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Equations with fractions

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Rocky Louis

#1 Posted : Friday, March 25, 2011 2:02:45 AM

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Hi,

I'm having problems with doing one or more step equations with fractions. I think I got the part about being able to solve the equations, but everytime I try to do an equation with fractions it gets me confused. The reason I'm asking this question is so I can skip an Algebra class at my community college.

Rocky Louis

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lisa

#2 Posted : Friday, March 25, 2011 11:17:44 AM

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It can be a bit tricky when there are fractions involved. A good thing to remember is that + and - are opposites whereas * and / are opposites. To "eliminate" a fraction you can multiply both sides of an equation with the denominator of the fraction.

$\\\frac{a}{b}=c \Rightarrow \frac{{\color{blue} b}*a}{b}={\color{blue} b}*c\Rightarrow \\\\\frac{{\color{red} \not}{b}*a}{{\color{red} \not}{b}}=b*c \Rightarrow a=b*c$

When you have an equation in more than one steps with a fraction involved you begin by "isolating" the fraction on one side of the equation by moving everything else to the other side and then multiplying both sides by the denominator.

You can read more here:

http://www.mathplanet.com/education/algebra-1/how-to-solve-linear-equations/properties-of-equalities

http://www.mathplanet.com/education/algebra-1/how-to-solve-linear-equations/fundamentals-in-solving-equations-in-one-or-more-steps

http://www.mathplanet.com/education/algebra-1/how-to-solve-linear-equations/ratios-and-proportions-and-how-to-solve-them

I hope this was what you wanted help with otherwise ask again and I or anyone else can continue explaning

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Rocky Louis

#3 Posted : Friday, March 25, 2011 11:07:47 PM

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Could you use at least three examples with real numbers ?

Thanks.

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lisa

#4 Posted : Saturday, March 26, 2011 6:16:25 PM

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Of course!

If look at an one-step equation we can see that if we've got

$\frac{x}{5}=2$

then if we want to know x we need to multiply both sides by 5 to get the x alone since 5/5=1

$\\\frac{x}{5}=2\Leftrightarrow \frac{{\color{blue} 5}*x}{5}={\color{blue} 5}*2 \\\\\frac{{\color{red} \not}{5x}}{{\color{red} \not}{5}}=10\Leftrightarrow x=10$

If we instead would have had a two-step equation like the one below

$\\\frac{x}{2}-3=7$

we first have to isolate the fraction (the part including the x) by "Moving" the 3 to the other side of the euality. This is done by adding 3 to both sides of the equation (hence not changing the equality since we do the same thing on both sides)

$\\\frac{x}{2}-3=7\Leftrightarrow \frac{x}{2}-3{\color{blue} \, +\, 3}=7{\color{blue} \, +\, 3} \\\\\frac{x}{2}=10$

Once here we just do the same thing as we did when we had the one-step equation i.e. multiply both sides by the denominator, in this case 2

$\\\frac{x}{2}=10\Leftrightarrow \frac{{\color{blue} 2}*x}{2}=10*{\color{blue} 2} \\\\\frac{{\color{red} \not}{2}*x}{{\color{red} \not}{2}}=10\Leftrightarrow x=20$

"When we want to solve an equation including one unknown variable, as x in the example above, we always aim at isolating the unknown variable. You can say that we put everything else on the other side of the equal sign. It is always a good idea to first isolate the terms including the variable from the constants to begin with as we did above by subtracting or adding before dividing or multiplying away the coefficient in front of the variable. As long as you do the same thing on both sides of the equal sign you can do whatever you want and in which order you want." - Mathplanet.com

Hope this helped to clearify it a bit more eitherwise please tell!

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Rocky Louis

#5 Posted : Sunday, March 27, 2011 1:09:28 AM

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I don't think this method works cause every time I try to use it I get the wrong answer. Is there another method ?

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lisa

#6 Posted : Sunday, March 27, 2011 12:39:27 PM

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Can you please show me one of your equations and i will see if i can solve it? It is much easier to see what the problem is if i can see the problem

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Rocky Louis

#7 Posted : Monday, March 28, 2011 2:43:55 AM

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1. x - 3/7 = 25/7

2. -2/3x + 7 = -1/3

3. 3x - 1/8 = 143/8

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lisa

#8 Posted : Monday, March 28, 2011 10:45:30 AM

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Hi!

Now i understand why it's not working for you. Thesw examples you've written are not at all (ok its two-step equations but still) the same as the ones i tries to explain. If we take these one at the time.

$x-\frac{3}{7}=\frac{25}{7}$

In this one you've already got the x by itself so if we just "move" the -3/7 to the other side of the equality by adding 3/7 to both sides. This is done to isolate the x. We wil then get:

$\\x-\frac{3}{7}=\frac{25}{7} \\\\x-\frac{3}{7}{\color{blue} \, +\, \frac{3}{7}}=\frac{25}{7}{\color{blue} \, +\, \frac{3}{7}} \\\\x=\frac{25}{7}+\frac{3}{7}=\frac{28}{7}=4 \\\\x=4$

since we have the same denominator we can just add the numerators to get the value of x

2. I think your example is written like this otherwise just tell med :)

$\\-\frac{2}{3}x+7=-\frac{1}{3}$

Begin the quest of isolating the x by moving the 7 to the other side of the equality by doing the opposite action (in this case subtracting 7 from both sides)

$\\-\frac{2}{3}x+7=-\frac{1}{3} \\\\-\frac{2}{3}x+7{\color{blue} \, -\, 7}=-\frac{1}{3}{\color{blue} \, -\, 7} \\\\-\frac{2}{3}x=-\frac{1}{3}-7$

To simplify the right hand side of the equation we can write the 7 as thirds. 7=21/3 which gives us

$\\-\frac{2}{3}x=-\frac{1}{3}-7=-\frac{1}{3}-\frac{21}{3}=-\frac{22}{3}$

To get the value of x we have to multiply both sides by 3

$\\-\frac{2}{3}x=-\frac{22}{3} \\\\-\frac{2}{3}x{\color{blue} \, \cdot\, 3}=-\frac{22}{3}{\color{blue} \, \cdot\, 3} \\\\-\frac{2}{{\color{red} \not}{3}}x\cdot {\color{red} \not}{3}=-\frac{22}{ {\color{red} \not}{3}}\cdot {\color{red} \not}{3} \\\\-2x=-22$

Mow you can just divide both side by -2 to get the value of x

$\\-2x=-22 \\\\\frac{-2x}{{\color{blue} -2}}=\frac{-22}{{\color{blue} -2}} \\\\\frac{{\color{red} \not}{-2}x}{{\color{red} \not}{-2}}=\frac{-22}{-2} \\\\x=11$

This could have been done in one step if we had senn -2/3 as a constant b

$\\bx=a \\\\\frac{bx}{b}=\frac{a}{b} \\\\x=\frac{a}{b}=\frac{-22\setminus 3}{-2\setminus 3}=11$

$\\3x-\frac{1}{8}=\frac{143}{8}$

Here you do basically the same thing as earlier. Begin by moving the constant (here -1/8) to the other side of the equality by doing the opposite action (adding 1/8) on both sides. Once the x-factor is alone on its side of the equality, divide both sides by 3 to get the value of x

$\\3x-\frac{1}{8}=\frac{143}{8} \\\\3x-\frac{1}{8}{\color{blue} \, +\, \frac{1}{8}}=\frac{143}{8}{\color{blue} \, +\, \frac{1}{8}} \\\\3x=\frac{144}{8} \\\\3x=18 \\\\\frac{{\color{red} \not}{3}x}{{\color{red} \not}{3}}=\frac{18}{3} \\\\x=6$

Hope this made it a bit clearer :) It gets easier once you get a hold of it. If you have any more questions. Please don't hesitate to ask!

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Rocky Louis

#9 Posted : Wednesday, March 30, 2011 1:27:31 AM

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Do you know any good sites other than kuta software, cool math or math.com where I can practice these problems, since this site doesn't have a practice area.

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lisa

#10 Posted : Wednesday, March 30, 2011 1:29:49 PM

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The once that you mentioned are pretty good. In the future we will probably have exercises here on mathplanet.com as well, but not at the moment.

I found this site http://www.teach-nology.com/worksheets/math/algebra/ It might be something :)

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