abs(16+t)=2t-3|16+t|=2t−3
abs(x) |x|is distance from the origin
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(16+t)=2t-3 or -(16t+t)=2t-3(16+t)=2t−3or−(16t+t)=2t−3
Take (16+t)=2t-3 (16+t)=2t−3
16+3=2t-t16+3=2t−t
19=t19=t
t=19t=19
Take (16+t)=-(2t-3)(16+t)=−(2t−3)
16+t=-2t+316+t=−2t+3
16-3=-2t-t16−3=−2t−t
13=-3t13=−3t
t=-13/3t=−133
plug t=19plugt=19 in the original equation
abs(16+19)=2(19)-3|16+19|=2(19)−3
abs(35)=35|35|=35
35=3535=35
So t=19 satisfies the original equation.
Put t=-13/3 in the original equation
abs(16-(13/3))=2(-13/3)-3∣∣∣16−(133)∣∣∣=2(−133)−3
abs ((48-13)/3)=-(26/3)-3∣∣∣48−133∣∣∣=−(263)−3
abs(35/3)=(-26-9)/3∣∣∣353∣∣∣=−26−93
35/3=-35/3353=−353Left and right hand side are not same
so t=-13/3 should not satisfies the original equation
so it is extraneous solution.
t=19
