Is X^2-Y^2=0 a quadratic equation ?
2 answers:
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3. First you should identify which line corresponds to which equation.
x + 2y = -8 ⇒ y = -x/2 - 4
is the line with slope -1/2 and intercept (0, -4), so the first inequality refers to the lower line in the graph.
For the inequality itself, the solution set that satisfies it is either the region above or below it. To decide which, pick any point in the plane and plug its coordinates into the inequality. The origin is a natural choice, since it's not on the line and working with 0s is easy.
In the first inequality, we have
x = 0 and y = 0 ⇒ 0 + 2•0 = 0 < -8
which is not true. So (0, 0) is not in the solution set, and it stands to reason that any point in the same region will not be a solution. In other words, the region above the line x + 2y = -8 does not satisfy the inequality, while the one below it does.
In the second inequality,
x = 0 and y = 0 ⇒ 2•0 + 0 = 0 ≥ -1
which is true. This means the region containing (0, 0) above the upper line solves the second inequality.
The solution to the system of inequalities is then the intersection of these two regions. (See attached; I've labeled the solution to the first inequality in blue, the solution to the second one in red, and the solution to both in green.)
All this work is kind of overkill for this particular question, though. All you really need to do is check if the given point satisfies the inequalities:
• (-5, 5) :
x = -5 and y = 5 ⇒ -5 + 2•5 = 5 < -8 ⇒ NO
• (2, -5) :
x = 2 and y = -5 ⇒ -2 + 2•5 = 8 < -8 ⇒ NO
• (-4, -6) :
x = -4 and y = -6 ⇒ -4 + 2•(-6) = -16 < -8 ⇒ MAYBE
x = -4 and y = -6 ⇒ 2•(-4) + (-6) = -14 < -8 ⇒ YES
• (5, -7) :
x = 5 and y = -7 ⇒ 5 + 2•(-7) = -9 < -8 ⇒ MAYBE
x = 5 and y = -7 ⇒ 2•5 + (-7) = 3 < -8 ⇒ NO
4. x is the number of candy boxes with chocolates and y is the number of boxes without chocolates. Each of the x boxes cost $27.50, so if you have x boxes, their total cost is $27.5x. Each of the y boxes cost $25.00, so y boxes cost $25y. The sponsor doesn't want to order more than 100 boxes total, so
x + y < 100
but wants to raise at least $2000, so
27.5x + 25y ≤ 2000
Now do the same thing as before; plug in the listed x- and y-coordinates and pick the point that satisfies both inequalities. You will find that (50, 30) is the correct choice.
5. Consult the "overkill" part of problem 3. The line y = -3x + 6 has a negative slope, so it's the downward sloping one. Check if the origin satisfies the inequality:
x = 0 and y = 0 ⇒ 0 ≤ -3•0 + 6 ⇒ 0 ≤ 6 ⇒ YES
This means Regions II and III solve the first inequality.
Do the same with the other inequality:
x = 0 and y = 0 ⇒ 0 ≥ 1/2•0 + 1 ⇒ 0 ≥ 1 ⇒ NO
This tells us that Regions I and II solve the second inequality.
The intersection of these regions is of course Region II.
6. One of the plotted lines is apparently y = x + 4, which has a positive slope, so this must be the upper line. The shaded region below it corresponds to the solution of either y < x + 4 or y ≤ x + 4 and we eliminate B.
The other line has negative slope, which eliminates A and D since
3x - 4y = 20 ⇒ y = 3/4x - 5
has positive slope. This leaves D.
