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Answer:
(x, y) = (3, 1)
2y +1 < x ≤ 5 -2y where y < 1
Step-by-step explanation:
a) We can add twice the second equation to the first to eliminate y.
(2x +4y) +2(x -2y) = (10) +2(1)
4x = 12 . . . . . simplify
x = 3
3 -2y = 1 . . . . substitute for x in the second equation
2 = 2y . . . . . add 2y-1 to both sides
1 = y . . . . . . divide by 2
The solution is (x, y) = (3, 1).
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b) The solution for a set of inequalities is generally the intersection of two half-planes. It is not just one point. The solution space is usually represented graphically, but we can write inequalities that express it.
The first inequality tells us that both x and y will be on or below the boundary line, which will have negative slope.
The second inequality tells us x will be above the boundary line with positive slope, and y will be below it. If we consider the boundary lines to cross in an X shape, the solution space will be in the bottom quadrant. This suggests the area can best be represented by expressing x as a range of values in terms of y.
Solving each inequality for x, we get ...
2x +4y ≤ 10 ⇒ x ≤ 5 -2y
x -2y > 1 ⇒ x > 2y +1
These give us the compound inequality ...
2y +1 < x ≤ 5 -2y
Of course, the relation between the limits must hold, so ...
2y +1 < 5 -2y
4y < 4 . . . . . . . add 2y-1
y < 1 . . . . . . divide by 4
The solution is 2y +1 < x ≤ 5 -2y where y < 1.
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The graph shows the region that is the solution to the system of inequalities. The vertex point (3, 1) is not part of that region. (It does not satisfy the left boundary condition.) However, that point is the solution to the set of equations. It is where the boundary lines meet.
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Additional comment
Above, we noted that the inequalities have boundary lines that intersect to divide the Cartesian plane into 4 parts. We called the intersecting lines an X shape. In general, if the solution space is in the left- or right-portion of this shape, the compound inequality defining the solution will be one that bounds y. If it is in the top or bottom of this shape, then the compound inequality will be one that bounds x (as in this problem).
A different generality may apply if the boundary lines have slopes of the same sign.
