3D - 2

3(1+d)-5=3d-2?

solve step by step

Public Comments

1. 3(1+d)-5=3d-2 3+3d-5=3d-2 3d-5=3d-5 -3 -3 d-5=-5 +5 +5 d=0
2. The rule is to apply the same function to both sides of the equtation. The only caveat is that the function applied must have an inverse, "that which you do, you must be able to undo". Here we go.... 3(1+d)-5=3d-2 apply the distributive property of mulitplication over addition to the left side of the equtation. 3*1 + 3d - 5 = 3d - 2 add -3d to both sides. (to undo, just add 3d to both sides.) 3*1 + 3d - 3d - 5 = 3d - 3d - 2 do some arithmetic, noting that 3d - 3d = 0 and 0 is a special number called the additive identity (meaning it can disappear) 3*1 - 5 = -2 -2 = -2 this is called a tautology. It is always true. Since the "d" factor dropped out, the equation is true for *every value* of d. Lets try a few 3(1+d)-5 = 3d - 2 substitute 42 for d (my age in earth years) 3(1+42)-5 = 3*42 - 2 3+42*3-5 = 3*42 - 2 124 = 124 and on and on and on. Truth rings eternal....
3. 3 (1 + d) - 5 = 3d - 2 3 + 3d - 5 = 3d - 2 3d + (-5 + 3) = 3d -2 3d - 2 = 3d - 2 3d = 3d 3/3 d = 3/3 d d = d d can be any number
4. 3 + 3d - 5 = 3d - 2 ; 3d - 3d = - 3 + 5 - 2 ; 0 = 0 . :-)