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Solve for d.
5d+2(2-d)=3(1+d)+1
a)no solution
b)d=4
c)infinitely many solutions
d)d=-3


Sagot :

dln
[tex]5d+2(2-d)=3(1+d)+1 \\ 5d+4-2d=3+3d+1 \\ 3d+4=3d+4 \\ 0=0 \\ \boxed{C.\ Infinitely\ many\ solutions}[/tex]
5d + 2(2 - d) = 3(1 + d) + 1
5d + 2(2) - 2(d) = 3(1) + 3(d) + 1
5d + 4 - 2d = 3 + 3d + 1
5d - 2d + 4 = 3d + 3 + 1
3d + 4 = 3d + 4
-3d       -3d       
       4 = 4
       d = 0

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