NationStates Jolt Archive

Alright, my friend is out of town for a few days (in the hospital for a minor knee surgery) and I'm letting his kid (15 years old) stay at my place until his dad gets back. He's got some homework he's having trouble with and I've tried to help him (I can help on some, but its been like 9 years since I last had to use the stuff, and I was stoned for the majority of classtime), but we can't figure out 16 of the 50 problems there are. If anyone can figure these out, please help us. Much thanks and respect to whoever can help.

Here's the problems (multiple choice).

1. Use substitution to solve the system of equations. x=2 and 3x + y = 5

A.) {(-1, 20)}
B.) {(2, -1)}
C.) None of These
D.) {(2, 11)}

=

2. Use elimination to find the value of 'x' in the system of equations. x - y = 5 and x + y =5

A.) 4
B.) 3
C.) 8
D.) 1/4

=

3. Use elimination to find the value of 'y' in the system of equations. x+6y=10 and x+5y=1

a.) 10
b.) 11
c.) 1
d.) 9

=

4. Bob is using substitution to solve the following system of equations r+s=4 and 3r+2s=15. Which of these expressions should he substitute for r in the second equation?

a.) 4-r
b.) s-4
c.) 4-s
d.) 4/s

=

5. What is the simplest form of
36b^4c^2 over 9b^-1c^5

a. 27b^3 over c^3
b. 4b^5 over c^3
c. 4b^4 over c^3
d. 27b^4 over c^3

=

6. Use elimination to solve the system of equations. 2x+5y=7 and 3x+6y=3

a. (5,-9)
b. (-1,1)
c. (-9, 5)
d. none of these

=

7. To eliminate the variable y in the system of equations 5x+4y=22 and 2x-y=1, multiply the second equation by

a. 9
b. -4
c. 4
d. 1/3

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8.What is the degree of 3xy-8x^2y^5+x^7y

a. 7
b. 2
c. 10
d. 8

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9. What is the simplest form of (5c^2d^3)(-2c^5d^2)

a. -10c^10d^6
b. 10c^10d^6
c. 10c^7d^5
d. -10c^7d^5

=

10. Find the solution of x(x+2)=x^2+x+2

a. -1
b. 0
c. 2
d. 1

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11. If 2n^2 is factored completely, one of the factors is

a. 2n-8
b. n-16
c. n+4
d. n+16

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12. Factor 14mn^2+2mn

a. 2mn(7n+1)
b. mn(7n+1)
c. 2n(7mn+m)
d. 7mn(2n+2)

=

13. Factor 14xy^2 - 2xy

a. 2xy(7y-1)
b. 2x(7y^2-y)
c. -2xy(-7y)
d. prime number

=

14. Factor 16-n^4

a. (4-n^2)(4+n^2)
b. (n+2)(n-2)(4+n^2)
c. (2-n)(2+n)(n^2+4)
d. (2-n)(2+n)(2-n)(2+n)

=

15. Factor 3n(n+4)-2(n+4)
a. (n+4)(3n-2)
b. (3n-2)(n+4)^2
c. (3n+2)(n-4)
d. None of these

=

16. Solve -3(ab^4)^3

a. 3ab
b. -3a^3b^4
c. 9a^3b^3
d. 27ab

END OF PROBLEMS

Jesus, I about went crazy typing those out. If anyone can answer those for me we'd appreciate it lots.

As I solve them..

1) Y = -1

4) 4-s

1 - B, substitute x=2 into the second equation.
2 - I am not getting any of those - I get x=5, y=0.
3 - D, subtract the second equation from the first
4 - C, just isolate r on the left and you have it
5 - B
8 - I think it is D, but I'm not sure
9 - C

Any help?

6) I'm going with None of These. I get (2.111111,.5555555555555)

For some reason though some of my answers don't match up to theirs, but they check out when testing them.

7) Multiple by -4

10) X = 0

Ah, Algebra.
The science of confusing letters with numbers and calling it logic.

Im sorry, Im just a chronic math-hater ;)

1) b
6) b
10) c
12) a
14) d

12) 2mn(7n+1)


13) 2xy(7y-1)

15. Factor 3n(n+4)-2(n+4)
a. (n+4)(3n-2)
b. (3n-2)(n+4)^2
c. (3n+2)(n-4)
d. None of these

=a (n+4)(3n-2)

...only one I felt like doing.

Ah, Algebra.
The science of confusing letters with numbers and calling it logic.

Im sorry, Im just a chronic math-hater ;)

Same here, thanks to everyone who has helped us out so far.

14) (4-n^2)(4+n^2)

6) b


Eh?

6. Use elimination to solve the system of equations. 2x+5y=7 and 3x+6y=3

a. (5,-9)
b. (-1,1)
c. (-9, 5)
d. none of these

Substitute -1,1 into 2x+5y = 7

2(-1)+5(1) = 7
-2+5 =/= 7, -2+5 = 3 no?

Now why don't you use the 1st and 2nd derivatives to find the critical points and tell me if they are concave up, concave down or a saddle point?

f(x,y) = 5x^2 + 6(x^2)(y) + y^2 + 12

God Bless Calculus?


If you feel like torturing yourself, solve this one

f(x,y) = 2x^3 + 6(x^2)(y^2) + y^3 + 12

Seeing as I can’t.


/changed the formula, again, and again, to make it easier... /

Eh?

6. Use elimination to solve the system of equations. 2x+5y=7 and 3x+6y=3

a. (5,-9)
b. (-1,1)
c. (-9, 5)
d. none of these

Substitute -1,1 into 2x+5y = 7

2(-1)+5(1) = 7
-2+5 =/= 7, -2+5 = 3 no?

You're right it's d. (1,-1)

1. Use substitution to solve the system of equations. x=2 and 3x + y = 5
B.) {(2, -1)}

2. Use elimination to find the value of 'x' in the system of equations. x - y = 5 and x + y =5

5's not a choice for this one? (x + y) + (x - y) = 5 + 5 so 2x = 10 and x = 5 unless I'm missing something fundamental.

3. Use elimination to find the value of 'y' in the system of equations. x+6y=10 and x+5y=1

d.) 9

4. Bob is using substitution to solve the following system of equations r+s=4 and 3r+2s=15. Which of these expressions should he substitute for r in the second equation?

c.) 4-s

5. What is the simplest form of
36b^4c^2 over 9b^-1c^5

b. 4b^5 over c^3

6. Use elimination to solve the system of equations. 2x+5y=7 and 3x+6y=3

c. (-9, 5)

7. To eliminate the variable y in the system of equations 5x+4y=22 and 2x-y=1, multiply the second equation by

c. 4

8.What is the degree of 3xy-8x^2y^5+x^7y

d. 8 (I think)

9. What is the simplest form of (5c^2d^3)(-2c^5d^2)

d. -10c^7d^5

10. Find the solution of x(x+2)=x^2+x+2

c. 2

11. If 2n^2 is factored completely, one of the factors is

I think something is missing from the question here, since 2n^2 can only be broken down into 2(n)(n).

12. Factor 14mn^2+2mn

a. 2mn(7n+1)

13. Factor 14xy^2 - 2xy

a. 2xy(7y-1)

14. Factor 16-n^4

a. (4-n^2)(4+n^2)

15. Factor 3n(n+4)-2(n+4)
a. (n+4)(3n-2)

16. Solve -3(ab^4)^3

I think this should be -3(a^3)(b^12), which isn't one of your choices listed.

Use elimination to solve the system of equations. 2x+5y=7 and 3x+6y=3

a. (5,-9)
b. (-1,1)
c. (-9, 5)
d. none of these

Substitute -1,1 into 2x+5y = 7

2(-1)+5(1) = 7
-2+5 =/= 7, -2+5 = 3 no?

2x + 5y = 7
3x + 6y = 3

Multiply the first equation by 3 and the second by -2.

6x + 15y = 21
-6x - 12y = -6

Sum them and you get 3y = 15 or y = 5.

Substitute y = 5 into the second equation.

3x + 6(5) = 3
3x = -27
x = -9

The answer is (-9, 5).

You're right it's d. (1,-1)

That doesn't work either.

I'll go with C lol.

Now someone go get cracking on the calculus problem.

Thanks a lot for helping us, I need to get some sleep so I can drop him off at school in the morning.

For anyone interested in calculus, here’s the solution, so far as I can find. Can’t be sure because I made up the equation instead of grabbing one from the book.

f(x,y) = 2x² + 6(x²)(y) + y³ + 12

(f/dx) = 4x + 12xy
(f/dy) = (6x²) + 3y

(6x²) + 3y = 0
(2x²) + y = 0
y = 2x²

Substitute y into the f/dx

4x + 12x(2x²) = 0 (part I’m not sure of, I don’t recall if this is right but (12x)(2x²) = 24x³ no?
4x + 24x³ = 0
X = 0, maybe something else as well but I am not solving for a cube.

Substitute X into f/dy

6(0)^2 + 3y = 0
Y = 0

Critical point = (0,0)

Find 2nd derivatives

(f²/dx²) = 4 + 12y
(f²/dy²) = 3
(f²/dx²dy²) = 12x²

use 2nd derivative test

f(0,0) = (4)(3)-0² > 0
(f²/dx²) = 4 + 12y > 0

Thus, The critical point (0,0) is a minimum

THere's a problem with your question 16. Maybe you didn't write it out properly?

The rest.. i double checked are correct to the rest of the answers the others gave.

Jesus, I about went crazy typing those out. If anyone can answer those for me we'd appreciate it lots.

I would but I can do grade 8 algebra in my sleep mainly because it puts me to sleep. :)

What is the point of giving multiple choice on this kind of problems. In a lot of cases, it's faster to test the possible answers than it is to find the solution; in the other cases, testing the possible answers is still fast enough for all purposes (but 'solving' the question only takes a second). It's just plain stupid from the teacher to use multiple choice for these problems as the best way of answering the questions does not involve exercising the algorithms used to solve these problems.


Compare it with this problem:

factorise n = 181843587030699605093598369085019

A: (179441839, 181154117, 181154161, 30880033)
B: (179441839, 181154117, 181154161, 30880049)
C: (179441839, 181154117, 181154161, 30880097)
D: (179441839, 181154117, 181154161, 30880133)

It would be stupid to attempt to find the answer by factorising n.
A more clever way would be checking the answers by calculating their product.
An even more clever way would be checking the answers by calculating only the last digit of the product - if this isn't enough to determine a single answer - calculate the next digit too, etc.

For anyone interested in calculus, here’s the solution, so far as I can find. Can’t be sure because I made up the equation instead of grabbing one from the book.

f(x,y) = 2x² + 6(x²)(y) + y³ + 12

(f/dx) = 4x + 12xy
(f/dy) = (6x²) + 3y

(6x²) + 3y = 0
(2x²) + y = 0
y = 2x²

Substitute y into the f/dx

4x + 12x(2x²) = 0 (part I’m not sure of, I don’t recall if this is right but (12x)(2x²) = 24x³ no?
4x + 24x³ = 0
X = 0, maybe something else as well but I am not solving for a cube.

Substitute X into f/dy

6(0)^2 + 3y = 0
Y = 0

Critical point = (0,0)

Find 2nd derivatives

(f²/dx²) = 4 + 12y
(f²/dy²) = 3
(f²/dx²dy²) = 12x²

use 2nd derivative test

f(0,0) = (4)(3)-0² > 0
(f²/dx²) = 4 + 12y > 0

Thus, The critical point (0,0) is a minimum
bI don't know what you did, but I would just use implicit differentiation tbh... i'll get the answer and tell you in a couple of minutes. EDIT: you used 6x^2.y this time, last time you used 6x^2.y^2.

If the point is zero, i'm not sure this should matter, but still.

second EDIT: I used the eqn f(x,y) = 2x^2 +6x^2.y^2 +y^3 + 12.

I also got the point (0,0) as a critical point, however my eqn was so messy I decided to leave it as a critical point, I know nothing about the nature of the point, and to be honest I don't care :O.

Noone is good at algebra! ;)

Lol, junior high school algebra. I can't be fucked, but what are you using the circumflex to represent? o0

I was hoping for some parabola's or something...

For anyone interested in calculus, here’s the solution, so far as I can find. Can’t be sure because I made up the equation instead of grabbing one from the book.

f(x,y) = 2x² + 6(x²)(y) + y³ + 12

(f/dx) = 4x + 12xy
(f/dy) = (6x²) + 3y

(6x²) + 3y = 0
(2x²) + y = 0
y = 2x²

Substitute y into the f/dx

4x + 12x(2x²) = 0 (part I’m not sure of, I don’t recall if this is right but (12x)(2x²) = 24x³ no?
4x + 24x³ = 0
X = 0, maybe something else as well but I am not solving for a cube.

Substitute X into f/dy

6(0)^2 + 3y = 0
Y = 0

Critical point = (0,0)

Find 2nd derivatives

(f²/dx²) = 4 + 12y
(f²/dy²) = 3
(f²/dx²dy²) = 12x²

use 2nd derivative test

f(0,0) = (4)(3)-0² > 0
(f²/dx²) = 4 + 12y > 0

Thus, The critical point (0,0) is a minimum

error: d/dy y^3 != 3*y

my answer:

> f:=(x,y)->2*x^2 + 6*(x^2)*(y) + y^3 + 12;
2 2 3
f := (x, y) -> 2 x + 6 x y + y + 12

> G:=grad(f(x,y),[x,y]);

[ 2 2]
G := [4 x + 12 x y, 6 x + 3 y ]

> solve({G[1]=0,G[2]=0},{x,y});

{y = 0, x = 0}, {y = 0, x = 0},

2
{x = 1/3 RootOf(2 _Z + 1, label = _L2), y = -1/3}

> fxx:=diff(f(x,y),x$2);
>

fxx := 4 + 12 y

> fyy:=diff(f(x,y),y$2);

fyy := 6 y

> fxy:=diff(diff(f(x,y),x),y);

fxy := 12 x

> Delta:=(subs(x=0,y=0,fxy))^2-subs(x=0,y=0,fxx)*subs(x=0,y=0,fyy);

Delta := 0

Which basicly means that for the critical point (0,0), the test I've used cannot draw a conclusion and i would have to consider higher-order terms in my approximation.

draw the function, and see that it's probably not an extremum.

proof: take the path x = 0
g(y) = f(0,y) = y^3 + 12

we see that g(y) can be greater and smaller than g(0) in any epsilon environment of y=0, thus f(x,y) can be greater and smaller than f(0,0) in any epsilon environment of (0,0) and (0,0) is not an extremum.