Solving Equations with Subtraction: Can You Help Me?

[xeon123]

Hi,

I would like to solve this equation but I can't do it.

2x+1-((2x+3)/2)

Can you help me?

Thanks,

 

[Doc Al]

2x+1-((2x+3)/2)

First simplify the stuff in the parentheses: carry out the division.

(Those are expressions, not equations. No equals sign!)

 

[xeon123]

This expression is the same as the following 2x+1-4x-6?

I've multiplied ((2x+3)/2) by 2, and removed the parenthesis.

 

[Doc Al]

This expression is the same as the following 2x+1-4x-6?

No.

I've multiplied ((2x+3)/2) by 2, and removed the parenthesis.

If you had an equation, you could multiply both sides by 2. But here you just have an expression. Divide by 2.

If you really mean to subtract two equations, please show those equations.

 

[HallsofIvy]

This expression is the same as the following 2x+1-4x-6?

I've multiplied ((2x+3)/2) by 2, and removed the parenthesis.

No, you can't just multiply by 2 and get the same result- and you surely can't multiply just part of an expression and expect to get any thing sensible. And, by the way, multiplying (2x+3)/2 by 2 would just cancel the 2 in the denominator to give 2x+ 3, not 4x+ 6.

You could write the first part as [itex](2/2)(2x+ 1)= (4x+2)/2[/tex] so you can add the fractions (remember getting a "common denominator" so you can add?):
$$\frac{4x+2}{2}- \frac{2x+3}{2}= \frac{4x+ 2- 2x- 3}{2}= \frac{2x- 1}{2}$$

 

[Mark44]

Fixed (?) LaTeX.

No, you can't just multiply by 2 and get the same result- and you surely can't multiply just part of an expression and expect to get any thing sensible. And, by the way, multiplying (2x+3)/2 by 2 would just cancel the 2 in the denominator to give 2x+ 3, not 4x+ 6.

You could write the first part as $(2/2)(2x+ 1)= (4x+2)/2$ so you can add the fractions (remember getting a "common denominator" so you can add?):
$$\frac{4x+2}{2}- \frac{2x+3}{2}= \frac{4x+ 2- 2x- 3}{2}= \frac{2x- 1}{2}$$

 

[Mark44]

Hi,

I would like to solve this equation but I can't do it.

2x+1-((2x+3)/2)

As already noted, this is not an equation, it's an expression. The best you can do with an expression is to rewrite it in a different and possibly more simplified form.

It's important to understand the difference between an equation (there's an =) and an expression, since there are many more things you can do to an equation, such as add a number to both sides, multiply both sides by a number, and so on.

With an expression, the only thing you can add is 0 (in some form) or multiply by 1 (in some form), or expand or factor to write the expression in a different form.

 

[BloodyFrozen]

In addition to Mark44's suggestion, try to get them with the same denominator.

 

[Mark44]

hi,
i write first equation than write second equation and less the second equation from first equation than get remainder this is our solution.

This makes no sense in the context of this thread. The OP's problem is to simplify an expression. There is no equation.

 

[Epic_Sarthak]

Hi,

I would like to solve this equation but I can't do it.

2x+1-((2x+3)/2)

Can you help me?

Thanks,

2x + 1 - ((2x + 3)/2)
2x + 1 - (x + 3/2)
2x + 1 - x - 3/2
x - 1/2


This is the step by step answer to your question.

 

[sankalpmittal]

2x+1-((2x+3)/2)

=(4x+2-2x-3)/2
=(2x-1)/2