Find the area bound by curves.

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.
2y=3x
y=5
and
2y+4x=7

Help please!!

Region enclosed by curves:
http://www.wolframalpha.com/input/?i=plo…

We find points of intersection:

Lines 2y=3x and y=5 --------> (10/3, 5)
Lines y=5 and 2y+4x=7 -----> (-3/4, 5)
Lines 2y=3x and 2y+4x=7 --> (1, 3/2)

Note that since this is a triangle with
base = distance from (-3/4, 5) to (10/3, 5) = 49/12 and
height = distance from point (1, 3/2) to line (y=5) = 7/2
then area = 1/2 (49/12) (7/2) = 343/48 = 7.1458333...

If you were to differentiate with respect to x, then you'd integrate from -3/4 to 10/3
But on interval -3/4 < x < 1, y is between lines y = 5 and 2y+4x = 7
and on interval 1 < x < 10/3, y is between lines y = 5 and 2y = 3x
This would require 2 integrals.

So we integrate with respect to y, from y = 3/2 to y = 5
On this interval, x is between lines 2y+4x = 7 and 2y = 3x

We solve these equations for x, so we can express x as a function of y:
2y + 4x = 7 ------> x = -1/2 y + 7/4
2y = 3x -----------> x = 2/3 y

A = ∫{y = 3/2 to 5} (2/3 y - (-1/2 y + 7/4)) dy
A = ∫{y = 3/2 to 5} (7/6 y - 7/4) dy
A = (7/12 y^2 - 7/4 y) |{y = 3/2 to 5}
A = (175/12 - 35/4) - (63/48 - 21/8)
A = 343/48 = 7.1458333...

I will refer to the equations as Y1, Y2, and Y3, as giver in order above. First use simple algebra or a graphing calculator to sketch the graph. It consists of a triangle. You could solve this using geometry but the question asks to integrate. Start by integrating Y2-Y3 from -1 to 1 and then add that answer to the integral of Y2-Y1 from 1 to 3.5. This covers the whole area of the triangle. The limits were determined by intersections of the lines. Make sure you graph so what I said makes sense. Good luck!