How do you combine #(x-1)/x-(t+1)/t#?

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Answer 1 · 2016-07-10T15:35:02

# -(t+x)/(tx)#

To combine them you must have a denominator that is divisible by both #t# and #x#. The most obvious one it #tx#

Consider #(x-1)/x#

Multiply by 1 but in the form of #1=t/t#

#(x-1)/x xx t/t = (t(x-1))/(tx) = (tx-t)/(tx)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Consider #(t+1)/t#

Multiply by 1 but in the form of #1=x/x#

#(t+1)/x xx x/x=(x(t+1))/(tx) = (tx+x)/(tx)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Putting it all together

#(tx-t)/(tx) - (tx+x)/(tx)" " =" " (cancel(tx)-t-cancel(tx)-x)/(tx)#

#(-t-x)/(tx)" " =" " (-(t+x))/(tx)" " =" " -(t+x)/(tx)#