Question

4(z-5)+2=4z-18 how many solutions does the equation have?

Answer 1

Answer by Mimiwhatsup:

`4\left(z-5\right)+2=4z-18\\4\left(z-5\right)+2\\4\left(z-5\right)\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b-c\right)=ab-ac\\a=4,\:b=z,\:c=5\\=4z-4\cdot \:5\\\mathrm{Multiply\:the\:numbers:}\:4\cdot \:5=20\\=4z-20\\=4z-20+2\\\mathrm{Add/Subtract\:the\:numbers:}\:-20+2=-18\\=4z-18\\4z-18=4z-18\\\mathrm{Add\:}18\mathrm{\:to\:both\:sides}\\4z-18+18=4z-18+18\\Simplify\\4z=4z\\\mathrm{Subtract\:}4z\mathrm{\:from\:both\:sides}\\4z-4z=4z-4z\\Simplify\\0=0\\`

`Both\:sides\:are\:equal\\True\:for\:all\:z`

Answer 2

4z-20+2=4z-18
4z-18=4z-18
Their answer is the same no answer for this .. or maybe you forgot to wrote an other thing... I hope that I could help you!! :)