer 2

6x - 10y = 12 ===> Eq. 1

-15x + 25y = -30 ==> Eq. 2

===================

SOlve by elimination

Multiply Eq. 1 by 5

5(6x - 10y = 12) =

30x - 50y =  60 ===> Eq. 3

Multuply Eq. 2 by 2

2(-15x + 25y = - 30)

-30x + 50y = -60 ===> Eq. 4

ADD Eq. 3 and Eq. 4

30x  - 50y  = 60 ===> Eq. 3

-30x + 50y = -60 ===> Eq. 4

--------------------------------------...

0 = 0

It means that there is no solution for this problem

Answer 3

(6x - 10y = 12)/2

3x - 5y = 6

(-15x+25y=-30)/5

-3x+5y=-6

3x - 5y + (-3x + 5y) = 6-6

3x - 5y - 3x + 5y = 0

0 + 0 = 0

lol

Answer 4

6x-10y=12

-15x+25y=-30

multiply 6x-10y=12 by - 5/2 you get -15x+25y = - 30

so the 2 equations are the same, you cannot solve it. you need another equation.

Answer 5

Hi there....Whenever you have to find a solution for two separate variables in a set of simultaneous equations you need at least 2 unique equations in the system (if there's 3 variables then you need 3 eqn's, 4 variables...4 eqn's etc).

Now, somewhat similarly to equivalent fractions i guess, you can have equivalent equations...that is, any multiple of an equation is equivalent to that equation...so since your second equation is -2.5 times your first equation then it is in fact the same equation as the first...thus you really only have one equation so cannot solve for the two variables x and y.

It may not be evident when you first look at the equations that one is a multiple of the other but when you go through one of the usual methods to solve the system and you find that something has gone wrong or looks iffy then that's usually a reason to suspect that there is no solution.

Since a solution to simultaneous equations gives the (x,y) coordinates of the point where the two lines (given by the equations) intersect, then if there is no solution then that just means that the two lines never cross...if the two lines never ever cross then they must infact be parallel...so another way to determine that there's no solution to a set of simultaneous equations is to notice that they have the same gradient...easier to realise this if the equations are in gradient-intercept form (y=mx+b), or if you suspect that there's no solution you can manipulate the equations into this form and see if they have the same gradient.

Hope it helps (^_^)