Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

3y+x=3 , y^2-x=1

Why are you switching names ??

find the intersection;
from the first:
x = 3 - 3y

into the 2nd:
y^2 - (3-3y) = 1
y^2 + 3y - 4 = 0
(y+4)(y-1) = 0
y = -4 or y = 1

in x = 3-3y ....
if y = 1, x = 0 ---->(0,1)
if y = -4 , x = 15 , ---> (15, -4)

Your sketch should look like this and my answers are confirmed
http://www.wolframalpha.com/input/?i=plot+3y%2Bx%3D3+%2C+y%5E2-x%3D1

judging from the graph I would integrate with respect to y
so the effective width of my horizontal slice
= 3-3y - (y^2 - 1)
= 4 - 3y - y^2

area = ∫(4 - 3y - y^2) dy from -4 to 1
= (4y - (3/2)y^2 - (1/3)y^3) | from -4 to 1
= (4 - 3/2 - 1/3) - (-16 -24 + 64/3)
= 13/6 - (-56/3)
= 125/6

To sketch the region enclosed by the given curves, we need to determine the intersection points of the two curves and determine the boundaries of the region. Then, we can decide whether to integrate with respect to x or y based on the orientation of the curves.

First, let's solve the system of equations to find the intersection points. We have:

Equation 1: 3y + x = 3
Equation 2: y^2 - x = 1

Solving Equation 1 for x, we get:
x = 3 - 3y

Substituting this value of x in Equation 2, we have:
y^2 - (3 - 3y) = 1
y^2 - 3 + 3y = 1
y^2 + 3y - 2 = 0

Now, we can solve this quadratic equation for y:

Using the quadratic formula, y = (-3 ± √(3^2 - 4 * 1 * -2)) / (2 * 1)
Simplifying, we get: y = (-3 ± √(9 + 8)) / 2
y = (-3 ± √17) / 2

So, we have two potential intersection points:
Point 1: (x1, y1) = (3 - 3((-3 + √17) / 2), (-3 + √17) / 2)
Point 2: (x2, y2) = (3 - 3((-3 - √17) / 2), (-3 - √17) / 2)

Now, let's look at the two curves separately and determine the boundaries of the region.

Curve 1: 3y + x = 3
To determine the boundaries, we set x = 0 and solve for y:
3y + 0 = 3
y = 1

To determine the other boundary, we can set y = 0 and solve for x:
3(0) + x = 3
x = 3

Curve 2: y^2 - x = 1
To determine the boundaries, we set x = 0 and solve for y:
y^2 - 0 = 1
y = ±1

To determine the other boundary, we can set y = 0 and solve for x:
0^2 - x = 1
x = -1

Now, let's see the orientation of the curves. In this case, Curve 1 (3y + x = 3) is a straight line and Curve 2 (y^2 - x = 1) is a parabola opening downwards. The intersection points help us determine which curve is on top at different sections.

The region enclosed by the curves appears to be bounded between x = -1 and x = 3, and between y = -1 and y = 1. It is better to integrate with respect to x in this case because the boundaries are defined in terms of x.

Therefore, to find the area of the region, we need to integrate the equation (Curve 2 - Curve 1) with respect to x from x = -1 to x = 3:

Area = ∫[x=-1 to 3] [(y^2 - x) - (3y + x)] dx

By integrating this expression, we can find the value of the area of the region enclosed by the curves.