How do you find the domain and range of #(3x+2)/(4x-5)#?

1 Answer
Aug 25, 2017

#x inRR,x!=5/4#
#y inRR,y!=3/4#

Explanation:

#"for "y=(3x+2)/(4x-5)#

The denominator cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the value that x cannot be.

#"solve "4x-5=0rArrx=5/4larrcolor(red)" excluded value"#

#rArr"domain is "x inRR,x!=5/4#

#"to find any excluded values in the range rearrange making"#
#"x the subject"#

#rArry(4x-5)=3x+2larrcolor(blue)" cross-multiplying"#

#rArr4xy-5y=3x+2#

#rArr4xy-3x=2+5ylarrcolor(blue)" collect terms in x"#

#rArrx(4y-3)=2+5ylarrcolor(blue)" factor out x"#

#rArrx=(2+5y)/(4y-3)#

#"the denominator cannot equal zero"#

#"solve "4y-3=0rArry=3/4larrcolor(red)" excluded value"#

#rArr"range is "y inRR,y!=3/4#