Answer
The solution set includes all real numbers \( x \) and \( y \) where:
- For \( -3 < x < 0 \): \( -\frac{x}{3} - 1 < y < \frac{2}{3}x + 2 \)
- For \( x = 0 \): \( -1 < y < 2 \)
- For \( 0 < x < 3 \): \( \frac{5x - 3}{3} < y < \frac{2}{3}x + 2 \)
Any point \( (x, y) \) that satisfies these conditions will solve the system of inequalities.
Solution
We wish to find all ordered pairs (x, y) that satisfy
(1) 2x – 3y > –6
(2) 5x – 3y < 3
(3) x + 3y > –3
One acceptable method is to “solve for y” in each inequality so that we can see the region in the xy‐plane.
Step 1. Write each inequality in the form y < (or >) an expression in x.
• For (1): 2x – 3y > –6
Subtract 2x: –3y > –6 – 2x
Multiply both sides by –1 (remember to reverse the inequality):
3y < 6 + 2x
Divide by 3: y < (2/3)x + 2
• For (2): 5x – 3y < 3
Subtract 5x: –3y < 3 – 5x
Multiply by –1 (and reverse the inequality): 3y > 5x – 3
Divide by 3: y > (5x – 3)/3
• For (3): x + 3y > –3
Subtract x: 3y > –3 – x
Divide by 3: y > (–3 – x)/3 or y > –(x/3) – 1
So the inequalities become
y < (2/3)x + 2 [Upper bound]
y > (5x – 3)/3 [Lower bound from (2)]
y > –(x/3) – 1 [Lower bound from (3)]
Step 2. Notice that we have two lower bounds for y. In order for a point (x, y) to satisfy both,
y must be greater than the larger of the two numbers (5x – 3)/3 and –(x/3) – 1.
That is, we must have
y > max { (5x – 3)/3, –(x/3) – 1 } and y < (2/3)x + 2.
Step 3. It is useful now to “compare” the two lower bounds.
Find x such that
(5x – 3)/3 = –(x/3) – 1.
Multiply both sides by 3:
5x – 3 = –x – 3.
Bring –x to the left:
5x + x – 3 = –3 ⟹ 6x – 3 = –3.
Add 3: 6x = 0, so x = 0.
Thus, for x = 0 the two lower bounds are equal. Now check a test value:
• If x > 0, say x = 1:
(5(1) – 3)/3 = 2/3 and –(1/3) – 1 = –4/3.
Since 2/3 > –4/3, for x > 0 the effective lower bound is (5x – 3)/3.
• If x < 0, say x = –1:
(5(–1) – 3)/3 = (–5 – 3)/3 = –8/3 ≈ –2.67,
–(–1/3) – 1 = 1/3 – 1 = –2/3 ≈ –0.67.
Now –0.67 > –2.67, so for x < 0 the effective lower bound is –(x/3) – 1.
So we must treat the answer in two cases:
Case 1. For x < 0:
The conditions become
• x < 0 (with x not too “small” – we will determine the x–range soon)
• Lower bound: y > –(x/3) – 1
• Upper bound: y < (2/3)x + 2
For these to be compatible we need
–(x/3) – 1 < (2/3)x + 2.
Multiply everything by 3 to clear fractions:
–x – 3 < 2x + 6.
Add x to both sides:
–3 < 3x + 6.
Subtract 6:
–9 < 3x
⟹ x > –3.
Thus, in Case 1 we require –3 < x < 0.
Case 2. For x > 0:
The conditions become
• x > 0,
• Lower bound: y > (5x – 3)/3,
• Upper bound: y < (2/3)x + 2.
For these to be compatible we need
(5x – 3)/3 < (2/3)x + 2.
Multiply by 3:
5x – 3 < 2x + 6.
Subtract 2x:
3x – 3 < 6.
Add 3:
3x < 9
⟹ x < 3.
Thus, in Case 2 we require 0 < x < 3.
At x = 0 the two lower bounds coincide (both equal –1), so x = 0 fits into either description (with y > –1 and also y < 2, since (2/3)(0)+2 = 2).
Step 4. Write the full answer. The solution set is all real numbers x and y satisfying
• For –3 < x < 0: –(x/3) – 1 < y < (2/3)x + 2.
• For x = 0: y satisfies –1 < y < 2.
• For 0 < x < 3: (5x – 3)/3 < y < (2/3)x + 2.
Any point (x, y) in the plane that satisfies these conditions will satisfy the original system.
An acceptable final answer is:
{x, y ∈ ℝ such that –3 < x < 3 and
if x < 0, then –(x/3) – 1 < y < (2/3)x + 2, and
if x ≥ 0, then (5x – 3)/3 < y < (2/3)x + 2.}
Note: The endpoints x = –3 and x = 3 are not included because when x = –3 or 3 the corresponding inequalities would force impossible conditions.
Any answer that is algebraically equivalent to the one above is correct.
Answered by UpStudy AI and reviewed by a Professional Tutor
Explain
Simplify this solution
