True or True #10
True or False: If a , b , c , d is a real number and a x + b = − c x − d has a solution such that 0 < c < a then x < − a + c b + d
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1 solution
I don't think that you can manipulate inequalities like that. For example, If A and B are non-negative, real-numbered variables then both A ≥ 0 and B ≥ 0 , but there can be instances where A − B is not necessarily positive, (e.g., when A = 2 and B = 3 ).
The answer is still false, however. I used the counterexample 2 x + 3 = − x . Then a = 2 , b = 3 , c = 1 , d = 0 , which makes − a + c ( b + d ) = − 1 , but the solution is actually − 1 , which is not less than itself.
a typo.. it should be x(a+c)+(b+d) > or = 0...
a x + b = − c x − d has a solution if and only if a x + b > o r = 0 ( 1 ) and − c x − d > o r = 0 ( 2 ) subtract equation 2 from 1 x ( a + b ) + ( b + d ) > o r = 0 So, x > o r = − ( b + d ) / ( a + c ) Therefore, the statement is FALSE