How do you subtract #\frac { 4r } { 5r + 3} - \frac { 2} { 4r }#?

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Answer 1 · 2017-09-23T08:22:32

#\frac { 16r^2 - 10r - 6 } { 20r^2 + 12r# ----you can stop here

Or factorise further to get #\frac {( 8 r + 3) (r - 1) } { 2r(5r^2 +3 )}#

#\frac { 4r } { 5r + 3} - \frac { 2} { 4r }#

making denominators equal

= #\frac { (4r)(4r) } {( 5r + 3)(4r)} - \frac { 2(5r +3)} { (4r)(5r + 3) }#

= #\frac { (16r^2) } {( 5r + 3)(4r)} - \frac { (10r +6)} { (4r)(5r + 3) }#

= #\frac { (16r^2) - (10r + 6) } {( 5r + 3)(4r)} #

=#\frac { 16r^2 - 10r - 6 } { 20r^2 + 12r# ----you can stop here

Or factorise further

=#\frac { 2(8r^2 - 5r - 3) } { 2(10r^2 +6r )}#

=#\frac { (8r^2 - 5r - 3) } { (10r^2 +6r )}#

=#\frac { (8r^2 - 8r + 3r - 3) } { 2r(5r^2 +3 )}#

=#\frac { 8 r (r - 1) + 3 (r - 1) } { 2r(5r^2 +3 )}#

=#\frac {( 8 r + 3) (r - 1) } { 2r(5r^2 +3 )}#