(x+3):(x11)=(x-2):(x+1) find vvalue of x
Answers
As you can see that the difficulty arises due to the terms |x-2| and |x-3|. So, we’ll solve it by simplifying our problem and breaking it down for various cases of x. Though this particular problem has an easy solution but I’m showing you the method that you can apply even when there are inequalities instead of equal signs or even if there are more than two modulus terms
R=(−∞,2)∪(2,3)∪(3,∞)∪{2,3}
Hence we have-
⋆Case-1–––––––x∈(−∞,2)Hence in this interval |x-2|<0 and |x-3|<0Thus the equation becomes-⟹−(x−2)−(x−3)=1⟹−x+2−x+3=1⟹−2x=−4⟹x=2which is not possible as x<2Hence no possible solution from this interval⋆Case-2–––––––x∈(2,3)Hence in this interval |x-2|>0 and |x-3|<0Thus the equation becomes-⟹(x−2)−(x−3)=1⟹x−2−x+3=1⟹1=1which is trueHence every x belonging to this interval is a solution⋆Case-3–––––––x∈(3,∞)Hence in this interval |x-2|>0 and |x-3|>0Thus the equation becomes-⟹(x−2)+(x−3)=1⟹x−2+x−3=1⟹2x=6⟹x=3which is not possible as x>3Hence no possible solution from this interval⋆Case-4–––––––x∈{2,3}Let first x=2, then |x-2|=0 and|x-3|=|2-3|=1Thus the equation becomes-⟹0+1=0which is trueHence one possible solution is x=2Now let x=3, then |x-2|=|3-2|=1 and|x-3|=0Thus the equation becomes-⟹1+0=0which is trueHence one possible solution is x=3
Hence possible solutions are x∈[2,3]