Elementary algebra: find the value of x

[chwala]

Homework Statement

This is my original question, i just want to know what i am missing here
find $x$ given;



$\frac {3}{2x+2}$+$\frac {4}{2x+2}$=$\frac {4x-3}{-2x-2}$

Relevant Equations

algebra addition and subtraction

$\frac {7}{2x+2}=\frac {4x-3}{-2x-2}$
$-7(2x+2)=(4x-3)(2x+2)$
$x^2+2x+1=0$
$x=1$ or $x=-1$

can we also have;

$-7=4x-3$ can the $2x+2$ cancel out? i am a bit mixed up on this very simple problem...and why am i getting false on my ti nsipre...

 

[DrClaude]

$x^2+2x+1=0$
$x=1$ or $x=-1$

Do you really have two solutions here?

$-7=4x-3$ can the $2x+2$ cancel out? i am a bit mixed up on this very simple problem...and why am i getting false on my ti nsipre...

You get the equality to
$$
-7=4x-3
$$
by dividing both sides by $2x+2$, which can can only do if $2x+2 \neq 0$.

 

[Mark44]

Homework Statement:: This is my original question, i just want to know what i am missing here
find $x$ given;
$\frac {3}{2x+2}$+$\frac {4}{2x+2}$=$\frac {4x-3}{-2x-2}$
Relevant Equations:: algebra addition and subtraction

$\frac {7}{2x+2}=\frac {4x-3}{-2x-2}$
$-7(2x+2)=(4x-3)(2x+2)$
$x^2+2x+1=0$
$x=1$ or $x=-1$

No.
The equation above has only one solution.

Also, the original equation can be written more simply as $\frac {7}{2x+2}=\frac {-4x+3}{2x+2}$.

With rational equations such as this, you should always note any solutions that must be disallowed, because they cause one or more denominators to be zero.

 

[Stephen Tashi]

$\frac {7}{2x+2}=\frac {4x-3}{-2x-2}$
$-7(2x+2)=(4x-3)(2x+2)$

There are two approaches to solving equations by algebraic manipulations.

In the careful approach, when you do any operation, you write a note stating what is being assumed. For example, suppose you are solving $x + \frac{1}{x-3} = 3 + \frac{1}{x-3}$ and you decide to subtract $\frac{1}{x-3}$ from both sides. That manipulation assumes $\frac{1}{x-3}$ is a number. It is only a number if $x \ne 3$. So when you do this manipulation, you write down "If $x \ne 3$" and remember it is assumed.

In the less careful approach (often taught by teachers who dispair of getting students to write words along with their algebraic manipulations) you perform all the manipulations as if they must work and then, at the end, you check each of your answers in the original equation. A thoughtful student will wonder why standard manipulations can fail to produce correct answers. Standard symbolic manipulations are not 100% reliable if you do them without remembering the assumptions that are needed.

 

[chwala]

yeah i just saw it...i did not counter check...$x=-1$

 

[chwala]

Do you really have two solutions here?You get the equality to
$$
-7=4x-3
$$
by dividing both sides by $2x+2$, which can can only do if $2x+2 \neq 0$.

can they cancel out...i guess so...with the condition given by stephen...

 

[epenguin]

Partially repeating what others have said:
if you bring the right-hand side term to the left hand side you see you have a sum of three fractions each one with the denominator $(2x + 2)$ .
The numerator, simplifying, is simply $4x + 4$,
To satisfy the equation this has to be equal to 0, which it will be only if $x = -1$ .
However in that case the denominator also is 0, so this is not a solution at all.
As for any other solution, you have been through a procedure involving a quadratic and found one ($x=1$).Substitute back and you easily see this does not satisfy your equation.
You might be thinking along the lines that there is something about when you get squares you can get solutions which are not really solutions of the original problem. But that is not what is happening here – you have simply made a mistake. You have confused with something else that is rather like it. As Dr Claude prodded you on, so get this clear.