How do you find the domain and range of #(5x-3) / (2x +1)#?

1 Answer
May 31, 2017

Answer:

#x inRR,x!=-1/2#
#y inRR,y!=5/2#

Explanation:

#"the domain is defined for all real values of x except for"#
#"values of x which make the denominator equal zero"#

#"to find the value that x cannot be, equate the denominator"#
#"to zero and solve"#

#"solve "2x+1=0rArrx=-1/2larrcolor(red)" excluded value"#

#rArr"domain is "x inRR,x!=-1/2#

#"to find any excluded values in the range, rearrange"#

#y=(5x-3)/(2x+1)" making x the subject"#

#rArry(2x+1)=5x-3larrcolor(blue)" cross-multiplying"#

#rArr2xy+y=5x-3#

#rArr2xy-5x=-3-y#

#rArrx(2y-5)=-(3+y)#

#rArrx=-(3+y)/(2y-5)#

#"the denominator cannot equal zero"#

#"solve "2y-5=0rArry=5/2larrcolor(red)" excluded value"#

#rArr"range is " y inRR,y!=5/2#