How do you solve #sqrt(2y + 7) + 4 = y#?

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Answer 1 · 2018-03-18T11:24:50

#y=9,y!=1#

#sqrt(2y+7)+4=y#

#:.(sqrt(2y+7))^2=(y-4)^2#

#:.2y+7=y^2-8y+16#

#:.2y+8y-y^2=16-7#

#:.10y-y^2=9#

#:.-y^2+10y-9=0#

#:.(-y+9)(y-1)=0#

#:.-y+9=0,y-1=0#

#:.-y=-9,y=1#

#:.y=9,y=1#

~~~~~~~~~~~~~

check:-

substitute #y=9#

#:.sqrt(2(9)+7)+4=(9)#

#:.sqrt(18+7)+4=9#

#:.sqrt25+4=9#

#:.5+4=9#

#:.9=9#

substitute y=1

#:.sqrt(2(1)+7)+4=(1)#

#:.sqrt(2+7)+4=1#

#:.3+4=1#

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