Solve for y

Question

I am given the following problem in Algebra 2. The directions are "Solve for y."

 

  m                   d

------ +  3a  -  --- = 0

c + x                y

 

I cannot decipher how to answer it. Can you help me?

Answer 1

(m/[c+x])+3a-(d/y)=0

(m/[c+x])+3a=(d/y)

(m/[c+x])+(3a[c+x]/(c+x)=(d/y)

(m+3a[c+x])/(c+x)=d/y

cross multiply

y(m+3a[c+x])=d(c+x)

y=d(c+x)/(m+3a[c+x])

Answer 2

To solve for "y," or any other variable means to isolate the variable on one side or the other of the equation.

 

m(c + x) + 3a - d/y = 0            Subtract 3a from both sides

m(c + x) - d/y = -3a                 Multiply both sides by y

ym(c + x) - d = -3ay                Add d to both sides

ym(c + x) = -3ay + d               Divide both sides by m(c + x)

y = (-3ay + d) / (m(c + x))

 

Answer 3

Hi James, the first thing I would do is to add (d/y) to both sides to get y on one side of the equation.

 

 m                  d

-----   + 3a = ----

c + x              y

 

Next, multiply both sides by y.

 

        m                  d
y•( ----- + 3a ) =( ----) • y     The y's on the right cancel. Leaves you with:

     c + x                y

 

       m                   
y•( ----- + 3a ) = d            
     c + x                

 

The same way you can divide 3x = 6 by 3 to isolate x,

you can divide each side by:

 

    m
( ----- + 3a )       This gives you the final answer:
  c + x

 

             d

y = ---------------

          m
      ( ----- + 3a ) 
       c + x

 

I hope this was clear enough :)