algebraic expressions

Printable View

January 18th 2009, 06:21 PM
Leona_Marie

algebraic expressions

What is the proper way to work this probelm...4cd =14...c=1/2d=7
What is the right way to wok this problem ?
January 19th 2009, 03:17 AM
darkness9375

Quote:

Originally Posted by Leona_Marie

What is the proper way to work this probelm...4cd =14...c=1/2d=7
What is the right way to wok this problem ?

the way i would solve it is by just replacing what we all ready know.

4(c)(d)=14 c=1 d=7/2=3.5

4(1)(3.5)=14

hope that helps.(Itwasntme)
January 19th 2009, 04:35 AM
ursa

Quote:

What is the proper way to work this probelm...4cd =14...c=1/2d=7
What is the right way to wok this problem ?

yup darkness9375 is right
there are 1 equation and 2 unknown variable
we have to assume 1 variable to get the value of another variable
January 19th 2009, 07:14 AM
Leona_Marie

puzzled

Quote:

Originally Posted by darkness9375

the way i would solve it is by just replacing what we all ready know.

4(c)(d)=14 c=1 d=7/2=3.5

4(1)(3.5)=14

hope that helps.(Itwasntme)

could you break it down to me like...
4(c)(d)
4(1)(d)
4(1)(7/2)
How did you get 1/2 to =1 and why are you x 7/2 and afterwards how are you x3.5 to get =14 I thought 4x7 is 28 -1/2=14
January 19th 2009, 08:58 AM
Leona_Marie

4cd=14

Quote:

Originally Posted by ursa

yup darkness9375 is right
there are 1 equation and 2 unknown variable
we have to assume 1 variable to get the value of another variable

Ursa when using frac, with algebra...1st equation is 4 sec is 1/2 which is(c)third is7 or(d). So 1/2 x7/1=1x7and2x1=7/2. 2divided by 7=3 1/2=3.5...3.5x4=14 yes???
January 19th 2009, 09:15 AM
masters

Quote:

Originally Posted by Leona_Marie

What is the proper way to work this probelm...4cd =14...c=1/2d=7
What is the right way to wok this problem ?

Hi Leona,

I'm not sure that you have stated your question properly. Try and be a little clearer. Use parentheses, separate lines, anything to set apart what it is you are trying to convey.

Looks like your original equation is $4cd=14$

Then, the second part is ambiguous.

Dose $c=\frac{1}{2}d$ and does $\frac{1}{2}d=7$ ?

If that's the case, then substitute 7 for c and solve for d. But I'm not sure that's the case.
January 19th 2009, 09:43 AM
Leona_Marie

Quote:

Originally Posted by masters

Hi Leona,

I'm not sure that you have stated your question properly. Try and be a little clearer. Use parentheses, separate lines, anything to set apart what it is you are trying to convey.

Looks like your original equation is $4cd=14$

Then, the second part is ambiguous.

Dose $c=\frac{1}{2}d$ and does $\frac{1}{2}d=7$ ?

If that's the case, then substitute 7 for c and solve for d. But I'm not sure that's the case.

4cd is the equation Im not sure how to use (c) which =1/2 to get the answer =14 Im guessing you x (c) which=1/2 x (d) which =7 . I just thought that you would use multiplying fractions ...1/2 x7/1 =3.5. 3.5 x 4 =14 4cd =14 please varify or change leona
January 19th 2009, 09:56 AM
masters

Quote:

Originally Posted by Leona_Marie

4cd=14 is the equation Im not sure how to use (c) which =1/2 to get the answer =14 Im guessing you x (c) which=1/2 x (d) which =7 . I just thought that you would use multiplying fractions ...1/2 x7/1 =3.5. 3.5 x 4 =14 4cd =14 please varify or change leona

Ok, Leona, I think you intend to state the problem this way.

$4cd=14$

$c=\frac{1}{2}$

Substituting, we get

$4\left(\frac{1}{2}\right)d=14$

$2d=14$

$\frac{2d}{2}=\frac{14}{2}$

$d=7$

Checking, we substitute the values we found for c and d back into the original equation and get:

$4\left(\frac{1}{2}\right)(7)=14$

$4(3.5)=14$

$14=14$
January 19th 2009, 10:40 AM
Leona_Marie

Quote:

Originally Posted by masters

Ok, Leona, I think you intend to state the problem this way.

$4cd=14$

$c=\frac{1}{2}$

Substituting, we get

$4\left(\frac{1}{2}\right)d=14$

$2d=14$

$\frac{2d}{2}=\frac{14}{2}$

$d=7$

Checking, we substitute the values we found for c and d back into the original equation and get:

$4\left(\frac{1}{2}\right)(7)=14$

$4(3.5)=14$

$14=14$

Thanks p.s how do you give thanks
January 19th 2009, 10:48 AM
masters

Quote:

Originally Posted by Leona_Marie

Thanks p.s how do you give thanks

Click on the "Thanks" button below the post.