C = 4P
P = 3 + W
Total = C + P + W
Plug in the second equation into the first to get
C = 4(3 + W) = 12 + 4W
Now plug in C = 12 + 4W and P = 3W into the Total equation
Total = (12 + 4W ) + 3W + W = 12 + 8W
We know that W is going to be a positive integer (1, 2, 3, 4…) so plug those into the Total equation
12 + 8(1) = 20
12 + 8(2) = 28
We can see already that 28 is a possibility, so (A)
I wouldn’t do all of that algebra for this one. I would simply take guesses for the number of cows.
If we guess that there are 2 cows, then there are 2 + 3 = 5 pigs, and (4)(5) = 20 chickens. So the total is 2 + 5 + 20 = 27, choice (B).
Of course, you may not be lucky enough to make 2 your first guess, but you will certainly know if your guess is too big or small when you do the computation.
@pillowspillows I agree with @DrSteve; you don’t need a lot of algebra for this one. Also, I think you accidentally put 3W instead of 3+W, which led to a wrong answer. The simpler, the better (usually).
If W is the number of cows, then the number of animals is W + (3+W) + (12+4W) = 15+6W. This number is necessarily a multiple of 3 (since W is an integer) and only choice B contains a number that is a multiple of 3.
@MITer94 ya I was typing it on my phone and switched 3+W and 3W somehow. I usually like having concrete equations and stuff to solve math problems instead for guess and check (probably because usajmo qualifier lol). But, I can see why taking guesses for cows would be much faster especially for an sat question.
@pillowspillows Same, I was a USAMO qualifier back in HS. I generally avoid guess and check unless I know other methods might take me more time.
However sometimes guessing at the solution (e.g. finding some invariant) can be a useful strategy on olympiad problems.
If x^2=9 means x= +3 and -3,
is sqrt(9)= +3 and -3 ?
if so, why does the equation
sqrt(x+4)= x-8
not have x=5 as a valid solution?
Wouldn’t sqrt(9) also be equal to -3?
NOTE: This question doesn’t actually appear on the SAT, but it’s the result of some confusion I’ve had doing a SAT question.
Every positive real number has two square roots, one negative and one positive. So -3 and 3 are both square roots of 9. The square root symbol, however, means to take the positive square root. So sqrt(9)=3. If we want the negative root, we would put a minus sign in front of the sqrt symbol.
In your example, x=5 is not a valid solution because sqrt(___) can never be negative, whereas x-8 would be negative in this case.
So when we do x^2= ___ , are we not taking the square root? I tried looking up the proof and it’s only confused me more. Why does the plus minus sign come before the square root symbol when we do x^2= ___?
I guess this is a bit too advanced for me. 
To solve an equation of the form x^2 = a requires the square root property, which says that the equation has two solutions: the positive and negative square roots of a.
Applying the square root property is different from taking the positive square root. The first generally gives two answers, the second just one.
@pizzatom999 as DrSteve said, for positive a, the equation x^2 = a has two distinct real solutions (the positive and negative of the square root of a).
For example, if x^2 = 9, then x = ±3. You cannot conclude that x = 3 without knowing more information about x. All you can conclude is that x = 3 or x = -3. This is a common mistake that a lot of students make. Another valid way to solve is to rewrite as x^2 - 9 = 0 and factor the LHS as (x-3)(x+3) = 0.
The converse is true; if x = 3, then x^2 = 9.
A slightly more involved example, if (x+1)^2 = 25 and you wish to find all solutions x, the common error is to “square root” both sides and claim that x+1 = 5 --> x = 4. But x = 4 is not the only solution!
y=3
y = ax^2 + b
In the system of equations above, a and b are constants. For which of the following values of a and b does the system of equations have exactly two real solutions?
A) a=−2,b=2
B) a=−2,b=4
C) a=2,b=4
D) a=4,b=3
this is a question from official college board test (#2). I read the explanation portion of the test, but I still am struggling in figuring this questions out. Can anyone help? Thanks
First, this problem seems to have 2 correct answers, so it is not a good SAT question. But I’ll give a solution anyway.
The quickest way is to just look at the discriminant of a quadratic equation.
The discriminant of ax^2 + bx + c = 0 is D = b^2 - 4ac
In this case, we have 3 = ax^2 + b, or equivalently ax^2 + (b + 3) =0, so that the discriminant is -4a(b + 3).
To have two real solutions, the discriminant must be positive. So it looks like both A and B are correct answers.
Note that the discriminant is just the expression under the square root in the quadratic formula.
It makes sense that when the discriminant is positive you should get 2 real solutions because of the + and - in front of the square root symbol in the formula.
When the discriminant is 0, the + and - become irrelevant and there is just one real solution.
And when the discriminant is negative, you get complex conjugate solutions.
@DrSteve the equation obtained should be ax^2 + (b - 3) = 0, in which the discriminant is -4a(b - 3).
@MITer94 Thanks for fixing my silly error. I believe that this means my original remark is incorrect as well. Looks like the problem has just one solution, and that it would be a perfectly reasonable SAT question.
@tiffym I solved this problem quickly by imagining the graphs. Imagine y=3; it’s a straight line. The graph y=ax^2 + b will always be a parabola where the vertext is located at (0, b). This means that, in order for the two equations to intersect at two locations, either the parabola has to be concave up (i.e. a>0) and b<3 OR the parabola has to be concave down (i.e. a<0) and b>3. (B) is the only answer that fits this criterion. This might not be the ideal way to view the question, but it was quick and easy that might come in handy.
@ObitoSigma I think that this is a really nice solution. Stronger students can probably follow your reasoning pretty easily. Students weaker in math can actually put each of the equations in their graphing calculator along with y=3, and see which one yields 2 points of intersection (if a calculator is allowed).
For a polynomial p x ( ), the value of p(3) is −2.
Which of the following must be true about p x ( ) ?
A) x − 5 is a factor of p x ( ).
B) x − 2 is a factor of p x ( ).
C) x + 2 is a factor of p x ( ).
D) The remainder when p x ( ) is divided
by x − 3 is −2.
Can someone simplify the answer and the way.
The answer is choice D by a direct application of the remainder theorem.
The remainder when the polynomial p(x) is divided by x - r is p®.