Please show me how to solve and what is the mane of this problem

Moving -3/2 to the other side of the equals sign changes the -ve sign to a +ve 3/2

(5/3)d = -1/2+3/2

(5/3)d = 1

d = 1/(5/3) = 3/5

Please show me how to solve and what is the mane of this problem

Tutors, sign in to answer this question.

Moving -3/2 to the other side of the equals sign changes the -ve sign to a +ve 3/2

(5/3)d = -1/2+3/2

(5/3)d = 1

d = 1/(5/3) = 3/5

Fractions are difficult for most people. A fraction is essentially a division problem. This question is relatively easy

(5/3)d-3/2=-1/2

First we must combine like terms; add 3/2 to both sides of the equation.

(5/3)d -3/2 + 3/2= -1/2 +3/2

(5/3)d= 2/2 or (5/3)d=1

now we must eliminate the fraction; we do this by multiplying 3 to both sides.

3(5/3)d= 1(3)

(5)d=3

now divide by 5

5d/5= 3/5

d= 3/5

Check your answer

5/3(3/5)-3/2= -1/2

15/15 -3/2 = -1/2

1-3/2= -1/2 (true)

d= 3/5 is correct.

d = 5/3 because 3(5/3d) = 1(3) and (5/d) = 3 so 5 = 3d and d = 5/3

We are saying the exact same thing. I solved the problem using the law of fractions while you gave a more algebraic solution. I did not disagree with your solution because it was also a viable way to solve the problem. I solved the problem using algebra, but proved the solution using simple mathematics.

James M. | James in Los AngelesJames in Los Angeles

Sorry Benjamin,

You do have a problem at step 3. 5/3(d) = 1 * 3 does not lead to 5d = 3.

It leads to 5 / d = 3.

Cancelling the 3 on the bottom does not bring the d to the top of the fraction.

Next you would multiply both sides by 1/5.

(1/5) 5 / d = 3 (1/5) which gives:

1 / d = 3 / 5

Then you are allowed to flip, or to invert both sides to get d, not 1 over d.

d = 5 / 3.

This is a syntax error. (5/3)d not 5/3d

Multiply both side of the equation with 2 then 2[(5/3d)-(3/2)=(-1/2)]

and you get (10/3d)-3 = -1

rearrange terms to (10/3d)= -1+3 and (10/3d)= 2

cross multiply and you get 6d = 10 and d = 10/6 = 5/3

Check answer by substituting d into the original equation [(5/3)x(3/5)]-(3/2)=(-1/2)

and it becomes 1-(3/2)=(-1/2) and -1/2 = -1/2or -1/2+1/2=0

and the solution checks for d=5/3

Your math is correct, but you are assuming that the student understands why each step you propose is needed. The object is not to simply solve the problem for the student, but to allow them to solve similar problems themselves. When you decided that my answer was incorrect what did you base it on? I combined terms adding 3/2 to both sides of the equation. this gave me an equation 5/3d=1

I did not need to cross multiply (which confuses some students). In math it is important that their is only one solution, not only one way to reach the solution. the more ways a student can learn to solve a problem the more likely they will be to retain the skill.

Already have an account? Log in

By signing up, I agree to Wyzant’s terms of use and privacy policy.

Or

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Your Facebook email address is associated with a Wyzant tutor account. Please use a different email address to create a new student account.

Good news! It looks like you already have an account registered with the email address **you provided**.

It looks like this is your first time here. Welcome!

To present the tutors that are the best fit for you, we’ll need your ZIP code.

Please try again, our system had a problem processing your request.

## Comments

Gentlemen:

I think we have to determine whether the problem statement is (5/3d) or (5/3)d. If it is the first then the solution for d is 5/3. If it is the latter then the solution for d is 3/5. I assumed that 5/3d is actually (5/3d) where d is in the denominator of the fraction. However, if it was in the numerator of the fraction such as (5/3)d or (5d/3) it would have been the reverse of my answer or d=3/5. Let us put an end to this argument and stress i=on the importance of presenting the problem statement accurately with proper paranthesis. For next time. we should require more information from the posting student to prevent ambiguity in understanding and solving the problem.