Ok, i have a math problem. Da problem iz 3x+5y=15 nd i have 2 solve 4 x, how do u do that?

Question

Ok, i have a math problem. Da problem iz 3x+5y=15 nd i have 2 solve 4 x, how do u do that?

Answer 1

3x + 5y = 15

3x = 15 - 5y

x = 5 - 5/3y

Answer: x = 5 - 5/3y or x = (15 - 5y)/3

Answer 2

3x + 5y = 15

Make y=0    3x + 0= 15   then divide 15 by 3x plug 3x under 3x and under 15=

x=5

And that is all there is to it.

                                            

Answer 3

x=1.5

y=2

Answer 4

3x + 5y = 15

3x = 15- 5y

x  15/ 3 -5y/3

x = 5 - 5/3y

Answer 5

With a single equation in 2 unknowns, there are an infinite number of X,Y pairs which are solutions. No matter what value you put in for X, there is a Y which makes the equation true. The best you can do is solve for X in terms of Y.

1. subtract 5Y from each side

3X = 15-5Y

3X= 5(3-y)

2. Divide both sides by 3

X= (5(3-y))/3

Which is as far as you can go...

3. Now, if I had a 2nd equation in X and Y, I could substitute this in for X and then solve for Y, substitute back in the other and solve for X.