Frenchfries13
Frenchfries13 Frenchfries13
  • 21-09-2017
  • Mathematics
contestada

How many solutions does the equation have?

4(z−5)+2=4z−18

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doinmybest
doinmybest doinmybest
  • 21-09-2017
Distribute the 4:
4z - 20 + 2 = 4z - 18
Subtract 4z
-20 + 2 = -18
-18 = -18
This is true so that means there are infinite number of solutions... z could be anything!
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Аноним Аноним
  • 21-09-2017
All real numbers is the solution of this equation

4z - 20 +2 = 4z - 18

4z - 18  = 4z - 18
      +18         +18

4z = 4z 


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