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June 11th, 2015, 03:14 PM   #1
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Simultaneous very hard problem

Can someone help me solve this simultaneous equation? We are to find a and b so that the three equations have a solution in common.

Thanx!
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 June 11th, 2015, 04:48 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,635 Thanks: 2620 Math Focus: Mainly analysis and algebra By writing $a$ and $b$ in terms of $x$ using the first two equations, you can find an equation (or a pair of equations) that give three possible values of $x$. You can use these values in the first two equations to find $a$ and $b$. My solutions are $(a,b) \in \big\{(0,0), (3,4), (0,-2)\big\}$
 June 11th, 2015, 07:59 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,469 Thanks: 2038 Adding the first and third equations gives 2x² + (b - 2)x = 0, so x = 0 or 1 - b/2. x = 0 implies a = b = 0. x = 1 - b/2 implies 1 - b + b²/4 - 3 + 3b/2 - b = 0, i.e. (b - 4)(b + 2)/4 = 0. It's easy to finish from there.
 June 13th, 2015, 08:15 PM #4 Senior Member   Joined: May 2015 From: Varanasi Posts: 110 Thanks: 5 Math Focus: Calculus I dont see there is a very big problem in that, there are 3 unknowns and 3 equations though it will need some manipulation.You must write your attempt.
June 14th, 2015, 04:49 AM   #5
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Quote:
 Originally Posted by manishsqrt I dont see there is a very big problem in that, there are 3 unknowns and 3 equations though it will need some manipulation.You must write your attempt.
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