Help needed in finding missing terms in mixed fractions equation

Hans

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Hi,

I came across this question , while going through the fractions chapter.Can anyone help me in how to approach and solve the problem (1) in a better way?

5. Find out the missing figures (denoted by stars) in the following equations, the fractions being given in their lowest terms:

. . .\(\displaystyle \mbox{i) }\, *\, \dfrac{3}{7}\, \times\, 2\, \dfrac{3}{*}\, =\, 14\, \dfrac{*}{14}\)

. . .\(\displaystyle \mbox{ii) }\, 7\, \dfrac{*}{3}\, -\, *\, \dfrac{5}{11}\, =\, 3\, \dfrac{*}{*3}\)

. . .\(\displaystyle \mbox{iii) }\, 8\, \dfrac{9}{**}\, \div\, *\, \dfrac{2}{27}\, =\, 7\, \dfrac{16}{17}\)

. . .\(\displaystyle \mbox{iv) }\, 2\, \dfrac{1}{2}\, -\, 3\, \dfrac{2}{3}\, +\, 1\, \dfrac{5}{6}\, -\, \dfrac{2}{*}\, =\, 0\)

. . .\(\displaystyle \mbox{v) }\, \dfrac{1}{2\,\dfrac{3}{4}}\, +\, \dfrac{1}{5\, \dfrac{1}{5}}\, +\, \dfrac{1}{*}\, +\, \dfrac{1}{9\, \dfrac{8}{15}}\, =\, \dfrac{144}{143}\)

Edit 1:
I have started the problem by replacing the asterisk sign by variables x,y,z.And I have found the missing denominator in the mixed fraction ( 2+ 3/* ) to be 4 as (LCM of 7,4,14 will be 28).Kindly let me know whether the approach is correct or not.


Ty.
 

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Hi,

I came across this question , while going through the fractions chapter.Can anyone help me in how to approach and solve the problem (1) in a better way?

5. Find out the missing figures (denoted by stars) in the following equations, the fractions being given in their lowest terms:

. . .\(\displaystyle \mbox{i) }\, *\, \dfrac{3}{7}\, \times\, 2\, \dfrac{3}{*}\, =\, 14\, \dfrac{*}{14}\)

. . .\(\displaystyle \mbox{ii) }\, 7\, \dfrac{*}{3}\, -\, *\, \dfrac{5}{11}\, =\, 3\, \dfrac{*}{*3}\)

. . .\(\displaystyle \mbox{iii) }\, 8\, \dfrac{9}{**}\, \div\, *\, \dfrac{2}{27}\, =\, 7\, \dfrac{16}{17}\)

. . .\(\displaystyle \mbox{iv) }\, 2\, \dfrac{1}{2}\, -\, 3\, \dfrac{2}{3}\, +\, 1\, \dfrac{5}{6}\, -\, \dfrac{2}{*}\, =\, 0\)

. . .\(\displaystyle \mbox{v) }\, \dfrac{1}{2\,\dfrac{3}{4}}\, +\, \dfrac{1}{5\, \dfrac{1}{5}}\, +\, \dfrac{1}{*}\, +\, \dfrac{1}{9\, \dfrac{8}{15}}\, =\, \dfrac{144}{143}\)

Edit 1:
I have started the problem by replacing the asterisk sign by variables x,y,z.And I have found the missing denominator in the mixed fraction ( 2+ 3/* ) to be 4 as (LCM of 7,4,14 will be 28).Kindly let me know whether the approach is correct or not.


Ty.
How did you get 4? Yes, LCM (7,4,14) = 28 but why can't the common denominator be larger than 28?

I would try y=2 (if you allow numbers like 12 8/5 -otherwise try y=4, like you suggested). In the end x will depend on z or if it is more convenient, z will depend on x. That is, the answer will not be unique.
 
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Hi,

I came across this question , while going through the fractions chapter.Can anyone help me in how to approach and solve the problem (1) in a better way?

5. Find out the missing figures (denoted by stars) in the following equations, the fractions being given in their lowest terms:

. . .\(\displaystyle \mbox{i) }\, *\, \dfrac{3}{7}\, \times\, 2\, \dfrac{3}{*}\, =\, 14\, \dfrac{*}{14}\)

. . .\(\displaystyle \mbox{ii) }\, 7\, \dfrac{*}{3}\, -\, *\, \dfrac{5}{11}\, =\, 3\, \dfrac{*}{*3}\)

. . .\(\displaystyle \mbox{iii) }\, 8\, \dfrac{9}{**}\, \div\, *\, \dfrac{2}{27}\, =\, 7\, \dfrac{16}{17}\)

. . .\(\displaystyle \mbox{iv) }\, 2\, \dfrac{1}{2}\, -\, 3\, \dfrac{2}{3}\, +\, 1\, \dfrac{5}{6}\, -\, \dfrac{2}{*}\, =\, 0\)

. . .\(\displaystyle \mbox{v) }\, \dfrac{1}{2\,\dfrac{3}{4}}\, +\, \dfrac{1}{5\, \dfrac{1}{5}}\, +\, \dfrac{1}{*}\, +\, \dfrac{1}{9\, \dfrac{8}{15}}\, =\, \dfrac{144}{143}\)

Edit 1:
I have started the problem by replacing the asterisk sign by variables x,y,z.And I have found the missing denominator in the mixed fraction ( 2+ 3/* ) to be 4 as (LCM of 7,4,14 will be 28).Kindly let me know whether the approach is correct or not.


Ty.
If I am understanding the problem correctly, not quite. Using your x, y, and z, I believe what we would have for (i) is
\(\displaystyle \dfrac{7x+3}{7}\, \dfrac{2y+3}{y}\, =\, \dfrac{196+z}{14}\)
So what we would want is 7y=14 or y=2. That would then lead to
\(\displaystyle \dfrac{(7x+3)\, 7}{14}\, =\, \dfrac{196+z}{14}\) or
\(\displaystyle 49x + 21\, =\, 196+z\)
etc.
 
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If I am understanding the problem correctly, not quite. Using your x, y, and z, I believe what we would have for (i) is
\(\displaystyle \dfrac{7x+3}{7}\, \dfrac{2y+3}{y}\, =\, \dfrac{196+z}{14}\)
So what we would want is 7y=14 or y=2. That would then lead to
\(\displaystyle \dfrac{(7x+3)\, 7}{14}\, =\, \dfrac{196+z}{14}\) or
\(\displaystyle 49x + 21\, =\, 196+z\)
etc.
If you use y=2, this will work. It will also make some of the fractions improper--which works for me!

But you can also let y= 4. It turns out that you can find x and z such that no fraction is improper.

There is also no reason why y can't be other values as well. After all, we have one equation with three unknowns.
 
If you use y=2, this will work. It will also make some of the fractions improper--which works for me!

But you can also let y= 4. It turns out that you can find x and z such that no fraction is improper.

There is also no reason why y can't be other values as well. After all, we have one equation with three unknowns.
Possibly why the OP used an LCM involving 7 and 4 (and 14) and what was meant in part by "fractions being given in their lowest terms". So my bad. Off to the corner for that extra 2 minutes.

That would also change my answer to the OP's original question to "Yes, that is the way to start". To expand on the question and the "fractions being given in their lowest terms", I would think that then meant that in
\(\displaystyle \dfrac{7x+3}{7}\, \dfrac{2y+3}{y}\, =\, \dfrac{196+z}{14}\)
x, y, and z must be integers larger than 0, y must be larger than 3, and z must be smaller than 14.
 
Possibly why the OP used an LCM involving 7 and 4 (and 14) and what was meant in part by "fractions being given in their lowest terms". So my bad. Off to the corner for that extra 2 minutes.

That would also change my answer to the OP's original question to "Yes, that is the way to start". To expand on the question and the "fractions being given in their lowest terms", I would think that then meant that in
\(\displaystyle \dfrac{7x+3}{7}\, \dfrac{2y+3}{y}\, =\, \dfrac{196+z}{14}\)
x, y, and z must be integers larger than 0, y must be larger than 3, and z must be smaller than 14.

\(\displaystyle \dfrac{7x+3}{7}\, \dfrac{11}{4}\, =\, \dfrac{196+z}{14}\)

Simplifying the above equation:
(7x+3)*11=(196+Z)*2
77x+33=(196+Z)*2
As Ishuda pointed out that z must be smaller than 14, X should be smaller than 7, then by hit and trail method, the values that satisfies the equation will be x=5 and z=13.

@Ishuda can you please explain me the sentence which I quoted below, is it a property of vulgar fractions, I didn’t quite get it …“x, y, and z must be integers larger than 0, y must be larger than 3, and z must be smaller than 14.

And if there is any method to find the unknowns x,z other than the hit and trail method,please let me know.

Ty
 
\(\displaystyle \dfrac{7x+3}{7}\, \dfrac{11}{4}\, =\, \dfrac{196+z}{14}\)

Simplifying the above equation:
(7x+3)*11=(196+Z)*2
77x+33=(196+Z)*2
As Ishuda pointed out that z must be smaller than 14, X should be smaller than 7, then by hit and trail method, the values that satisfies the equation will be x=5 and z=13.

@Ishuda can you please explain me the sentence which I quoted below, is it a property of vulgar fractions, I didn’t quite get it …“x, y, and z must be integers larger than 0, y must be larger than 3, and z must be smaller than 14.

And if there is any method to find the unknowns x,z other than the hit and trail method,please let me know.

Ty
The hit and trial method (never heard it expressed that way before!) was not too difficult.

I'll try to explain what Ishuda meant (I hate explaining what others said--but students ALWAYS come first)
You want nice mixed numbers, not \(\displaystyle 3 \dfrac{.2}{5} or 13 \dfrac {23}{13} or 12 \dfrac{-2}{4}\). The 1st mixed number is not nice since it has .2 in the numerator (that is way we want y and z to be integers and surely you want x to be an integer as well-why??) The 2nd fraction is not nice since the fractional part is improper (that is why you want y>3 and z<14). The 3rd number is not nice since the numerator of the fraction of a mixed number should not be negative. Same for the denominator (that is why y and z should be positive. Personally I have no problem with x being negative.
 
\(\displaystyle \dfrac{7x+3}{7}\, \dfrac{11}{4}\, =\, \dfrac{196+z}{14}\)

Simplifying the above equation:
(7x+3)*11=(196+Z)*2
77x+33=(196+Z)*2
As Ishuda pointed out that z must be smaller than 14, X should be smaller than 7, then by hit and trail method, the values that satisfies the equation will be x=5 and z=13.

@Ishuda can you please explain me the sentence which I quoted below, is it a property of vulgar fractions, I didn’t quite get it …“x, y, and z must be integers larger than 0, y must be larger than 3, and z must be smaller than 14.

And if there is any method to find the unknowns x,z other than the hit and trail method,please let me know.

Ty
Non-trivial proper mixed fractions (mixed fractions in their lowest terms), that is something like \(\displaystyle a\frac{b}{c}\) are, by definition, composed of positive integers and, again by definition, if they are in their lowest terms, a, b and c are greater than zero and b is less than c. If a were zero, you would have a trivial case of zero for the mixed fraction; if b were zero you would have the trivial case of an integer; c can not be zero since it occurs in the denominator. If b were equal to c then a could be increased by 1 and b replaced by zero which is the integer trivial case. If b were greater than c, then a could be increased by 1 and b reduced by c (and this could be repeated until b was less than or equal to c).

As far as finding x and z so that the other conditions are satisfied, yes there is. The reasoning goes something like this: Given
77x + 33 = (196 + z)*2 = 392 + 2 z
or
77x = 359 + 2z
we can start with, since 2z is even, 77x-359 must be even which implies x is odd [if x were even, 77x-359 would be odd]. Since 77x-359 = 77(x- \(\displaystyle 4\frac{51}{77}[/tex) must be positive (that is z is positive), x must be at least 5. Since it is smaller than 7 and odd, x is 5. So y=4, x=5 and z=13.\)
 
Non-trivial proper mixed fractions (mixed fractions in their lowest terms), that is something like \(\displaystyle a\frac{b}{c}\) are, by definition, composed of positive integers and, again by definition, if they are in their lowest terms, a, b and c are greater than zero and b is less than c. If a were zero, you would have a trivial case of zero for the mixed fraction; if b were zero you would have the trivial case of an integer; c can not be zero since it occurs in the denominator. If b were equal to c then a could be increased by 1 and b replaced by zero which is the integer trivial case. If b were greater than c, then a could be increased by 1 and b reduced by c (and this could be repeated until b was less than or equal to c).

As far as finding x and z so that the other conditions are satisfied, yes there is. The reasoning goes something like this: Given
77x + 33 = (196 + z)*2 = 392 + 2 z
or
77x = 359 + 2z
we can start with, since 2z is even, 77x-359 must be even which implies x is odd [if x were even, 77x-359 would be odd]. Since 77x-359 = 77(x- \(\displaystyle 4\frac{51}{77}[/tex) must be positive (that is z is positive), x must be at least 5. Since it is smaller than 7 and odd, x is 5. So y=4, x=5 and z=13.\)
\(\displaystyle
May I ask why you do not think of numbers like \(\displaystyle -4\dfrac{2}{5}\) as a non-trivial fraction?\)
 
I am also facing the problem and I am not so sound. So please can anybody explain and solve the 1st equations for me so that I could grab th idea of doing it and think over it.
 
Hi, the way it works here is that you try to solve the problem, show us your work and we help you if you get stuck.
 
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