Solving Polynomial Equations

[adr8]

I dont know if I have this one right


x-1=cx+d (c can't equal to 1)

This is what I did:

I moved the x to the other side so its going to be -1=(cx-x)+d

-1=bx+d

-d-1/b=x

 

[mmm4444bot]

I dont know if I have this one right

Neither do I. You forgot to post the instructions that came with the given equation.

Did they ask you to solve for x? To solve for c? Something else?


-d-1/b=x

From where did symbol b come?

.

 

[adr8]

It just says to solve the equations mentally if possible. In other words, in this problem we will not have a specific number like x+1=0 x=-1. Like the previous one that I did was ax+b=c (a cannot equal to 0).

I moved the b to the other side so I had b-c=ax. Then I moved the a to the other side by canceling it out with another a. So my final answer was (b-c)/a=x.

I got the b by subtracting cx-x. I thought it would be b since its alphabetical order and since it has a negative it would go one backwards. Dont know if I explained myself correctly lol.

 

[wjm11]

x-1=cx+d (c can't equal to 1)
This is what I did:
I moved the x to the other side so its going to be -1=(cx-x)+d
-1=bx+d
-d-1/b=x

…the previous one that I did was ax+b=c (a cannot equal to 0). I moved the b to the other side so I had b-c=ax. Then I moved the a to the other side by canceling it out with another a. So my final answer was (b-c)/a=x.
I got the b by subtracting cx-x. I thought it would be b since its alphabetical order and since it has a negative it would go one backwards. Dont know if I explained myself correctly lol.

JeffM beat me to it, but I'll add my two cents, in case that helps any.

First of all, it is incorrect to insert new variables (b) into this type of problem. Secondly, there is no rule/tradition in math regarding “alphabetical order and since it has a negative it would go one backwards.”

More to the point, let’s discuss some fundamentals. When we talk about moving something from one side of an equation to the other, we are being a little lax in stating what we are actually doing. What we are doing is performing a mathematical operation, in this case subtracting the same thing from both sides of the equation. I think it would be very useful to you to stop saying you are “moving” a variable from one side to the other. Instead say you are subtracting/adding/multiplying/dividing both sides of the equation in the same way. For example,

ax+b=c
First, subtract b from both sides of the equation and simplify:
ax + b – b = c – b
ax = c - b
Second, divide both sides of the equation by “a” and simplify:
(ax)/a = (c – b)/a
x = (c – b)/a

You will notice that I got a different answer than you did. Mine has “c – b” in the numerator while yours has “b – c”. This is likely because you “moved b over", rather than “subtracted b from both sides of the equation.” Your imprecise way of thinking about the mathematical operation resulted in an incorrect sign change.

Now, on to your current problem:

x-1=cx+d
-1=(cx-x)+d

You have correctly subtracted x from both sides of the equation (though there are other sequences of steps that will work). You have two terms – cx and x – that both have x as a factor. Therefore you can factor x out and rewrite your equation:

-1 = x(c – 1) + d

Next, subtract d from both sides. Then divide both sides by (c – 1):

-1 – d = x(c – 1)
(-1 – d)/(c – 1) = x

You have now solved for x. This result can also be written in the following ways:

x = (-1 – d)/(c – 1) = -(1 + d)/(c – 1) = (1 + d)/(1 – c)

 

[mmm4444bot]

It just says to solve the equations mentally if possible.


When an equation contains more than one symbolic number, it is ambiguous to simply instruct to "solve it".

I'll guess that you're supposed to solve for the symbolic number x in terms of symbols c and d.

Solve it mentally? Ugh. I think that you'd be better off using paper and pencil, on this one.


I got the b by subtracting cx-x. I thought it would be b since its alphabetical order and since it has a negative it would go one backwards.

I don't know what this means; maybe your class is doing something different.

x - 1 = cx + d

If we want to solve for x, we first need to get all of the terms containing x on one side of the equals sign AND get everything that does not contain x on the other side.

There's more than one way to do this first step; here's one way: subtract x from both sides and subtract d from both sides.

-1 - d = cx - x

See? Every term on the righthand side contains x, and every term on the lefthand side does not contain x.

Now factor each side.

-(d + 1) = x(c - 1)

Do you know the last step, to solve for x?

EDIT: YOU SHOULD KNOW, IF YOU JUST READ THE RESPONSES FROM JEFFM AND WJM11.