ohioleslie

10-03-2008, 07:42 PM

Here's my query: How do you know when an equation has infinitely many solutions AND how do you know when an equation has no solution?

View Full Version : Algebraic inequality: how to know when equations have infini

ohioleslie

10-03-2008, 07:42 PM

Here's my query: How do you know when an equation has infinitely many solutions AND how do you know when an equation has no solution?

Mrspi

10-03-2008, 08:25 PM

Here's my query: How do you know when an equation has infinitely many solutions AND how do you know when an equation has no solution?

Let's say that you have THIS equation:

2(x - 5) = 2x + 3

Multiply each term inside the parentheses by the 2 which is outside:

2x - 10 = 2x + 3

Now, to eliminate the variable term from the right side of the equation, subtract 2x from both sides:

2x - 10 - 2x = 2x + 3 - 2x

Combine like terms:

-10 = 3

Now...the variable has "disappeared" from the equation, and we are left with

-10 = 3

That's a statement which is NEVER true, and that tells us that the original equation is NEVER TRUE regardless of what value we use for x. So, this equation has NO SOLUTIONS.

Next, let's consider this equation:

2(x - 5) - 3 = 2x - 13

2x - 10 - 3 = 2x - 13

2x - 13 = 2x - 13

Now, if you subtract 2x from both sides, you get

2x - 13 - 2x = 2x - 13 - 2x

-13 = -13

When is this statement true? All of the time, right? So, no matter what value x has, the original statement will be true. And the equation has infinitely many solutions.

You did not give us an example of a problem you're having trouble with, so this was my best guess at what you're dealing with.

If your equation simplifies to something that is NEVER true, the equation has no solutions. If it simplifies to something which is ALWAYS true, then the equation has infinitely many solutions.

As I said, I had to guess at what you were asking about because you didn't post an example problem.

Let's say that you have THIS equation:

2(x - 5) = 2x + 3

Multiply each term inside the parentheses by the 2 which is outside:

2x - 10 = 2x + 3

Now, to eliminate the variable term from the right side of the equation, subtract 2x from both sides:

2x - 10 - 2x = 2x + 3 - 2x

Combine like terms:

-10 = 3

Now...the variable has "disappeared" from the equation, and we are left with

-10 = 3

That's a statement which is NEVER true, and that tells us that the original equation is NEVER TRUE regardless of what value we use for x. So, this equation has NO SOLUTIONS.

Next, let's consider this equation:

2(x - 5) - 3 = 2x - 13

2x - 10 - 3 = 2x - 13

2x - 13 = 2x - 13

Now, if you subtract 2x from both sides, you get

2x - 13 - 2x = 2x - 13 - 2x

-13 = -13

When is this statement true? All of the time, right? So, no matter what value x has, the original statement will be true. And the equation has infinitely many solutions.

You did not give us an example of a problem you're having trouble with, so this was my best guess at what you're dealing with.

If your equation simplifies to something that is NEVER true, the equation has no solutions. If it simplifies to something which is ALWAYS true, then the equation has infinitely many solutions.

As I said, I had to guess at what you were asking about because you didn't post an example problem.

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