1. ## system of equations

im not sure how to solve problem 1, need a lil help.
solve the following:
1)

1/x + 1/y = 1/19
x/y = 12

2)solve the variable v

3 = 25x + 5v
xv = v + x

im not sure if im doing 2 right, but i times the 1st equation by (v) and 2nd equation by (25). so got:
-25xv - 5v^2 +3x = 0
25xv -25v -25x = 0
so:
-5v^2 -22v -25x=0, if thats right do i use the quadratic equation to solve for v? if not please explain.

2. Hello,
Originally Posted by viet
im not sure how to solve problem 1, need a lil help.
solve the following:
1)

1/x + 1/y = 1/19
x/y = 12
In the second equation, write x=12y and substitute in the first

2)solve the variable v

3 = 25x + 5v
xv = v + x

im not sure if im doing 2 right, but i times the 1st equation by (v) and 2nd equation by (25). so got:
-25xv - 5v^2 +3x = 0
25xv -25v -25x = 0
so:
-5v^2 -22v -25x=0, if thats right do i use the quadratic equation to solve for v? if not please explain.
Not very correct... The aim of it is to leave only v or x in the equation.

The second equation looks more "difficult" to isolate x or v than the first one. I'll show if you take the 1st equation, you'll do the second one

3=25x+5v ----> 5v=3-25x ----> v=(3-25x)/5

In the first question, you will just have to add 2 fractions. In the second question, you will have to use the quadratic formula.

3. im working on the first question.
after plugging $\displaystyle x = 12y$ into the first equation, i got:
1/12y + 1/y = 1/19
=
1/13y = 1/19
at this point do i solve for y?

4. Originally Posted by viet
im working on the first question.
after plugging $\displaystyle x = 12y$ into the first equation, i got:
1/12y + 1/y = 1/19
=
1/13y = 1/19
at this point do i solve for y?

$\displaystyle \frac{1}{12y}=\frac1{12} \cdot \frac 1y+\frac 1y$

Be careful ! $\displaystyle \frac 1a+\frac 1b \neq \frac{1}{a+b}$

And after that, you can solve for y by taking the inverse

5. what happen to the $\displaystyle \frac{1}{19}$?