Need help how do you do theses???? 3x+4-4x=2(3x-5)

Chick07Blondie

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How do you do these problems:

3x+4-4x=2(3x-5)

3x-6=4(x-2)-x

3(2x-5)/2 <8


Find the substitution
4x+2y=3
Y=6-2x
 
Re: Need help how do you do theses????

Chick07Blondie said:
How do you do these problems:

3x+4-4x=2(3x-5)

3x-6=4(x-2)-x

3(2x-5)/2 <8


Find the substitution
4x+2y=3
Y+6-2x<<< Something wrong - does not make sense

You had posted a similar problem before:

viewtopic.php?f=16&t=31779&p=122981#p122981

Follow the same principle.

Please show your work, indicating exactly where you are stuck - so that we would know where to begin to help you.

In thee meantime go to :

http://www.purplemath.com/modules/solvelin.htm

to review example problems worked out (your text-book would have some of those too)
 
Re: Need help how do you do theses????

It is always best to show your work.

How are we to know where you are stuck?

I will solve one of the questions and provide the prove (if possible).

3x - 6 = 4(x-2) - x

The first thing to do is clear the parentheses using the distributive rule.

Everytime you see parentheses in an equation, clear them using the distributive rule.

If you do not know what this rule is, visit the link provided below:

http://www.themathpage.com/alg/distributive-rule.htm

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On the right side of the equation, we apply the distributive rule and combine like terms to further simplify.

4(x - 2) - x becomes 4x - 8 - x, which then becomes 3x - 8.

We now the following linear equation:

3x - 6 = 3x - 8

Next, we get the terms with x together on one side and the numbers without x on the other.
Do not forget to change the signs infront of the numbers and terms when crossing the equal symbol.

We get this:

3x - 3x = 6 - 8

Stop right here!

Notice that when we place the terms with x together on the left side, they cancel each other out because one is positive and the other is negative.

So, what does this mean?

It means there is no solution to this equation.
 
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