Don't think too hard on this one

Algebra Level 4

If the two system of equations

{ a x + b y = 1 3 x y = 2 \begin{cases} ax+by=1\\ 3x-y=2\\ \end{cases}

{ x + 2 y = 1 a x b y = 2 \begin{cases} x+2y=1\\ ax-by=-2\\ \end{cases}

has the same solution ( x , y ) (x,y) , then the value of a + b a+b is m n \frac{m}{n} for relatively prime m m and n n . Find m + n m+n .


The answer is 54.

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2 solutions

since both set of equations have same solutions as stated. so lets solve 3x-y=2 and x+2y=1 first so that we get x=5/7 and y=1/7. After that solve ax+by=1 and ax=by=-2 by replacing x and y values obtained previously. Then after we get a=-7/10 and b=21/2. Now finally stated a+b=m/n so that (-7/10+21/2=49/5). So that final answer m+n=49+5 = 54....

Used the same method.

Niranjan Khanderia - 6 years, 9 months ago

Great question @Zhou ZeHao

Jayakumar Krishnan - 6 years, 9 months ago

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But I didnt understand how we had to solve the equations with coefficients before the ones without (with algebraic) them. Also, does it means all four equations have the same solution?? If yes, then good, else how?

Jayakumar Krishnan - 6 years, 9 months ago

What if a=(6) and b= -(2/3) ? We get a+b = 16/3 Hence m+n = 19?

Aayush Todi - 6 years, 9 months ago
Esha Aslam
Sep 29, 2014

first solve 3x-y=1 & x+2y= 1 S.S =(5/7 ,1/7)

now add ax+by=1 & ax-by= -2
=> 2ax= -1 , ax=-1/2 , a(5/7)= -1/2 => a= - 7/10

now subtract ax+by = 1 & ax-by = -2
=> 2by =3 => by =3/2 , b(1/7)= 3/2 => b =21/10

a+b= - 7/10 +21/10 = 98/10 = 49/5 m= 49 n= 5
m+n = 49+5 =54

i also used the same method

Parv Maurya - 6 years, 6 months ago

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