To solve f'(x) = 7x+5....(1) and

x+3y = 6 .....................(2).

Solution:

We presume y = f(x).

Then from the 2nd equation x+3y =6, we get:

3y = 6-x

y =(6-x)/3

y = 2-(1/3)x.

y' = {(2-(1/3)x}' = (2)' -(1/3)(x)' = 0 -1/3.

y' = -1/3

So f'(x) = -1/3.

Since f'(x) = 7x+5,

-1/3 = 7x+5

-1/3 -5 = 7x.

-(1+15)/3 = 7x. Divide by 7.

-16/21 = x

Or

x = -16/21.

From the second equation,

x+3y =6,

-16/21+3y = 6.

3y = 6+16/21 = (126+16)/21

y = 142/21*3

y = 142/63.

f’(x)= 7x+ 5 , x+3y =6

1. 7x + 5 = y

2. x + 3y = 6

Multiply 2. by 7:

7x + 21y = 42

Subtract number 1. from this

7x + 21y = 42

7x - y = -5

22y = 47

y = 47/22

Substitute back into one of the equations.

x+3y=6 3y=6-x y=f(x)=1/3*(6-x)=2-1/3x

y'=f'(x)=2'-1/3x'=0-1/3*1=-1/3

we have f'(x)=7x+5

-1/3=7x+5

7x=-1/3-5

7x=-16/3

x=-16/3*1/7=-16/21