Do I need to determine their values first?

Algebra Level 3

$\large 2a^2 + 3b^2 = 7ab, \ \ \ \ \ 0 < a <b$

If $a,b$ satisfy the conditions given above, and that $\frac{a-3b}{a+3b}$ can be expressed as $-\frac qs$ for coprime positive integers $q,s$ , find $q+s$ .

This problem is not original.

The answer is 12.

2 solutions

Rafid Bin Mostofa
Nov 8, 2014

If 2a^2+3b^2=7ab then (a-3b)=0 or (2a-b)=0.but if( a-3b) happens to be 0;the fraction never happens.So you have to go for 2a-b=0;b=2a.Input the value of b in the fraction & you will get this: (a-6a)/(a+6a)=-5a/7a=-5/7 .So then q=5,s=7 & q+s=12

$2a^{2} + 3b^{2} + 6ab + ab$

$a(2a - b) - 3b(2a - b) = 0$

$(a - 3b)(2a - b) =0$

$a -3b\neq 0$ $2a = b , \frac{a}{b} = \frac{1}{2}$

$\frac{\frac{a}{b} - 3}{\frac{a}{b} + 3} = \frac{-5}{7}$

@Rafid Bin Mostofa very nice voted up

U Z - 6 years, 7 months ago

Another reason NOT to take $(a-3b)=0$ would be that we would get $0\lt b\lt a$ when we take values $a,b\gt 0$ . But it is clearly stated that $0\lt a\lt b$ in the problem, so we go with the second possibility that $(2a-b)=0$

Prasun Biswas - 6 years, 6 months ago

U Z
Nov 8, 2014

$2a^{2} + 3b^{2} = 7$

$\frac{2a}{b} + \frac{3b}{a} = 7$

$\frac{a}{b} = x$

$2x^{2} - 7x + 3 =0$

$x = \frac{1}{2} , 3$

Since $b > a$ thus $\frac{a}{b} = \frac{1}{2}$

$\frac{a - 3b}{a + 3b} = \frac{\frac{a}{b} - 3}{\frac{a}{b} + 3}$

$= \frac{ -5}{7}$

I also done it the same way!!!!!

Piyush Maheshwari - 6 years, 7 months ago

See Rafid's approach a clever and beautiful approach , i think he is a new genius emerging out!

U Z - 6 years, 7 months ago

Yeah! I agree you. That's the genius way to approach the problem.

Piyush Maheshwari - 6 years, 7 months ago

First line is wrong. Right side should be 7ab.

谦艺伍 - 4 years, 9 months ago

Do I need to determine their values first?

This problem is not original.

The answer is 12.

2 solutions

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