1.0 Solve for x:
(a) 2x ¨C 3(x + 2) ¡Ý 5x ¨C( x- 5)
(b) Solve simultaneously x - 5¡Ü 2 and x ¨C 1 ¡Ý 3
2.0 Find solution to:
(a) 5/(x-1) > 0
(b) (3x+3)/(3x-x2 ) < 0
1.0 (a) To solve the inequality 2x - 3(x + 2) ≥ 5x - (x - 5), follow the steps below:
First, distribute the -3 and -1 to the terms within the brackets:
2x - 3x - 3(2) ≥ 5x - x + 5
Simplify the expression:
2x - 3x - 6 ≥ 5x - x + 5
Combine like terms:
-x - 6 ≥ 4x + 5
Isolate the variable:
-x - 6 - 4x ≥ 4x + 5 - 4x
-5x - 6 ≥ 5
Move constants to the other side of the inequality:
-5x ≥ 11
Divide both sides by -5. Since we divide by a negative number, the inequality is reversed:
x ≤ 11/-5
x ≤ -11/5
(b) To solve the simultaneous inequalities x - 5 ≤ 2 and x - 1 ≥ 3, follow the steps below:
x - 5 ≤ 2
Add 5 to both sides:
x - 5 + 5 ≤ 2 + 5
x ≤ 7
x - 1 ≥ 3
Add 1 to both sides:
x - 1 + 1 ≥ 3 + 1
x ≥ 4
So the solution to the simultaneous inequalities is 4 ≤ x ≤ 7.
2.0 (a) To find the solution to the inequality 5/(x-1) > 0, follow the steps below:
First, find the critical points by setting the denominator equal to zero:
x - 1 = 0
x = 1
Since the denominator cannot be zero, we conclude that x ≠ 1.
Next, determine the sign of the expression (x - 1). To do this, choose a test value from each interval and substitute it into the expression:
For x < 1, choose x = 0:
(0 - 1) = -1, which is negative.
For x > 1, choose x = 2:
(2 - 1) = 1, which is positive.
Thus, we have:
Case 1: For x < 1 (x < 1 does not include 1), (x - 1) < 0
Case 2: For x > 1, (x - 1) > 0
Now, consider the inequality 5/(x-1) > 0. We have to determine when the expression is positive:
Case 1: (x - 1) < 0 and 5 > 0, implies the inequality is not satisfied.
Case 2: (x - 1) > 0 and 5 > 0, implies the inequality is satisfied.
Therefore, the solution is x > 1 (excluding 1).
(b) To find the solution to the inequality (3x+3)/(3x-x^2) < 0, follow the steps below:
First, factor the denominator:
3x - x^2 = x(3 - x)
Now, find the critical points by setting the numerator and denominator equal to zero:
3x + 3 = 0
x = -1
x(3 - x) = 0
x = 0 or x = 3
Next, set up a sign chart and choose test values from each interval:
Interval (-∞, 0):
Choose x = -2, plug it into (3x + 3)/(3x - x^2):
(-6 + 3)/(-6 - 4) = -3/-10 = 3/10, which is positive.
Interval (0, 1):
Choose x = 0.5, plug it into (3x + 3)/(3x - x^2):
(1.5 + 3)/(1.5 - 0.25) = 4.5/1.25 = 18/5, which is positive.
Interval (1, 3):
Choose x = 2, plug it into (3x + 3)/(3x - x^2):
(6 + 3)/(6 - 4) = 9/2 = 4.5, which is positive.
Interval (3, ∞):
Choose x = 4, plug it into (3x + 3)/(3x - x^2):
(12 + 3)/(12 - 16) = 15/-4, which is negative.
From the sign chart, we can conclude that the inequality is satisfied when x < 0, and when (3 - x) > 0 and x > 3. However, we must exclude the critical point x = -1 since it makes the denominator zero.
Therefore, the solution is x < 0 and x > 3.