forming and solving a quadratic equation

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LEONA123

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Hi I'm a year 9 student however this question below is from my tutor and im not sure what year level it is. I have learnt only 1 way to solve quadratic equations and its the method when A and B multiply to get 0.

Question -
The numerator of a certain fraction is 3 less than its denominator. If 6 is added to the numerator and 5 to the denominator, the value of the fraction is doubled. Find
the fraction.

Someone told me that x is what you are trying to find so in this case it's the fraction. I've tried x is the numerator. And x is the denominator but I'm stuck. I've only been taught this a few days ago and this is my first time trying to form and solve a quadratic equation.
Help is appreciated.
Thank you!
 
LEONA123 said:
Hi I'm a year 9 student however this question below is from my tutor and im not sure what year level it is. I have learnt only 1 way to solve quadratic equations and its the method when A and B multiply to get 0.

Question -
The numerator of a certain fraction is 3 less than its denominator. If 6 is added to the numerator and 5 to the denominator, the value of the fraction is doubled. Find
the fraction.

Someone told me that x is what you are trying to find so in this case it's the fraction. I've tried x is the numerator. And x is the denominator but I'm stuck. I've only been taught this a few days ago and this is my first time trying to form and solve a quadratic equation.
Help is appreciated.
Thank you!
Click to expand...

Assume that the numerator = N

and the denominator = D

Fraction = F = N/D

The numerator of a certain fraction is 3 less than its denominator → N = D-3 → F = (D-3)/D .......................................(1)

f 6 is added to the numerator and 5 to the denominator, the value of the fraction is doubled → [(D-3)+6]/[D+5] = 2*F = 2*(D-3)/D

[(D-3)+6]/[D+5] = 2*(D-3)/D

D * [D + 3] = 2*[D-3]*[D+5]...... [edited]

Above will give you a quadratic equation → solve for D → use equation and calculate F


 
Last edited by a moderator:
Subhotosh Khan said:
Assume that the numerator = N

and the denominator = D

Fraction = F = N/D

The numerator of a certain fraction is 3 less than its denominator → N = D-3 → F = (D-3)/D .......................................(1)

f 6 is added to the numerator and 5 to the denominator, the value of the fraction is doubled → [(D-3)+6]/[D+5] = 2*F = 2*(D-3)/D

[(D-3)+6]/[D+5] = 2*(D-3)/D

D * [D + 6] = 2*[D-3]*[D+5]

Above will give you a quadratic equation → solve for D → use equation and calculate F
Click to expand...

I understand until
D * [D + 6] = 2*[D-3]*[D+5]

How do you go from this
[(D-3)+6]/[D+5] = 2*(D-3)/D

to this
D * [D + 6] = 2*[D-3]*[D+5]

I understand that you cross multiplied but doesn't the two have to be multiplied at well? or is "2*(D-3)/D" one term?
Sorry if I am confusing you.

Thank you for helping!
 
LEONA123 said:
I understand until
D * [D + 6] = 2*[D-3]*[D+5]

How do you go from this
[(D-3)+6]/[D+5] = 2*(D-3)/D

to this
D * [D +3] = 2*[D-3]*[D+5] ... [edited]

I understand that you cross multiplied but doesn't the two have to be multiplied at well? or is "2*(D-3)/D" one term?
Sorry if I am confusing you.

Thank you for helping!
Click to expand...

Yes ... and it is there ....

Go through the steps with paper and pencil - don't just stare at the screen....
 
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