Inequalities | Jamb Mathematics

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Hey scholar, time to power up for your exam! Dive deep into the mechanics of inequality—think of it as a system with structural bugs in economics, society, and health. Optimize your knowledge by analyzing key frameworks, real-world data, and policy patches that aim to debug disparities. 🚀

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Are you preparing for your JAMB Mathematics exam and feeling a bit uncertain about how to approach the topic of Inequalities? Don’t worry—you’re in the right place! This lesson is here to break it down in a simple, clear, and engaging way, helping you build the strong foundation you need to succeed. Whether you're struggling with complex questions or just seeking a quick refresher, this guide will boost your understanding and confidence. Let’s tackle Inequality together and move one step closer to achieving your exam success! Blissful learning.

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Calculation problem involving linear Inequalities

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1. Solve for X

3x - 7 > 5

Solution:

Add 7 to both sides:
$3x > 12$
Divide by 3:
$x > 4$

2. Solve for x

5x + 2 \leq 17

Solution:

Subtract 2 from both sides:
$5x \leq 15$
Divide by 5:
$x \leq 3$

3. Solve for y

2(y - 4) \geq 10

Solution:

Expand:
$2y - 8 \geq 10$
Add 8 to both sides:
$2y \geq 18$
Divide by 2:
$y \geq 9$

4. Solve for x

\frac{x}{3} - 2 < 4

Solution:

Add 2 to both sides:
$\frac{x}{3} < 6$
Multiply by 3:
$x < 18$

5. Solve for x

4 - 3x \geq -5

Solution:

Subtract 4 from both sides:
$-3x \geq -9$
Divide by -3 (flip inequality):
$x \leq 3$

6. Solve for a

2a - 5 > a + 1

Solution:

Subtract $a$ from both sides:
$a - 5 > 1$
Add 5 to both sides:
$a > 6$

7. Solve for x

\frac{5x}{2} + 1 \leq 6

Solution:

Subtract 1 from both sides:
$\frac{5x}{2} \leq 5$
Multiply by 2:
$5x \leq 10$
Divide by 5:
$x \leq 2$

8. Solve for y

6 - \frac{y}{4} > 2

Solution:

Subtract 6 from both sides:
$-\frac{y}{4} > -4$
Multiply by -4 (flip inequality):
$y < 16$

9. Solve for x

7x + 3 \leq 4x + 9

Solution:

Subtract $4x$ from both sides:
$3x + 3 \leq 9$
Subtract 3 from both sides:
$3x \leq 6$
Divide by 3:
$x \leq 2$

10. Solve for x

2(3x - 1) \geq 4x + 6

Solution:

Expand:
$6x - 2 \geq 4x + 6$
Subtract $4x$ from both sides:
$2x - 2 \geq 6$
Add 2 to both sides:
$2x \geq 8$
Divide by 2:
$x \geq 4$

11. Solve for x

\frac{x - 1}{3} > 2

Solution:

Multiply by 3:
$x - 1 > 6$
Add 1 to both sides:
$x > 7$

12. Solve for x

5 - \frac{x}{2} \leq 3

Solution:

Subtract 5 from both sides:
$-\frac{x}{2} \leq -2$
Multiply by -2 (flip inequality):
$x \geq 4$

13. Solve for y

\frac{y}{5} + 3 < 6

Solution:

Subtract 3 from both sides:
$\frac{y}{5} < 3$
Multiply by 5:
$y < 15$

14. Solve for x

2x - 3(1 - x) > 7

Solution:

Expand:
$2x - 3 + 3x > 7$
Combine like terms:
$5x - 3 > 7$
Add 3 to both sides:
$5x > 10$
Divide by 5:
$x > 2$

15. Solve for a

4a + 2 \geq 3a + 8

Solution:

Subtract $3a$ from both sides:
$a + 2 \geq 8$
Subtract 2 from both sides:
$a \geq 6$

16. Solve for x

\frac{3x - 5}{2} > 4

Solution:

Multiply by 2:
$3x - 5 > 8$
Add 5:
$3x > 13$
Divide by 3:
$x > \frac{13}{3}$

17. Solve for x

5x + 7 < 2x + 10

Solution:

Subtract $2x$ from both sides:
$3x + 7 < 10$
Subtract 7 from both sides:
$3x < 3$
Divide by 3:
$x < 1$

18. Solve for x

\frac{x}{4} - 3 \geq 1

Solution:

Add 3 to both sides:
$\frac{x}{4} \geq 4$
Multiply by 4:
$x \geq 16$

19. Solve for y

8 - 2y \leq 4y + 6

Solution:

Subtract 6 from both sides:
$2 - 2y \leq 4y$
Add $2y$ to both sides:
$2 \leq 6y$
Divide by 6:
$\frac{2}{6} \leq y$
Simplify:
$\frac{1}{3} \leq y$
or equivalently,
$y \geq \frac{1}{3}$

20. Solve for x

3(x + 2) > 2(x + 5)

Solution:

Expand both sides:
$3x + 6 > 2x + 10$
Subtract $2x$ from both sides:
$x + 6 > 10$
Subtract 6 from both sides:
$x > 4$
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Calculation problem involving quadratic inequalities

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1. Solve for x

x^2 - 4x - 5 > 0

Solution:

Factorize:
$(x - 5)(x + 1) > 0$
Find critical points:
$x = 5, x = -1$
Test intervals:
- For $x < -1$ , choose $x = -2$ : $(-2 - 5)(-2 + 1) = (-7)(-1) = 7 > 0$
- For $-1 < x < 5$ , choose $x = 0$ : $(0 - 5)(0 + 1) = (-5)(1) = -5 < 0$
- For $x > 5$ , choose $x = 6$ : $(6 - 5)(6 + 1) = (1)(7) = 7 > 0$
Solution:
$x \in (-\infty, -1) \cup (5, \infty)$

2. Solve for x

x^2 - 9 \leq 0

Solution:

Factorize:
$(x - 3)(x + 3) \leq 0$
Find critical points:
$x = -3, x = 3$
Test intervals:
- For $x < -3$ , choose $x = -4$ : $(-4 - 3)(-4 + 3) = (-7)(-1) = 7 > 0$
- For $-3 \leq x \leq 3$ , choose $x = 0$ : $(0 - 3)(0 + 3) = (-3)(3) = -9 \leq 0$
- For $x > 3$ , choose $x = 4$ : $(4 - 3)(4 + 3) = (1)(7) = 7 > 0$
Solution:
$x \in [-3, 3]$

3. Solve for x

x^2 + 2x - 8 > 0

Solution:

Factorize:
$(x + 4)(x - 2) > 0$
Find critical points:
$x = -4, x = 2$
Test intervals:
- For $x < -4$ , choose $x = -5$ : $(-5 + 4)(-5 - 2) = (-1)(-7) = 7 > 0$
- For $-4 < x < 2$ , choose $x = 0$ : $(0 + 4)(0 - 2) = (4)(-2) = -8 < 0$
- For $x > 2$ , choose $x = 3$ : $(3 + 4)(3 - 2) = (7)(1) = 7 > 0$
Solution:
$x \in (-\infty, -4) \cup (2, \infty)$

4. Solve for x

x^2 - 6x + 8 \geq 0

Solution:

Factorize:
$(x - 4)(x - 2) \geq 0$
Find critical points:
$x = 4, x = 2$
Test intervals:
- For $x < 2$ , choose $x = 0$ : $(0 - 4)(0 - 2) = (-4)(-2) = 8 > 0$
- For $2 \leq x \leq 4$ , choose $x = 3$ : $(3 - 4)(3 - 2) = (-1)(1) = -1 < 0$
- For $x > 4$ , choose $x = 5$ : $(5 - 4)(5 - 2) = (1)(3) = 3 > 0$
Solution:
$x \in (-\infty, 2] \cup [4, \infty)$

5. Solve for x

x^2 - 5x + 6 < 0

Solution:

Factorize:
$(x - 3)(x - 2) < 0$
Find critical points:
$x = 3, x = 2$
Test intervals:
- For $x < 2$ , choose $x = 1$ : $(1 - 3)(1 - 2) = (-2)(-1) = 2 > 0$
- For $2 < x < 3$ , choose $x = 2.5$ : $(2.5 - 3)(2.5 - 2) = (-0.5)(0.5) = -0.25 < 0$
- For $x > 3$ , choose $x = 4$ : $(4 - 3)(4 - 2) = (1)(2) = 2 > 0$
Solution:
$x \in (2,3)$

6. Solve for x

x^2 - 3x - 10 \leq 0

Solution:

Factorize:
$(x - 5)(x + 2) \leq 0$
Find critical points:
$x = 5, x = -2$
Test intervals:
- For $x < -2$ , choose $x = -3$ : $(-3 - 5)(-3 + 2) = (-8)(-1) = 8 > 0$
- For $-2 \leq x \leq 5$ , choose $x = 0$ : $(0 - 5)(0 + 2) = (-5)(2) = -10 \leq 0$
- For $x > 5$ , choose $x = 6$ : $(6 - 5)(6 + 2) = (1)(8) = 8 > 0$
Solution:
$x \in [-2,5]$

7. Solve for x

2x^2 - 3x - 2 > 0

Solution:

Factorize:
$(2x + 1)(x - 2) > 0$
Find critical points:
$x = -\frac{1}{2}, x = 2$
Test intervals:
- For $x < -\frac{1}{2}$ , choose $x = -1$ : $(2(-1) + 1)(-1 - 2) = (-2 + 1)(-3) = (-1)(-3) = 3 > 0$
- For $-\frac{1}{2} < x < 2$ , choose $x = 0$ : $(2(0) + 1)(0 - 2) = (1)(-2) = -2 < 0$
- For $x > 2$ , choose $x = 3$ : $(2(3) + 1)(3 - 2) = (6 + 1)(1) = 7 > 0$
Solution:
$x \in (-\infty, -\frac{1}{2}) \cup (2, \infty)$

8. Solve for x

x^2 + 4x + 3 \geq 0

Solution:

Factorize:
$(x + 1)(x + 3) \geq 0$
Find critical points:
$x = -1, x = -3$
Test intervals:
- For $x < -3$ , choose $x = -4$ : $(-4 + 1)(-4 + 3) = (-3)(-1) = 3 > 0$
- For $-3 \leq x \leq -1$ , choose $x = -2$ : $(-2 + 1)(-2 + 3) = (-1)(1) = -1 < 0$
- For $x > -1$ , choose $x = 0$ : $(0 + 1)(0 + 3) = (1)(3) = 3 > 0$
Solution:
$x \in (-\infty, -3] \cup [-1, \infty)$

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- Jamb Mathematics- Lesson notes on progression for utme Success

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Inequalities | Jamb Mathematics

Calculation problem involving linear Inequalities

1. Solve for X

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Calculation problem involving quadratic inequalities

1. Solve for x

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