Changing the Subject (1 Viewer)

[Paj20]

In -3x-8y-12=0

Make y the subject, would that =

12+3x
OVER
8 (or would it be over -8)?

 

[tommykins]

-3x - 8y - 12 = 0

-8y = 12+3x
y = 12+3/-8 =

So -8.

 

[watatank]

i could've sworn i saw this exact question asked before

-3x-8y-12=0

add 8y to both sides

8y = -3x -12

divide both sides by 8.

y = -3/8 x -12/8
= -3/8 x -3/2

 

[Paj20]

i could've sworn i saw this exact question asked before

-3x-8y-12=0

add 8y to both sides

8y = -3x -12

divide both sides by 8.

y = -3/8 x -12/8
= -3/8 x -3/2

thats good you should b good at doing it then.

 

[bling05]

In -3x-8y-12=0

Make y the subject, would that =

12+3x
OVER
8 (or would it be over -8)?

-3x - 8y -12 = 0
8y = -3x - 12
y = -3x - 12/8

 

[michael1990]

x=x-y/xy

Make y the subject of the equation.

 

[watatank]

x=x-y/xy

Make y the subject of the equation.

serious question?

x=x-y/xy

x^2.y = x - y

x^2.y + y = x

y(x^2 + 1) = x

y = x/(x^2 + 1)

 

[m&ss2008]

x=x-y/xy

Make y the subject of the equation.

assuming you mean
x=(x-y)/xy ?
otherwise
x=x - y/xy
is a whole different problem

 

[michael1990]

assuming you mean
x=(x-y)/xy ?
otherwise
x=x - y/xy
is a whole different problem

was in the test as.

x= x-y over xy

 

[notaaron]

serious question?

x=x-y/xy

x^2.y = x - y

x^2.y + y = x

y(x^2 + 1) = x

y = x/(x^2 + 1)

correct.