How do you solve for #r# in #r/n+s=t#?

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Answer 1 · 2018-03-07T10:11:09

#r=nt-ns#

The algebraic equation we have is:

#r/n+s=t#

We can solve this by:

1) Subtract #s# from each side

#->color(blue)r/n+scolor(red)(-s)=tcolor(red)(-s)#

#->color(blue)r/n=t-s#

2) Multiply each side by #n#

#->color(blue)r/n color(red)(*n) = (t-s)color(red)(*n)#

#->color(blue)r=(t-s)n#

#color(blue)r=nt-ns#

Thus, we have the answer:

#r=nt-ns#