find an algebraic expression equivalento to....

eddy2017

eddy2017

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Hi, I am not understanding well this problem. I have to find an equivalent solution.
I know that two fractions are called equivalent if they have the same numeric value.
I solve the addition by cross multiplying,

[math]y * 2y =2y^2[/math][math]3 * x = 3x[/math]So, I know the right choice has to be D because the rest does not have 3x, but I am not getting why the denominator is 2xy??. I don't understand how to get that denominator.
thanks for any help,

1645239430399.png
 
eddy2017 said:
Hi, I am not understanding well this problem. I have to find an equivalent solution.
I know that two fractions are called equivalent if they have the same numeric value.
I solve the addition by cross multiplying,

[math]y * 2y =2y^2[/math][math]3 * x = 3x[/math]So, I know the right choice has to be D because the rest does not have 3x, but I am not getting why the denominator is 2xy??. I don't understand how to get that denominator.
thanks for any help,

View attachment 31197
Click to expand...
How would you add:

(1/2) + (2/7) = .........?

Follow the same procedure - step-by-step
 
This is a somewhat different take from that of the Khan of Khans.

There are equations that are true for all elements in a set. These equivalences are sometimes called identities and shown with [imath]\equiv[/imath] rather than [imath]=[/imath]. What is interesting about them is reasoning frontwards is no better in logical terms than reasoning backwards if they are true equivalences.

I just got finished telling you not to work backwards from the proposed answers in a multiple choice you are unless desperate. It will almost always waste time. Furthermore, SK gave you the intuitive answer for those with some mathematical experience. However, your experience is not extensive enough for you just to “see” it. In this case, however, looking at the proposed answers gives you a clue. Each proposed answer has xy in the denominator. You have two fractions, one with x in the denominator and one with y in the denominator. How do you get y in the denominator of the first and x in the denominator of the second?

You take advantage of two general principles that come up all the time.

[math]a \equiv a * 1\\ b \ne 0 \implies \dfrac{b}{b} \equiv 1\\ \therefore a \equiv a * 1 \equiv a * \dfrac{b}{b} \equiv \dfrac{ab}{b}.[/math]
Here a common feature of the answers gives you a clue. Frequently, math problems are like a mystery novel: you are the detective looking for clues.
 
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\(\displaystyle \frac{y}{x} + \frac{3}{2y}\)

\(\displaystyle =\frac{2y^2}{2xy} + \frac{3x}{2xy}\) getting a common denominator

\(\displaystyle =\frac{2y^2+3x}{2xy}\) just like \(\displaystyle \frac{2}{7} + \frac{3}{7} = \frac{5}{7}\)

When you add fractions the common denominator remains.

I think you are using the "cross-multiplying" concept incorrectly. What you are really doing in the first step is obtaining a common denominator.

When two fractions are equal eg \(\displaystyle \frac{y}{x} = \frac{3}{2y}\) then you can conclude that \(\displaystyle y*2y = 3*x\) by cross-multiplying.
 
Harry_the_cat said:
\(\displaystyle \frac{y}{x} + \frac{3}{2y}\)

\(\displaystyle =\frac{2y^2}{2xy} + \frac{3x}{2xy}\) getting a common denominator

\(\displaystyle =\frac{2y^2+3x}{2xy}\) just like \(\displaystyle \frac{2}{7} + \frac{3}{7} = \frac{5}{7}\)

When you add fractions the common denominator remains.

I think you are using the "cross-multiplying" concept incorrectly. What you are really doing in the first step is obtaining a common denominator.

When two fractions are equal eg \(\displaystyle \frac{y}{x} = \frac{3}{2y}\) then you can conclude that \(\displaystyle y*2y = 3*x\) by cross-multiplying.
Click to expand...
Good meow to you.

I pedagogically hate cross multiplication. Obviously

[math]\dfrac{a}{b} = \dfrac{p}{q} \implies b * \dfrac{a}{b} = b * \dfrac{p}{q} \implies a = \dfrac{bp}{q} \implies q * a = q * \dfrac{p}{q} \implies aq = bp.[/math]
That of course shows why the short-cut of “cross multiplication” works, but if the student does not firmly grasp that logic, the “short-cut” frequently causes errors.
 
Subhotosh Khan said:
How would you add:

(1/2) + (2/7) = .........?

Follow the same procedure - step-by-step
Click to expand...
I find a common denominator to 2 and 7 which 14. Got it. Have to follow the same procedure
Let's give it a try.
I posted this late last night. I saw your responses around 5am this morning. Allow me some time to study all your info and I'll get back to you with my work. Thanks to all.
 
Harry_the_cat said:
\(\displaystyle \frac{y}{x} + \frac{3}{2y}\)

\(\displaystyle =\frac{2y^2}{2xy} + \frac{3x}{2xy}\) getting a common denominator

\(\displaystyle =\frac{2y^2+3x}{2xy}\) just like \(\displaystyle \frac{2}{7} + \frac{3}{7} = \frac{5}{7}\)

When you add fractions the common denominator remains.

I think you are using the "cross-multiplying" concept incorrectly. What you are really doing in the first step is obtaining a common denominator.

When two fractions are equal eg \(\displaystyle \frac{y}{x} = \frac{3}{2y}\) then you can conclude that \(\displaystyle y*2y = 3*x\) by cross-multiplying.
Click to expand...
I went the easy road or at least what for me was the easy way. Harry the cat's way.
I saw that a common denominator to both fractions was"
CD=2xy
then I cross-multiplied
[math]y*2y =2y^2 and 3*x =3x=[/math]so,
[math](2y^2 +3x )/2xy[/math]
Now, I kind of felt my way through this like jeff said taking a look at the answer cohices
what I wish i was certain about:
how to get x as part of the common denominator. fail on that!.

SK said:
How would you add:

(1/2) + (2/7) = .........?

Follow the same procedure - step-by-step

That's is more or less what HTC advised . how to get x in the denominator. I know it because it was given to me but not because I know the operation that put the x in the denominator.
Jeff's way; it went over my head. I know it is a a good principle, but first I need to know how to apply.
My doubt in applying the principles is what variable represents a ( I suppose x, but not sure)
what variable represents b ( i suppose y, but i am not sure)
and overall how to go back to the to fractions and apply it.
 
Finding the cd is the easiest.
Denominators

___ +____= ____
x 2y 2xy
 
If the denominators are x and 2y then the common denominator has to be/ no other option but 2xy.
 
JeffM said:
Good meow to you.

I pedagogically hate cross multiplication. Obviously

[math]\dfrac{a}{b} = \dfrac{p}{q} \implies b * \dfrac{a}{b} = b * \dfrac{p}{q} \implies a = \dfrac{bp}{q} \implies q * a = q * \dfrac{p}{q} \implies aq = bp.[/math]
That of course shows why the short-cut of “cross multiplication” works, but if the student does not firmly grasp that logic, the “short-cut” frequently causes errors.
Click to expand...
Thank you, Jeff. I would love to get to know how to apply this principle to the problem. I have an idea but do not know to apply that.
 
Subhotosh Khan said:
How would you add:

(1/2) + (2/7) = .........?

Follow the same procedure - step-by-step
Click to expand...
I' ll follow the same procedure as you have advised and youvwill see where I am stuck.
To add
1/2 + 2/7
Now I find a CD =14
Now, I do this
1/2(7/7)=7/14
2/7(2/2)=4/14
7/14 + 4/14=11/14
This is how I would add up the example you gave me. Then you asked me to follow the same procedure.

Ok, I'll try to follow it.
y/x + 3/2y
CD = 2xy
How to multiply the CD as a fancy form of one and get the same denominator 2xy.
I tried to do it and I am not getting the same CD.
 
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I do believe that you should know how to solve this problem.
However, if on an exam you feel that you can't solve this problem, then work backwards by choosing numbers for x and y.

Suppose x=y=1. Then y/x + 3/(2y) = 1/1 + 3/2 = 1 + 3/2 = 5/2.

Now check to see which choices give 5/2 when x=y=1
A) 4/2=2 wrong
B) 5/2 maybe
C) 5/2 maybe
D) 5/2 maybe
So only A is ruled out.

Now let x=2 and y=1. Then y/x + 3/(2y) = 1/2 + 3/2 = 2
B) 5/4 wrong
C) 5/4 wrong
D) So D must be correct, but evaluating yields 8/4=2
 
Steven G said:
I do believe that you should know how to solve this problem.
However, if on an exam you feel that you can't solve this problem, then work backwards by choosing numbers for x and y.

Suppose x=y=1. Then y/x + 3/(2y) = 1/1 + 3/2 = 1 + 3/2 = 5/2.

Now check to see which choices give 5/2 when x=y=1
A) 4/2=2 wrong
B) 5/2 maybe
C) 5/2 maybe
D) 5/2 maybe
So only A is ruled out.

Now let x=2 and y=1. Then y/x + 3/(2y) = 1/2 + 3/2 = 2
B) 5/4 wrong
C) 5/4 wrong
D) So D must be correct, but evaluating yields 8/4=2
Click to expand...
Great. Thanks, Steve. Let me study this and try to apply it to solve. Will get back to you.
 
eddy2017 said:
Thank you, Jeff. I would love to get to know how to apply this principle to the problem. I have an idea but do not know to apply that.
Click to expand...
Let’s straighten things out.

First, I presume you are talking about my post # 3 rather than my post # 5, which explains when and why cross multiplication is valid. That post is not about how to solve your problem.

Second, instead of memorizing specific procedures, rely on general principles if you want to develop your mathematical muscles.

Here is the specific problem. You are asked which fraction is equivalent to the sum of two other fractions. All three fractions differ in their denominators. NO COMMON DENOMINATOR. However, you notice that the denominator of the single fraction is the product of the denominators of the fractions being added together.

Consider the first fraction being added, y / x. You want to change the denominator from x to 2xy without changing the value of the fraction so lets see how you do that using basic principles. Our first logical step is to multiply the fraction by one.

[math]\dfrac{y}{x} \equiv \dfrac{x}{y} * 1 \text { BECAUSE } a = a * 1 \text { for ANY number } a.[/math]
Furthermore

[math]\dfrac{2y}{2y} = 1 \text { BECAUSE } \dfrac{b}{b} = 1 \text { for ANY number } b \ne 0.[/math]
[math]\therefore \dfrac{y}{x} = \dfrac{y}{x} * 1 = \dfrac{y}{x} * \dfrac{2y}{2y} = \dfrac{2y^2}{2xy}.[/math]
When you are taught arithmetic, you are taught a seemingly weird way rule to find a common denominator for two fractions. Make the product of the numerator of the first and the denominator of the second the numerator of a new first fraction. Then make the product of the numerator of the second and the denominator of the first the numerator of a new second fraction. The denominator of both new fractions will be the product of the original denominators. They may show you a little cross to help you remember (a cross that will then confuse you when you get to cross-multiplication of ratios). But they never explain why they are doing it or why it works.

It is a necessary way to learn for 8 year olds. But it is horrible for adults. The whole process is based on very simple, very obvious facts about numbers in general. If you understand those general principles, you will never have any difficulty remembering the rules of arithmetic because you will understand their logic. Algebra strengthens arithmetic just as calculus strengthens algebra.

EDIT: The logic of a number of procedures starts by multiplying by one or by adding zero because those operations leave values unchanged.
 
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Harry_the_cat said:
\(\displaystyle \frac{y}{x} + \frac{3}{2y}\)

\(\displaystyle =\frac{2y^2}{2xy} + \frac{3x}{2xy}\) getting a common denominator

\(\displaystyle =\frac{2y^2+3x}{2xy}\) just like \(\displaystyle \frac{2}{7} + \frac{3}{7} = \frac{5}{7}\)

When you add fractions the common denominator remains.

I think you are using the "cross-multiplying" concept incorrectly. What you are really doing in the first step is obtaining a common denominator.

When two fractions are equal eg \(\displaystyle \frac{y}{x} = \frac{3}{2y}\) then you can conclude that \(\displaystyle y*2y = 3*x\) by cross-multiplying.
Click to expand...
Excellent!. Yes, it wa
JeffM said:
Let’s straighten things out.

First, I presume you are talking about my post # 3 rather than my post # 5, which explains when and why cross multiplication is valid. That post is not about how to solve your problem.

Second, instead of memorizing specific procedures, rely on general principles if you want to develop your mathematical muscles.

Here is the specific problem. You are asked which fraction is equivalent to the sum of two other fractions. All three fractions differ in their denominators. NO COMMON DENOMINATOR. However, you notice that the denominator of the single fraction is the product of the denominators of the fractions being added together.

Consider the first fraction being added, y / x. You want to change the denominator from x to 2xy without changing the value of the fraction so lets see how you do that using basic principles. Our first logical step is to multiply the fraction by one.

[math]\dfrac{y}{x} \equiv \dfrac{x}{y} * 1 \text { BECAUSE } a = a * 1 \text { for ANY number } a.[/math]
Furthermore

[math]\dfrac{2y}{2y} = 1 \text { BECAUSE } \dfrac{b}{b} = 1 \text { for ANY number } b \ne 0.[/math]
[math]\therefore \dfrac{y}{x} = \dfrac{y}{x} * 1 = \dfrac{y}{x} * \dfrac{2y}{2y} = \dfrac{2y^2}{2xy}.[/math]
When you are taught arithmetic, you are taught a seemingly weird way rule to find a common denominator for two fractions. Make the product of the numerator of the first and the denominator of the second the numerator of a new first fraction. Then make the product of the numerator of the second and the denominator of the first the numerator of a new second fraction. The denominator of both new fractions will be the product of the original denominators. They may show you a little cross to help you remember (a cross that will then confuse you when you get to cross-multiplication of ratios). But they never explain why they are doing it or why it works.

It is a necessary way to learn for 8 year olds. But it is horrible for adults. The whole process is based on very simple, very obvious facts about numbers in general. If you understand those general principles, you will never have any difficulty remembering the rules of arithmetic because you will understand their logic. Algebra strengthens arithmetic just as calculus strengthens algebra.

EDIT: The logic of a number of procedures starts by multiplying by one or by adding zero because those operations leave values unchanged.
Click to expand...
Excellent. And yes, it was # 3 post I was talking about.
 
eddy2017 said:
Excellent!. Yes, it wa

Excellent. And yes, it was # 3 post I was talking about.
Click to expand...
I studied what you have written. I parsed every bit of info down. Still, I fail to see its application to the solution of the problem!. I do not see it at all.

1. I do not understand what effect has multiplying for 1 in our problem,.How does it help to see the corect solution
2. 2y/2y Where dos that come from:
Because b/b=1 for any number b not equal to 0 does not tell me anything. I do not see how it relates or applies to the solution.

I am full of thanks, to you and every one I read it, i divided into segments, still I see it as kind of arbitrary because i do not see how that helps me solve the equation.
i know you ca not give real examples, like, you multiply times 1 because when you do that the fraction because this and that....I am not expecting that but to someone new to a principal a detailed example and a reason why we're using this and here take a look this is how it is done, without that there is no help at all. That is a principle in pedagogy which I know I can ask for here.
Teach,
Explain
give an example
and then give an exercise to be done where what has bee taught has to be put in practice.
it is difficult without that understand mathematical principles.
I repeat, I am full of thanks!.
 
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eddy2017 said:
[math]y*2y =2y^2 and 3*x =3x=[/math]so,
[math](2y^2 +3x )/2xy[/math]
Click to expand...

That needs the extra grouping symbols, such as \(\displaystyle \ (2y^2 +3x )/(2xy) \), because of the Order of Operations
when you write this out in horizontal style.
 
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