How do you solve for a in #5 + ab = c#?

3 Answers
May 26, 2018

Answer:

#a=(c-5)/b#

Explanation:

#"isolate "ab" by subtracting 5 from both sides"#

#cancel(5)cancel(-5)+ab=c-5#

#ab=c-5#

#"divide both sides by "b#

#(a cancel(b))/cancel(b)=(c-5)/brArra=(c-5)/b#

May 26, 2018

Answer:

#a = (c - 5)/b #

Explanation:

As there are 3 variables in this equation, i.e #a#, #b# and #c#.

Therefore, the value of #a# would be in terms of #b# and #c#.

Now, as per the question, we have

#5 + ab = c#

#:. ab = c - 5#

#:. a = (c - 5)/b#

Hence the value of #a# in terms of #b# and #c# is #(c - 5)/b#.

May 26, 2018

#color(blue)(5+ab=c#

To, find a, we need to isolate it in the left hand side of the equation.

We can do that by adding or subtracting or multiplying or dividing the same number on both sides of the equation

So, subtract #5# from both sides

#rarrcancel5+ab-cancel5=c-5#

#rarrab=c-5#

Divide both sides by #b#

#rarr(acancelb)/cancelb=(c-5)/b#

#color(green)(rArra=(c-5)/b#

Hope that helps.... #phi#