Math help

bjbj1991 · January 20, 2006

I need to find a faster method of solving these tipe of problems because they normaly take about 20 minuts to solve and take half a page

solve for A,B,C,D

A+20=4*B

B/2=D

C=3*A

D*C=5

DONT SOLVE THE PROBLEM I GAVE YOU! I WAS LOOKING FOR A GENRAL METHOD TO SOLVE THAT FORM OF A PROBLEM.

sorry to everyone who tried to solve it.

Phyllarski · January 20, 2006

I don't believe there is an answer to that

I've been trying many substitutions

Here is what I got, correct me if i got an error

A + 20 = 4B

B/2 = D

C = 3A

DC = 5

B = 2D

C = 3A = 5/D

3AD = 5

AD = 5/3

3/C X D = 5/3

3 X DC = 5/3 X C

SINCE DC = 5 AS GIVEN

3 X 5 = 5/3 X C

5/3 X C = 15

5C = 45

C = 9

Using that result

D = 5/9

B = 18/5 and A is negative which isnt possible

Phyllarski · January 20, 2006

Working another way I get the quadratic

c^2 - 2c - 120 = 0

Hence (c+10)(c=12) = 0

c = -10 or 12

Again doesn't work

runesmithie · January 20, 2006

I cheated and used my calculator

a= (-2(sqrt(255)+15))/3

b=-(sqrt(255)-15)/6

c=-2(sqrt(255)+15)

d=-(sqrt(255)-15)/12

or

a= 2(sqrt(255)-15)/3

b=(sqrt(255)+15)/6

c=2(sqrt(255)-15)

d=(sqrt(255)+15)/12

sizzlorr1 · January 20, 2006

AD = 5/3

3/C X D = 5/3

3 X DC = 5/3 X C

That parts wrong I'm fraid, you put A= 3/C on the second line, where it's actually A = C/3... which cancels it all out and gives DC = 5 again... I'm working through this now, and have so far got nothing!

sizzlorr1 · January 20, 2006

OK, I've got an answer!!!

A+20=4B {the given equations}

B/2=D

C=3A

DC=5

B = 2D {from B/2 = D}

D = 5/C {from DC = 5}

A+20 = 4B {given equation}

A+20 = 8D {substitute B for 2D}

A+20 = 40/C {substitute D for 5/C}

A+20 = 40/3A {substitute C for 3A}

3(A^2+20A)=3(40) {multiply by 3A}

3(A^2+20A-40) = 0 {minus 3(40) from both sides}

3A^2+60A-120 = 0 {multiply out the 3}

-b(+-)(root)(b^2-4ac) / 2a {quadratic equation}

-60(+-)(root)(3600+480) / 6 {substitute in the values}

A = 0.645812... or a negative, which we'll ignore

If A = 0.645:

C = 3A = 3*0.645 = 1.935

D = 5/C = 5/1.935 = 2.584

B = 2D = 2*2.584 = 5.168

If you put those values:

A = 0.645

B = 5.168

C = 1.935

D = 2.584

into the first equations, it all works... Well, it gives almost the correct answers, due to rounding from the quadratic equation above... Wow, that was kinda complicated :)

And to answer your initial question... I don't believe there is a simpler method!

mad4u689 · January 20, 2006

A+20=4*B

B/2=D

C=3*A

D*C=5

I solved (I think :D) it with two different methods of substitution and both came down to the quadratic equation, and all will, unfortunately, because of the way its set up with three equations defining them as multiples of each other, and the last equation involving an addition with an extra term of 20.

Anyway, this seems like a messier problem than they'd be likely to give you - are you sure it's right? Mmm.

C=3A so D*3A=5 so D=5/(3A) so B/2=5/(3A) so B=5/(6A) SO

A+20 = 4*(5/(6A)) = 10/(3A)

3A^2 + 60A - 10 = 0

Then you can realize it's not solvable by the guess'n'check method, so you complete the square (or use quadratic equation.)

A^2 + 20A -10/3 = 0

A^2 + 20 A = 10/3

(A+10)^2 = 10/3 + 100 = 44/3

(A+10) = + or - (44/3)^(1/2)

A = + or - (44/3)^(1/2) - 10

Then C = 3A so C = + or - (3(44/3)^(1/2)) - 30

and since D*C = 5, D = 5/C, so

D= 5 /{+ or - [3(44/3)^(1/2)] - 30}

So, another way of solving it.

A = 4B-20

C = 12B-60

12BD-60D=5

B=2D

24D^2 - 60D - 5 = 0

D^2 - (5/2)D - 5/24 = 0

D^2 - (5/2)D = 5/24

(D-(5/4))^2 = 5/24 + (25/16) = 85/48

D - (5/4) = + or - (85/48)^(1/2)

D = + or - (85/48)^(1/2) + (5/4)

Since these two solutions don't appear to jive, it means I've done something wrong somewhere and you can't rely on anything I've said :D Erm... sorry :x

wahoo · January 20, 2006

OK, I've got an answer!!!

A+20=4B {the given equations}

B/2=D

C=3A

DC=5

B = 2D {from B/2 = D}

D = 5/C {from DC = 5}

A+20 = 4B {given equation}

A+20 = 8D {substitute B for 2D}

A+20 = 40/C {substitute D for 5/C}

A+20 = 40/3A {substitute C for 3A}

3(A^2+20A)=3(40) {multiply by 3A}

3(A^2+20A-40) = 0 {minus 3(40) from both sides}

3A^2+60A-120 = 0 {multiply out the 3}

-b(+-)(root)(b^2-4ac) / 2a {quadratic equation}

-60(+-)(root)(3600+480) / 6 {substitute in the values}

A = 0.645812... or a negative, which we'll ignore

If A = 0.645:

C = 3A = 3*0.645 = 1.935

D = 5/C = 5/1.935 = 2.584

B = 2D = 2*2.584 = 5.168

If you put those values:

A = 0.645

B = 5.168

C = 1.935

D = 2.584

into the first equations, it all works... Well, it gives almost the correct answers, due to rounding from the quadratic equation above... Wow, that was kinda complicated :)

And to answer your initial question... I don't believe there is a simpler method!

not quite right, you made the mistake of saying you were multiplying by 3a, but on the right hand side you multiplied 3a but then 3 again?. It should be 3A^2+60A-40=0 shouldn't it?

Duke_Freedom · January 20, 2006

sizzlorr1 got it right..

Exact answers are:

A = -10 + 2/3 root(255)

B = 5/2 + 1/6 root(255)

C = -30 + 2 root(255)

D = 5/4 + 1/12 root(255)

A = -10 - 2/3 root(255)

B = 5/2 - 1/6 root(255)

C = -30 - 2 root(255)

D = 5/4 - 1/12 root(255)

Both are possible.

vladmoney · January 20, 2006

A+20=4B

B=2D <--changed a bit

C=3A

Plug In: (1/3)C+20=8D --> C=24D-60 --> DC=5 --> 24D^2 - 60D=5

Just solve the quadratic for D, and then its all just plug'n'chug. Pretty simple :? , probably just did something wrong :oops: .

bjbj1991 · January 20, 2006

i ment that type of problem not that one problem. I dont think an answer to that one even exists.

runesmithie · January 20, 2006

i ment that type of problem not that one problem. I dont think an answer to that one even exists.

*points at the last....10? posts*

It exists all right :P

I think I have claim to first answer :D (But I used my calculator)

EDIT: As for how to solve, get each variable isolated in the equations and substitute in like this:

2A=4B

C^2=B/2

2C=.6476

Get your variables out

A=2B

B=2(C^2)

C=.3373

Then substitute

A=2(2(C^2)) --> A=2(2(.3373^2)) --> A=.455

B=2(.3373^2) --> B=.2275

C=.3373

bjbj1991 · January 21, 2006

ok well thank you. lol i hate higher maths i guess i doomed myself :lol:

vladmoney · January 21, 2006

ok well thank you. lol i hate higher maths i guess i doomed myself :lol:

As long as there are as many forumals as there are variables (like in this one there are 4 variables [A,B,C,D] and there are 4 formulas with all the variables represented) than the problem is solvable.

Math help

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