Easy way to solve equations in which all terms are under 1?

Test Preparation SAT Preparation
1

<p>I need an easy way to solve an equation such as this 1/x+1/2x=1/20 for combined rate problems on the SAT</p>

<p>I know there's probably an easy way, but for some reason it's giving me trouble..</p>

2

<p>Common denominators. :D</p>

3

<p>As LoseYourself said, you must have common denominators. The easiest way to get common denominators is to multiply the denominators:</p>

<p>x * 2x = 2x^2 (new denominator) </p>

<p>Then, cross multiply the denominator of one fraction with the numerator of another fraction and add them. Put the result on top of your new denominator. </p>

<p>x * 1 = x
2x * 1 = 2x </p>

<p>x + 2x = 3x</p>

<p>thus, </p>

<p>1/x + 1/2x = 3x/2x^2 = 1/20</p>

<p>I also have the problem worked out on paper:</p>

<p><a href=“http://i.min.us/ij5LJM.jpg[/url]”>http://i.min.us/ij5LJM.jpg&lt;/a&gt;&lt;/p&gt;

4

<p>The first thing I do is get rid of the fractions. </p>

<p>1/x + 1/2x = 1/20
20x * (1/x+1/2x) = (1/20) * 20x
20 + 10 = x
30 = x
x = 30</p>

5

<p>Oh sorry I get it …</p>

<p>1(2)/x(2)+1/2x=1/20</p>

<p>3/2x=1/20
then cross multiply</p>

<p>haha i forgot about common denominators for some reason when variables came into play… thanks man</p>

6

<p>^No problem :). I’m always free to help :D.</p>

7

<p>Okay since I feel like I’m spamming SAT prep I’ll just ask this in here haha…</p>

<p>Has anyone ever seen a problem on the SAT where you have to add a certain amount of liquid to a solution to dilute or concentrate it to a certain percent? It’s a “tip” in my Dr. Chungs SAT book, but it seems too complicated for the SAT…</p>

<p>Here’s a couple examples, feel free to share your personal strategies, cause his are pretty freaking vague haha…</p>

<p>1)how many gallons of water must be added to 40 gallons of 10% alcohol solution to produce an 8% solution?</p>

<p>2)how many gallons of a 20% salt solution must be added to 10 gallons of a 50% salt solution to produce a 30% salt solution?</p>

<p>3)How many quarts of alcohol must be added to 10 quarts of a 25% alcohol solution to produce a 40% alcohol solution?</p>

<p>4)how many gallons of acid must be added to G gallons of a k% acid solution to bring it up to an m% solution?</p>

<p>5)M gallons of a p% salt solution must be mixed up with G gallons of a q% salt solution to produce an r% solution, how can we represent the value of r?</p>

8

<p>1) Here’s the lowdown. It’s a cross-multiplication problem. Don’t be fazed by the percent signs - just convert all the percents to decimals. </p>

<p><a href=“http://i.min.us/ijAk66.jpg[/url]”>http://i.min.us/ijAk66.jpg&lt;/a&gt;&lt;/p&gt;

<p>2) I’m down for the count. I have no clue :(.</p>

9

<p>IceQube- thanks for the replies man. I SERIOUSLY doubt #2-5 would show up on an actual SAT due to their difficulty, but I’ll post Chung’s solutions if you would like.(even though they didn’t really help me understand it fundamentally)</p>

<h1>2- Solution A has x amount of mixture and .2x amount of salt, solution B has 10 units of mixture and 5 units of salt. Since (0.2x+5)/(x+10)=30/100, cross multiply to get 20x+500=30x+300, thus x = 20</h1>

<h1>3- Solution A has x amount of mixture and x amount of alcohol, Solution B has 10 units of mixture and 25 units of alcohol, since (2.5+x)/(10+x)=40/100, x=2.5.</h1>

<h1>4 - Solution A has x amount of mixture and x amount of acid, Solution B has G amount of mixture and kG/100 amount of acid, therefore (x+kG/100)/(x+g)=m/100. Now cross multiply to get 1000x+kG=mx+mG ----> x(100-m)=(m-k)G thus, x=(m-k)G/(100-m)</h1>

<p>I’m not even going to post #5… lmao it’s like half a page…</p>

10

<p>I finally figured out a way for #2. It’s pretty much the same way as Chung’s. Those problems are hard, but that’s what makes them so gratifying (if you get them right) :)! </p>

<p>.2x + .5(10) = .3(x+10), where x = amount of solution</p>

<p>.2x + 5 = .3x + 3</p>

<p>0.1x = 2</p>

<p>x = 20.</p>