Thread: A few equstions, i mean confusions...

1. A few equstions, i mean confusions...

ex - 2y = 3z x=?

ay + z = am - ny y=?

a(y + 1) = b y=?

(3/5)y + a = b y=?

What I'm trying to figure out is well... How to do these equations, I am no less than stuck in doing any equations..

So, would anyone please explain to me how to get started on these equations?

2. Originally Posted by Blakester
ex - 2y = 3z x=?

ay + z = am - ny y=?

a(y + 1) = b y=?

(3/5)y + a = b y=?

What I'm trying to figure out is well... How to do these equations, I am no less than stuck in doing any equations..

So, would anyone please explain to me how to get started on these equations?
These are called literal equations, since you are trying to solve for a particular variable when you have more than one variable in the equation.

$\displaystyle ex - 2y = 3z$, solve for $\displaystyle x$

First, add $\displaystyle 2y$ to both sides:

$\displaystyle ex=3z+2y$

Finally, divide by $\displaystyle e$ in order to isolate the variable x:

$\displaystyle x=\frac{3z+2y}{e}$

Now you try the next one, and let's see how you do.

3. Originally Posted by Blakester
$\displaystyle ay + z = am - ny$ Solve for $\displaystyle y$

So, would anyone please explain to me how to get started on these equations?
Well, that second one is a little bit different, so I'll get you started and see if you can finish.

First, get the $\displaystyle y$ terms together. Add $\displaystyle ny$ to both sides of the equation:

$\displaystyle ay+ny+z=am$

Now subtract $\displaystyle z$ from both sides.

$\displaystyle ay+ny=am-z$

Factor $\displaystyle y$ from the left side and that leaves you with one more step.

4. $\displaystyle (a + n) = (am - z)$

To?

$\displaystyle y = (am - z)/(a + n)$?

5. Originally Posted by Blakester

$\displaystyle (a + n){\color{red}y} = (am - z)$

To?

$\displaystyle y = (am - z)/(a + n)$?
You forgot the y...

You should be able to do it now.

--Chris

6. Originally Posted by Blakester
$\displaystyle a(y + 1) = b$ Solve for $\displaystyle y$

So, would anyone please explain to me how to get started on these equations?
Very good!

Now, for the third one. This time you will have to expand the left side in order to free up the variable $\displaystyle y$.

$\displaystyle ay+a=b$

I think you can finish from here.

$\displaystyle \frac{3y}{5} + a = b$ Solve for $\displaystyle y$
First, multiply each term by $\displaystyle 5$ to eliminate the fraction.
$\displaystyle 3y+5a=5b$