Problem: | x + 3 | – | 4 - x | = | 7 + x | how many solutions will this equation have ?
Rationale: you have | x + 3 | - | 4 - x | = | 8 + x | first , look at the three values independently of their absolute value sign , in other words : | x + 3 | - | 4 - x | = | 8 + x | ( x + 3 ) - ( 4 - x ) = ( 8 + x ) now , you ' re looking at x < - 8 , s ố x is a number less than - 8 . let ' s pretend x = - 10 here to make things a bit easier to understand . when x = - 10 i . ) ( x + 3 ) ( - 10 + 3 ) ( - 7 ) ii . ) ( 4 - x ) ( 4 - [ - 10 ] ) ( double negative , s ố í t becomes positive ) ( 4 + 10 ) ( 14 ) iii . ) ( 8 + x ) ( 8 + - 10 ) ( - 2 ) in other words , when x < - 8 , ( x + 3 ) and ( 8 + x ) are negative . to solve problems like this , we need to check for the sign change . here is how i do it step by step . i . ) | x + 3 | - | 4 - x | = | 8 + x | ii . ) ignore absolute value signs ( for now ) and find the values of x which make ( x + 3 ) , ( 4 - x ) and ( 8 + x ) = to zero as follows : ( x + 3 ) x = - 3 ( - 3 + 3 ) = 0 ( 4 - x ) x = 4 ( 4 - 4 ) = 0 ( 8 + x ) x = - 8 ( 8 + - 8 ) = 1 c
annotated_formula: divide(multiply(add(4, 3), const_2), 7)