How do you solve #-\frac { 2} { 3} a + 2a = \frac { 6} { 5} q + \frac { 6} { 5}#?

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How do you solve #-\frac { 2} { 3} a + 2a = \frac { 6} { 5} q + \frac { 6} { 5}#?

↳Redirected from "Suppose that I don't have a formula for #g(x)# but I know that #g(1) = 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"

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Answer 1 · 2018-01-20T07:12:35

#q=10/9a-1#
or
#a=9/10 * (q+1)#

Explanation:

#4/3a=6/5 * (q+1)#

#20/18a=q+1#

#10/9a = q+1#

q=10/9a-1

or

#q+1=10/9a#
#a=9/10 * (q+1)#

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How do you solve #-\frac { 2} { 3} a + 2a = \frac { 6} { 5} q + \frac { 6} { 5}#?

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