1. 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 ?

2. 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.

3. 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

4. puzzled

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.
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

5. 4cd=14

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???

6. 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 $\displaystyle 4cd=14$

Then, the second part is ambiguous.

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

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

7. 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 $\displaystyle 4cd=14$

Then, the second part is ambiguous.

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

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

8. 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.

$\displaystyle 4cd=14$

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

Substituting, we get

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

$\displaystyle 2d=14$

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

$\displaystyle d=7$

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

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

$\displaystyle 4(3.5)=14$

$\displaystyle 14=14$

9. Originally Posted by masters
Ok, Leona, I think you intend to state the problem this way.

$\displaystyle 4cd=14$

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

Substituting, we get

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

$\displaystyle 2d=14$

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

$\displaystyle d=7$

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

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

$\displaystyle 4(3.5)=14$

$\displaystyle 14=14$
Thanks p.s how do you give thanks

10. Originally Posted by Leona_Marie
Thanks p.s how do you give thanks
Click on the "Thanks" button below the post.