A fraction with two variables

[window789456]

How do I find a correct 'a' and 'b' combination so the result will be the same as the output above?


There is an easier one and it got solved:

so after many inputs of trial and errors, I got the a = 1 and b = 2. But this one seem too easy out of luck.

Can anyone solve the '3/10' above?

 

[Cubist]

You posted this in the arithmetic section, therefore I'll assume that you don't know any algebra (please tell me if this is wrong).

Are you comfortable with negative numbers? If so, then it might help you to know that there IS an answer. Just be systematic in your trials! Please post back if still stuck.

 

[JeffM]

You have two unknowns. You need two equations. The numerator gives one; the denominator gives the other.

[MATH]a + b = 3 \text { and } a + 2b = 5 \\ \therefore a = 3 - b \implies 3 - b + 2b = 5 \implies \\ 2b - b + 3 = 5 \implies b + 3 = 5 \implies b + 3 - 3 = 5 - 3 \implies\\ b = 2 \implies a = 3 - 2 = 1.[/MATH]Make sense?

No guessing is needed with algebra.

Now see how far you can get on the harder one.

 

[Deleted member 4993]

View attachment 23237

How do I find a correct 'a' and 'b' combination so the result will be the same as the output above?


There is an easier one and it got solved:
View attachment 23238
so after many inputs of trial and errors, I got the a = 1 and b = 2. But this one seem too easy out of luck.

Can anyone solve the '3/10' above?

What are the constraints on 'a' and/or 'b' - rational and/or integers and/or positive, etc. ?

 

[HallsofIvy]

From \(\displaystyle \frac{a+ b}{a+ 2b}= \frac{3}{5}\)
we get \(\displaystyle 5(a+ b)= 3(a+ 2b)\)
\(\displaystyle 5a+ 5b= 3a+ 6b\)
\(\displaystyle 2a= b\).

So a= 1, b= 2 is one solution but there are infinitely many solutions!
Choose any number at all for a then b is twice that number.

If a=2 then b= 4: \(\displaystyle \frac{a+ b}{a+ 2b}= \frac{2+ 4}{2+ 8}= \frac{6}{8}= \frac{3}{5}\).

If a= -5 then b= -10: \(\displaystyle \frac{a+ b}{a+ 2b}= \frac{-5-10}{-5- 20}= \frac{-15}{-25}= \frac{3}{5}\).

As Suubhotosh Kahn JeffM (response #3) showed, you can get a= 1, b= 2 setting the the numerators equal and the denominators equal. But that does not have to be true for two fractions to be equal. (.....edited - corrected attribution)

 

[Steven G]

a + b = 3
a+ 2b = a+b + b = (a+b) + b = 10. So b =7

a+b =3, a+7 = 3 so b a =-4. [edited]