1. z=(x+y/3)w; solving for y 2. A=1/2bcd+bc; solving for d

Question

I just started Algebra. I am in the unit called "Literal Equations and Formulas". These problems are difficult for me, and I just need a bit of help! Thanks so much if you do help me :) please tutor me!

Answer 1

1.  z = (x+y/3)w

First thing we can do is to move the w to the other side by dividing w on both sides. So

z/w = (x+y/3)w /w 

 

The w's cancel out on the right side and we now have

z/w = (x+y/3)

We can take out the parenthesis since it's not being multiplied by anything and we have

 

z/w = x + y/3

 

Subtract x on both sides

z/w - x = x - x +y/3

 

The x's on the right side cancel. So we are left with

z/w - x = y/3

we multiply 3 on both sides and we have 

3*(z/w - x) = y/3 * 3

So y = 3(z/w - x)

Answer 2

Hi Chloe. 

 

The first equation: z=(x+y/3)w, solving for y.  First, I want to say that when you are asked to solve for something, that means that you want whatever you're solving for on one side of the equal sign and everything else on the other side of the equal sign. So for the first equation, we want y on one side of the equal sign and everything else on the other side.

We are going to move everything over one by one. When you want to move something across the equal sign, you do the opposite operation to it. I'll explain as I go along.

 

1. Lets move the w first since it is not in the parentheses. The w is multiplied to the rest of the terms. We know this because if it was added or subtracted, there would be a + or - sign, and if it was divided there would be a division sign. Multiplication will either have a * sign, or it will just be a two terms attached to each other or a term attached to a parenthesis. Since the w is a term attached to a parenthesis, we know we are multiplying. To move it we are going to do the opposite of multiplication- division. We are going to do this to both sides of the equal sign because whatever you do to one side, you have to do the same thing to the other side. So we will divide both sides by w.

 

z/w = (x+y/3)w /w   -----> z/w = (x+y/3)

 

2. Now we can get rid of the parentheses. Nothing is next to them, so they aren't necessary. 

(Note. I'm assuming your original question had both x and y divided by 3, not just y divided by 3. If this is incorrect, just reply and let me know.)

 

z/w = x+y / 3

 

3. Now multiply both sides by 3. (Three is divided originally, multiplication is the opposite operation)

 

z/w * 3 = x+y/3  * 3   ----->  z/w * 3/1 = x+y  -----> 3z/w = x+y

 

4. Now subtract x from both sides (Opposite of addition)

 

3z/w - x = x + y - x   ----->   3z/w - x = y

 

So y is by itself, which means we solved for it. So our final answer is: y=3z/w - x

 

 

 

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Second equation, we will solve like the first. I won't go into as much detail because the steps are the same as in the first equation- get d by itself.

A=1/2bcd + bc

 

1. Subtract bc from both sides. (Treat bc as a unit. If you don't have to separate something, don't.)

A - bc = 1/2bcd + bc - bc   ----->  A - bc = 1/2bcd

 

2. Divide both sides by 1/2. (Usually, you can divide everything that is multiplied to what you're solving for at once. However, if a fraction is attached to what you're solving for, do it separately so that you don't get confused.)

 

(A - bc)/ 1/2 = 1/2bcd / 1/2  -----> (A - bc / 1) * 2/1 = bcd   ---->   2(A-bc) = bcd

(Note: When dividing a fraction by a fraction (left side of equal sign) change to multiplication and flip the fraction on the right of multiplication sign, then multiply.)

 

3. Now, you can divide by the bc.

2(A-bc) / bc = bcd/bc  ----> 2(A-bc)/bc = d (We can't divide the bc's on the left because one is part of an equation so it can't be separated out and divided)

 

This means that we are done. Our final answer is: d = 2(A-bc)/bc

Answer 3

Chloe,

 

All we're trying to do in these types of problems is "isolate" the variable we're solving for. In other words we would like to have it appear by itself on one side of the equation (right side or left side either one, it doesn't matter which, as long as it's by itself).

 

1)  z = (x+y/3)w

 

Since w on the right is outside of the ( ... ) we can divide both sides by w

 

z/w = (x+y/3)w/w =(x+y/3)

 

Since we got rid of the w on the right we can remove the ( ... )

 

z/w = x + y/3

 

Now subtract x from both sides:

 

(z/w) - x = x -x + y/3  (I enclosed z/w to avoid any possible confusion:  we're subtracting x from the quantity z/w, not just from w!)

 

(z/w) - x = y/3

 

We're almost there, we just need to multiply both sides by 3 to get y by itself, but since we need to multiply everything on the left side by 3 lets enclose that whole side to avoid confusion:

 

3·[(z/w) - x] = 3·(y/3)

 

Finally:

 

3·[(z/w) - x] = y

 

 

2)  A = (1/2)bcd + bc  ( I'm guessing that this is the way the factor ½ is effecting the right-hand side, not 1/(2bcd)+bc or 1/(2bcd + bc), either of which would give a different result).

 

subtract bc from both sides:

 

A - bc = (1/2)bcd + bc - bc = (1/2)bcd

 

To avoid confusion:

 

(A - bc) = (1/2)bcd

 

multiply both sides by 2:

 

2·(A - bc) = (1/2)bcd 2·(1/2)bcd = bcd

 

Finally divide both sides by bc:

 

2·(A - bc)/(bc) = bcd/bc

 

2·(A - bc)/(bc) = d