<p>Im really confused on this one guys.</p>
<p>If x+2y=a , and x-2y=b, which of the following is an expression for xy?</p>
<p>The answer is a^2-B^2 / 8 .</p>
<p>I just don't understand how to arrive at that answer! help!</p>
<p>bump- Somebody help</p>
<ol>
<li>Using x+2y = a, get x on one side of the equation.</li>
<li>Then set b in terms of a and y.</li>
<li>Then, using that expression, get y in terms of a and b.</li>
<li>Use the expression for y in terms of a and b and substitute into the first equation to get x as a function of a and b.</li>
<li>Now you have x as a function of a and b, and y as a function of a and b.</li>
<li>Multiply them to get xy as a function of a and b.</li>
</ol>
<p>Solution:</p>
<p>1st Equation:</p>
<p>x - 2y = b
[x = b + 2y] --> I took this and plugged it in to the second equation for x</p>
<p>2nd Equation:</p>
<p>x + 2y = a --> Since we know that x = b +2y plug that instead of x, u have:</p>
<p>(b + 2y) + 2y = a
b + 4y = a
4y = a -b
[[y = (a - b)/4]]</p>
<p>Now you know what y is. Now take what we solved for the first equation, we found that x = b + 2y, right, so plug in (a - b)/4 instead of y in that equation, since we found out that y = (a-b)/4. </p>
<p>x = b + 2y
x = b + 2((a - b)/4)
x = b + (a - b)/2
When you solve this...
x = (a + b)/2
Now you know both x and y just multiply the values. </p>
<p>x = (a + b)/2 & y = (a - b)/4 </p>
<p>X<em>Y = (a + b)/2 * (a - b)/4
X</em>Y = (a^2 - b^2)/8</p>
<p>Hope it helps.</p>
<p>dchow08,i appreciate your method.....but what about a simpler but more efficient one?here it goes:
1.first add up these two equation and you will get 2x=(a+b),which means x=(a+b)/2
2.now substract the 2nd equation from the first equation to get 4y=a-b,therefor y=(a-b)/4
3.now you find xy=(a+b)/2 X (a-b)/4,these means xy=(a+b)(a-b)/8,so you get xy=(a^2-b^2)/8</p>
<p>that gives you the answer,doesn't it?</p>
<p>You guys rock.
Im currently scoring around the high 600s in math and im hoping to push it to the 700s, so i'm just trying to iron out the types of questions i have trouble with! Only a month left =(</p>
<p>In response to harvard bounds method, whenever there are two unknowns, we must always find one and plug it into the other? As you did with X=b+2y, then subbing X on the 2nd equation with B+2y.</p>
<p>Another way:
"Square" both equations:
(x+2y)^2=a^2 , and (x-2y)^2=b^2
x^2+4xy+4y^2=a^2, and x^2-4xy+4y^2=b^2
subtract one from another
8xy=a^2-b^2
xy=(a^2-b^2)/8</p>
<p>Quix, Gluttony's way is faster but If you have a minute to spare you can apply this "plugging in" method to most math questions with two unknowns. Alternatively, if you dont like dealing with variables, you can plug in numbers for a, b, x, and y and compare the answer you get after plugging into the question to that of the answer choices. If one of the answer choices gives you the same value as the value you got for the original question, then that would be your answer.</p>
<p>^Thanks harvardbound</p>
<p>Gluttony, I guess its more of what method "you see" when you encounter a problem. I practiced plugging in a lot so thats what I "see" when I encounter these types of problems. But your way is good, no doubt.</p>