<p>For all those genius math people, here is an SAT question on steroids:</p>
<p>Find a,b,c,d:</p>
<p>25=a+b+c+d
40=8a+4b+2c+d
60=27a+9b+3c+d
95=64a+16b+4c+d</p>
<p>Anyone know the answer? It'll be great if you could give an explanation also. If it's too long, then don't worry about the explanation.</p>
<p>Are a, b, c and d integers?</p>
<p>What are the answer choices?</p>
<p>yes, a,b,c,d are integers</p>
<p>Here are the answers
a:5/3
b:-7.5
c:25.833333
d:5</p>
<p>This is simple AlgebraI (4 systems, 4 variables).</p>
<p>I didn't get integers for x,y,z,t and there should only be one answer.</p>
<p>EDIT:
[quote]
yes, a,b,c,d are integers
[/quote]
^^Contradicted yourself there.</p>
<p>just use TI 89- algebraic solver haha...</p>
<p>but it can also be done pretty easily by hand</p>
<p>I believe you solve through elimination.
4 - 1
4 - 1 - 2
4 - 2 - 3
4 - 3
First, take the first two, eliminate one variable, and leave 3 variables.
Next, take the second and the third, eliminate one variable and leave 3.
Finally, take the last two and eliminate one, and leave 3 variables.</p>
<p>Now, you have a system of 3 variable and 3 equations.
Continute eliminating.
3 - 1
3 - 1 - 2
3 - 2
You eliminate one more variable after completion.</p>
<p>And then finally, you solve for the last two variables through a combination of elimination and substitution.
2 - 1
2 - 1</p>
<p>Hope that helped! :)</p>
<p>use matrices...</p>
<p>Cooldude speaks the truth.</p>
<p>Matricies should have been covered in Algerbra 2</p>
<p>Or matrices. :p</p>
<p>Very simple. Substitution. Just very TEDIOUS.
a = 25 - b - c - d
now plug into equation 2
40 = 8 (25 - b - c - d) + 4b + 2c + d
40 = 200 - 8b - 8c - 8d + 4b + 2c + d
-160 = -4b - 6c - 7d
I am going to take out a - 1 and solve for B
4b = 160 - 6c - 7d
now dont divide this by 4 and plug it into equation 4. (you' see why)
95 = 64 (25 - b -c -d) + 4 (160 - 6c - 7D) <-- may have confused you + 4c + d
95 = 1600 - 64 b - 64c - 64d + 640 - 24 c - 28D + 4c + d
-2145 = -84 C - 91 D - 64 B
Now remember B = (160 - 6c - 7d) / 4 (from equation 2)
Plug this in
-2145 = -84C - 91D - 16 (160 - 6c - 7d)
-2145 = -84C - 91D - 2560 + 96C + 112D
415 = 12 C + 21D
415 - 21D = 12C
c = 415 - 21D / 12
Now use equation 3..
It simplifies ALL the way to (took me 10 minutes.. to do. Is this an SAT problem?)
.25 D = 1.25
D = 5
And if you plug back in you get A B C which equal 5/3 , -7.5, 25.8333. I checkehd them. They all work. ARe these correct answers? also You said they had to be integers... but can you post correct answers?
5/3
-7.5
25.8333
and 5</p>
<p>Oh nm You posted AFTER I figured it out lol. Wow I can't believe I did this one. It took me 30 minutes (went to go eat and left screen on half the time). HOw is this an SAT problem. WAY too complex (not everyone has calc's to do matrixes etc.). But wow I am so happy i got it.</p>
<p>Why don't you substitute the numbers and see which ones work?</p>
<p>^ That's what I would have done. Or matrices.</p>
<p>I think it was grid-in.. or that would be too easy. And you said integers onl.. that through me off for a good 10 mins. because i was just guessing numbers up to then..</p>
<p>hehehe, just joking guys, this was one of my IB Math internal assessment questions, not an SAT question. At first I couldn't find the answer, but later I solved it myself. But, hey, good practice, right?</p>
<p>yea it doesn't take long but trying to do it when typing... takes a LONG time. I did it like 34 times on typing. Once in paper. Anyhow whats IB math internal assessment?</p>
<p>how would you do it with matrices algebra II seems like a distant time :(</p>
<p>Or you can just use the TI-89's solve function. :-)</p>
<p>But yes, matrices is the way most people would do it.</p>