<p>This ones a doozy: If x doesnt equal zero and x is inversely proportional to y, which of the following is directly proportional to 1/(x^2)?</p>
<p>2: (x-8)(x-k) = x^2 - 5kx + m
In the equation above, k and m are constants. If the equation is true for all values of x, what is the value of m?</p>
<p>I got this by plugging in the choices but Im sure thats not how ur sposed to</p>
<p>For 2, I would do this:</p>
<p>m= (-8)(-k)
-5k= -8-k</p>
<p>Tada?</p>
<p>1) Is the answer y/x ?</p>
<p>2) Here’s the algebraic way, but plugging in choices is faster and easier</p>
<p>(x-8)(x-k) = x^2 - 5kx + m
x^2 - 8x - kx + 8k = x^2 - 5kx + m
-8x - kx + 8k = -5kx + m</p>
<p>At this point you should realize that the two ‘x’ terms on the left must equal the one ‘x’ term on the right, and the two constants must be equal. Get two equations and solve them</p>
<p>Equation 1: -8x - kx = -5kx
(-8 - k)x = -5kx
-8 - k = -5k
-8 = -4k
k = 2</p>
<p>Equation 2: 8k = m
and we know the value of k…
8(2) = m
m = 16</p>
<p>^thanks for the explanation, I had trouble figuring out that question.</p>
<ol>
<li><p>x=1/y (equals in this case implies proportionality). 1/(x^2)=1/((1/y)^2)=1/(1/(y^2))=y^2. Right?</p></li>
<li><p>(x-8)(x-k) = x^2 - 5kx + m. x^2-8x-kx+8k=x^2-5kx+m. Looking at which terms have x^2, which have x, and which are just constants, we realize that x^2=x^2, -8x-kx=-5kx, and 8k=m. So, -8x-kx=-5kx so 8+k=5k (since you are an 800, I’m not gonna do the basic math) so 8=4k so k=2. Since 8k=m, m=8*2=16. So, m=16. Basically the exact same way as Jamesford (realized after I wrote it), but much less organized :).</p></li>
</ol>
<p>for the second one, set up a quick system. first, let x=8, that’s allowed since its true for all x. get an expression.</p>
<p>then, let x=k. get an expression.</p>
<p>i chose these since that will make the left side=0</p>
<p>Number 1 is fairly simple. If x is inversely proportional to y, then y = k/x, where k is some constant. Square both sides of the equation to get y^2 = (k^2)/(x^2), and rewrite to get (y^2)<em>(1/(k^2)) = 1/(x^2). Now, 1/(k^2) is just a constant, so you can just replace it with another k and get k</em>y^2 = 1/(x^2). And this equation tells you that y^2 is directly proportional to 1/(x^2).</p>
<p>I guess y/x or y^2 would work since y/x = y/(1/y) = y^2</p>
<p>If you want to know how I did it:</p>
<p>x is inversely proportional to y means
x = k/y</p>
<p>and we want 1/x^2 * k = something.
xy = k from the above
something = xy/x^2
something = y/x or y^2</p>
<p>I find it funny that a usual “800’er” found these questions hard. These are trivial.</p>
<p>^But it’s not like it’s near-impossible to get an 800 on SAT math.</p>
<p>
</p>
<p>Individuals look at the same problem in many different ways. What appears trivial to most might remain a complete mystery to another person. Everyone gets a few questions that do not yield an immediate click, and this happens often with the … simplest of questions. </p>
<p>The beauty of this forum, especially when it is very active, is that one can read many DIFFERENT approaches and realize that there is almost always a very simple solution to what appears to be a tricky question. </p>
<p>And, fwiw, the more “trivial” questions people pose, the better this forum works for all. The last thing anyone should do is question why a question is brought here. It is good to remember that people learn more by teaching others than by any other method. :)</p>
<p>
The SAT math section as a whole is trivial.</p>
<p>I don’t mind people asking questions, but be modest. Don’t mention how you’re normally an “800’er”. Just ask the questions.</p>
<p>Speaking of questions, I started a help thread for all SAT math questions (“Math Help Center”) :)</p>
<p>x^2-xk-8x+8k=x^2-5kx+m
[-xk-8x=-5kx]-1
xk+8x=5kx
4kx=8x
4k=8
k=2</p>
<p>m=8k
m=8[2]
m=16</p>
<p>Hehe i just took this practice test.</p>
<p>The first one was really easy, the second was slightly harder, but then i realized i had a ti 89 lol. Solve() function is a godsend </p>
<p>And whenever it says “all values of x”, just go ahead and plug in 2 or something nice for x.</p>
<p>^How did you use your ti-89 when there are two variables (k&m) left? Or did you simplify the equation enough to reduce it to one var?</p>
<p>^I’m not sure, but it sounds like he used arbitrary x-values. But he would have to have done it twice to get two equations which the ti89 can then solve.</p>
<p>Making up arbitrary x values is totally legit. If you are going to do it that way, you should really go all in on lazy: use x = 8 to get 0 = 64 - 40k +m </p>
<p>and then use x = 0 to get 8k = m. Then solve the system by hand or ti89. </p>
<p>But I still think post #3 has the quickest way. And this shows up regularly on the SAT. So it’s worth knowing: if they give you two expressions, one in polynomial form and one that is factored, and they tell you the expressions are equal for all values of x, then when you foil out the mess, each polynomial coefficient has to match…</p>
<p>You dont have to use the solve feature just to find exact values with one variable. When you type solve(…,x) you are telling it to solve for x. If there are 2 variables it will give x in terms of the other variable.</p>
<p>Solve(a+b+c = 10,c) </p>
<p>Output: c = -a-b+10</p>
<p>Just saves time instead of having to expand and simplify yourself.</p>
<p>For the first one: The question says x is inversly prop. to y. Let x=2 and therefore y=1/2. 1/x^2 will be 1/4. y/x=1/4. This seems to be the simplest approach.</p>
<p>^Wow, I did not know that was possible. That would help a lot since there’s always a few where they ask you to find x in terms of y. Thanks =D</p>