How do you add, simplify and state the domain of #(t^2s)/(rs)+(rs^2)/(rt)#?

Question

Julia Adams · Accepted Answer

#{t^3+s^2r}/{tr}# You need to use "Lowest Common Denominator": #{t^2color(green)(s)}/{rcolor(green)(s)}+{color(red)(r)s^2}/{color(red)(r)t}=?# First step, if #s!=0# and #r!=0# you may cancel #color(green)(s)# and #color(red)(r)# #=># #{t^2cancel(color(green)(s))}/{rcancel(color(green)(s))}+{cancel(color(red)(r))s^2}/{cancel(color(red)(r))t}={t^2}/{r}+{s^2}/{t}=?# Now we can see that #rt# is the lowest common denominator: #{t^2color(orange)(t)}/{rcolor(orange)(t)}+{s^2color(orange)(r)}/{tcolor(orange)(r)}={t^3+s^2r}/{tr}#

Eva Adamson · Answer

undefined To add, simplify, and state the domain of the expression (t^2s)/(rs) + (rs^2)/(rt), we first need to find a common denominator. The common denominator in this case is rst.

Next, we can rewrite the expression with the common denominator: (t^2s * t) / (rs * t) + (rs^2 * s) / (rt * s).

Simplifying further, we get (t^3s) / (rst^2) + (rs^3) / (rst^2).

To add these fractions, we combine the numerators over the common denominator: (t^3s + rs^3) / (rst^2).

Finally, we state the domain of the expression. The domain is the set of all real numbers that make the expression defined. In this case, the expression is defined for all values of t, r, and s except when the denominator becomes zero. Therefore, the domain is t ≠ 0, r ≠ 0, and s ≠ 0.