#algebraic manipulation

[pwnthesat]

Which of the following statements is true for all real numbers c ? A) 8+c > 4+c B) 8+4c > 4-4c C) 4c > 8c D) 8c > 4c E) 8c^2 > 4c^2

The correct answer is A. Subtract c from both sides of the inequality and you’re left with 8 > 4. That’s always true!

[pwnthesat]

If x+y+y=y and y not equal to 0, then x/y =

Rearrange that equation:

x + 2y = yx = –yx/y = –1

[pwnthesat]

If p and q are nonzero integers and 81p = 27q , what is the value of p÷q? A)3÷4 B)4÷3 C)27÷7 D)7÷27

81 is 3(27) so you can rewrite: 

27(3p) = 27q3p = q3p/q = 1p/q = 1/3

[pwnthesat]

if x(x+y)=3 and y(x+y)=10, which of the following is a possible value of x+y? a)√7 b)√13 c)6.5 d)7 e)13

Let’s do a little clever combination. 

x(x + y) + y(x + y) = 3 + 10

Whoa, looks we can factor (x + y) out of the left hand side:

(x + y)(x + y) = 13

If that’s not obvious to you, you can also expand the original combined equation

x^2 + xy + yx + y^2 = 13x^2 + 2xy + y^2 = 13

Either way, what you have on the left is (x + y)^2. If (x + y)^2 = 13, then x + y = ±√13.

[maths-help]

Okay, so I have this question that's asking me to solve for s, and the equation is R = ts² + 3 and I'm not even sure where to start? My only answer choices have square roots in them. Help please?

Hey! The expression contains s², so of course your answers have square roots in them when you’re supposed to solve for s ;)

I got s = +/- [(R-3) / t]^(½) as an answer. And in this case I wouldn’t know how to solve for s /without/ square roots….

-sorrel

[leannedodge]

SAT math: algebraic manipulation

I’ve posted a lot of critiques of the new SAT, but--at the end of the day--we all know that juniors will be taking the PSAT, and some juniors and seniors will be taking the SAT.

In addition to continuing to post about the status of the exam (and to reiterate my current preference for the ACT over the SAT!), I’ll also be posting some more strategies for the SAT.

That starts with algebraic manipulation. You may know that one of the categories on the SAT, Heart of Algebra (name given by the College Board), is frequently going to require you to switch around equations to solve for a particular variable. Here’s how to approach a question:

1. Look at how the equation is written and what you need to do to isolate the variable they are asking you about.

2. Work to get the variable you are solving for alone on one side of the equation. You will most commonly do this in one of two ways:

2a. Cross multiplication. If you have a fraction on one (or both) sides, you may find it helpful to cross multiply as a first step.

2b. Opposite operations. Do the same thing to both sides of the equations. If you need to get rid of something that is added to one side, subtract it from both sides. If you need to isolate something that is multiplied by something else, divide both sides by it.

Here’s an example:

S = (2av - 5b) / 3c

Let’s say we need to solve for c. Right now, c is on the bottom of a fraction on the right side of the equation. Let’s cross multiply as a first step:

3cS = 2av - 5b

That looks better! Now, what’s in the way of the c being alone? 3 and S. Let’s divide both sides of the equation by 3S. We are left with:

c = (2av - 5b) / 3S

Not as bad as it first appeared, right? Just don’t forget to divide the whole right side by 3S in the last step.

[maths-help]

Hi there, I’m having trouble with the above questions. I’ll let you know what I’m thinking first and please tell me where I’m going wrong haha

A. T will decrease as you are increasing the value of denominator

B. T will increase as you are increasing the value of the numerator

C. (I have no idea about this one 😳)

D. I got very confused with this one!! The second picture (handwritten) relates to this question ( no idea if it’s right 😳)

I would be so so grateful if you are able to help me 😊 Thanks, Britt

Hello!

A. Yes, that’s correct - since everything else in the equation is supposed to stay the same nd you increase the denominator, T will decrease.

B. You kind of overlooked the fact that sqrt(m1) is also present in the denominator. Actually, you can cancel that. The equation above is exactly the same as T=2/(1+sqrt(m2))! (You can also check by inserting some values for m1 and m2 ...). So if you increase m1, absolutely nothing about the value of T will change.

C. m2 is written under a square root. If you want your result to stay rational instead of complex, the lowest value you can put under a square root is zero, so your limiting value is m2=0.

D. It’s almost right! You made a small mistake when you subtracted 1. You need to subtract 1 from the whole fraction, not just from the numerator. Also, it’d be easier if you cancelled sqrt(m1) from the fraction first (like I did in B.), and then make m2 the subject. My result is m2=(2/T  -1)², which is the same as m2=[(2-T)/T]²=[(T-2)/T]².

I hope that helped!

-sorrel

[pwnthesat]

3(x+y)=y If (x,y) is a solution to the equation above and y cannot equal 0, what is the ratio x/y?

3x + 3y = y3x = -2y3x/y = -2x/y = -2/3