# x + y = 1. f'(x) = 3x + 2. Which are the values of x, y.

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We'll note f(x) = y and we'll express y from the relation x + y = 1.

We'll isolate y to the left side. For this reason, we'll subtract x both sides:

y = 1 - x

We'll differentiate with respect to x;

dy = -1*dx

But f'(x) = -1 and, from enunciation, f'(x) = 3x + 2.

So,

3x + 2 = -1

We'll subtract 2 both sides:

3x = -1-2

3x = -3

We'll divide by 3:

x = -3/3

**x = -1**

We'll substitute x in the first relation and we'll calculate y:

x+y = 1

-1 + y = 1

y = 1+1

**y = 2**

x+y = 1............(1)

f'(x) = 3x+2.......................(2)

To find the value of x and y.

Solution:

We assume f(x) = y.

So from eq (1),

x+y = 1

y = 1-x

Differentiating, we get:

dy/dx = f'(x) = (1-x)' = -1....(3)

But from f'(x) = 3x+2.

Therefore from (3) and (2) , we get: 3x+2 = -1

3x = -1-2

3x = -3

x = -3/3 = -1.

So x = -1 and y = 1-x = 1- - 1 = 2.

So x = -1 and y = 2 .

x+y=1 y=f(x)=1-x

f'(x)=(1-x)'=1'-x'=0-1=1

We have f'(x)=3x+2

1=3x+2

3x=1-2

3x=-1

x=-1/3

y=1-x=1-(-1/3)=1+1/3=4/3