5.1 Introduction to equation and inequalities

Equations and inequalities are fundamental concepts in algebra and mathematics that express relationships between variables and constants. They are used to model real-world situations and solve for unknown values.

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1. Equations

An equation is a mathematical statement that asserts the equality of two expressions. It contains an equals sign (“=”) and typically includes variables, constants, and operations like addition, subtraction, multiplication, or division.

General Form of an Equation

$$\text{Expression 1}=\text{Expression 2}$$

An equation is true when both expressions on either side of the equals sign are equal.

Types of Equations

1. Linear Equations: A linear equation is an equation in which the highest power of the variable is 1. Its graph is a straight line.
   • General Form: $ax+b=0$, where $a$ and $b$ are constants.
   • Example: $2x+3=7$.
2. Quadratic Equations: A quadratic equation is a second-degree polynomial equation where the highest power of the variable is 2.
   • General Form: ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0, where $a$, $b$, and $c$ are constants and a≠0a \neq 0a=0.
   • Example: x2−4x+4=0x^2 – 4x + 4 = 0x2−4x+4=0.
3. Cubic Equations: A cubic equation involves a variable raised to the power of 3.
   • General Form: ax3+bx2+cx+d=0ax^3 + bx^2 + cx + d = 0ax3+bx2+cx+d=0.
   • Example: 2×3−3×2+5x−1=02x^3 – 3x^2 + 5x – 1 = 02x3−3x2+5x−1=0.
4. Rational Equations: Rational equations contain variables in the numerator or denominator of fractions.
   • Example: 2xx−1=3\frac{2x}{x – 1} = 3x−12x​=3.
5. Exponential Equations: An exponential equation has the variable in the exponent.
   • Example: 2x=82^x = 82x=8.

Solving Equations

The goal of solving an equation is to find the value(s) of the variable that make the equation true.

• Linear Equation Example: $$2x+3=7$$ Subtract 3 from both sides: $$2x=4$$ Divide by 2: $$x=2$$

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2. Inequalities

An inequality is a mathematical statement that compares two expressions and shows that one is either greater than, less than, greater than or equal to, or less than or equal to the other.

Symbols of Inequality

• $<$: Less than
• $\le$: Less than or equal to
• $>$: Greater than
• $\ge$: Greater than or equal to
• ≠\neq=: Not equal to

Types of Inequalities

1. Linear Inequalities: A linear inequality is similar to a linear equation but uses inequality symbols instead of an equals sign.
   • Example: $3x+5>2$.
2. Quadratic Inequalities: These involve quadratic expressions with inequality symbols.
   • Example: x2−4x+3≤0x^2 – 4x + 3 \leq 0x2−4x+3≤0.
3. Rational Inequalities: Rational inequalities involve fractions and require careful handling of the numerator and denominator.
   • Example: x+1x−2≥0\frac{x + 1}{x – 2} \geq 0x−2x+1​≥0.

Solving Inequalities

The process of solving inequalities is similar to solving equations, but you must be mindful of special rules, especially when multiplying or dividing by negative numbers (which reverses the inequality symbol).

• Linear Inequality Example: $$3x+5>2$$ Subtract 5 from both sides: $$3x>-3$$ Divide by 3: $$x>-1$$

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3. Graphical Representation

• Equations: When graphing an equation, you typically plot the solution set on a coordinate plane. For example, the graph of a linear equation $y=2x+3$ is a straight line.
• Inequalities: Inequalities can be represented on a number line or coordinate plane. For example, $x>2$ would be a ray starting from $x=2$ and extending to the right.

For two-variable inequalities like $y>2x+3$, the solution is represented by shading the region above or below the corresponding line, depending on the inequality.

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4. Key Concepts in Equations and Inequalities

• Solution Set: The set of all values of the variable that satisfy the equation or inequality.
• Solving by Substitution: In systems of equations or inequalities, substituting the value of one variable into another equation.
• Multiplying or Dividing Inequalities by Negative Numbers: When multiplying or dividing both sides of an inequality by a negative number, reverse the inequality sign.

 

5.2 Roots of linear and quadratic equations

Roots of an equation refer to the values of the variable that satisfy the equation, making the expression equal to zero. For linear and quadratic equations, finding the roots involves different approaches depending on the degree of the polynomial.

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1. Roots of Linear Equations

A linear equation is an equation of the form:

$$ax+b=0$$

where $a$ and $b$ are constants, and a≠0a \neq 0a=0.

Solving for the Root of a Linear Equation

The root of a linear equation is straightforward to find. To solve for $x$:

1. Isolate $x$ by moving $b$ to the other side of the equation: $$ax=-b$$
2. Divide by $a$ to solve for $x$: x=−bax = \frac{-b}{a}x=a−b​

Example

For the linear equation:

$$2x+4=0$$

1. Subtract 4 from both sides: $$2x=-4$$
2. Divide by 2: $$x=-2$$

Thus, the root of the linear equation is $x=-2$.

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2. Roots of Quadratic Equations

A quadratic equation is a second-degree polynomial equation of the form:

ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0

where $a$, $b$, and $c$ are constants, and a≠0a \neq 0a=0.

Methods for Solving Quadratic Equations

There are several methods to find the roots of quadratic equations:

1. Factoring Method

If the quadratic can be factored into two binomials, the roots are the values of $x$ that make each binomial zero.

• Example: Solve x2−5x+6=0x^2 – 5x + 6 = 0x2−5x+6=0 by factoring: x2−5x+6=(x−2)(x−3)=0x^2 – 5x + 6 = (x – 2)(x – 3) = 0x2−5x+6=(x−2)(x−3)=0 Set each factor equal to zero: $$x-2=0\text{or}x-3=0$$ So, the roots are $x=2$ and $x=3$.

2. Quadratic Formula

The quadratic formula is a general method for solving any quadratic equation. It is given by:

x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}x=2a−b±b2−4ac​​

where $a$, $b$, and $c$ are the coefficients from the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0.

• Example: Solve 2×2−4x+1=02x^2 – 4x + 1 = 02x2−4x+1=0 using the quadratic formula.

  Here, $a=2$, $b=-4$, and $c=1$. Substituting these into the formula:

  x=−(−4)±(−4)2−4(2)(1)2(2)x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(2)(1)}}{2(2)}x=2(2)−(−4)±(−4)2−4(2)(1)​​ x=4±16−84x = \frac{4 \pm \sqrt{16 – 8}}{4}x=44±16−8​​ x=4±84x = \frac{4 \pm \sqrt{8}}{4}x=44±8​​ x=4±224x = \frac{4 \pm 2\sqrt{2}}{4}x=44±22​​Simplifying:

  x=1±22x = 1 \pm \frac{\sqrt{2}}{2}x=1±22​​Thus, the roots are x=1+22x = 1 + \frac{\sqrt{2}}{2}x=1+22​​ and x=1−22x = 1 – \frac{\sqrt{2}}{2}x=1−22​​.

3. Completing the Square

This method involves rewriting the quadratic equation in the form (x−h)2=k(x – h)^2 = k(x−h)2=k, and then solving for $x$.

• Example: Solve x2+6x−7=0x^2 + 6x – 7 = 0x2+6x−7=0 by completing the square.
  1. Move the constant term to the other side: x2+6x=7x^2 + 6x = 7x2+6x=7
  2. Take half the coefficient of $x$, square it, and add to both sides: x2+6x+9=7+9x^2 + 6x + 9 = 7 + 9x2+6x+9=7+9 (x+3)2=16(x + 3)^2 = 16(x+3)2=16
  3. Solve by taking the square root of both sides: $$x+3=\pm 4$$
  4. Finally, solve for $x$: $$x=-3\pm 4$$ So, $x=1$ or $x=-7$.

4. Graphical Method

The roots of a quadratic equation correspond to the points where the parabola (graph of the quadratic function) intersects the $x$-axis. These points are called the x-intercepts or zeroes of the function.

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3. Discriminant and Nature of Roots

The discriminant is the part of the quadratic formula under the square root:

Δ=b2−4ac\Delta = b^2 – 4acΔ=b2−4ac

The value of the discriminant determines the nature of the roots:

• If $\Delta >0$: The equation has two real and distinct roots.
• If $\Delta =0$: The equation has two real and equal roots (repeated root).
• If $\Delta <0$: The equation has two complex roots (no real solutions).

 

5.3 Roots of linear and quadratic inequalities of one variable

Inequalities involve finding the range of values that satisfy a given condition, rather than just specific points as in equations. For linear and quadratic inequalities involving one variable, the solution is typically a range of values rather than a single root.

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1. Roots of Linear Inequalities

A linear inequality is an inequality of the form:

$$ax+b<0,ax+b>0,ax+b\le 0,ax+b\ge 0$$

where $a$ and $b$ are constants, and $x$ is the variable.

Steps to Solve Linear Inequalities

1. Isolate the variable $x$ by performing algebraic operations similar to solving a linear equation.
2. Perform operations carefully: When multiplying or dividing both sides by a negative number, reverse the inequality sign.
3. Graph the solution on a number line or express it in interval notation.

Example 1: Solve $3x-5>1$.

1. Add 5 to both sides: $$3x>6$$
2. Divide by 3: $$x>2$$
3. The solution is all $x$-values greater than 2: $$x\in (2,\infty )$$

The solution can also be represented graphically on a number line, with an open circle at 2 and an arrow extending to the right.

Example 2: Solve $-4x+3\le 7$.

1. Subtract 3 from both sides: $$-4x\le 4$$
2. Divide by -4 (and reverse the inequality sign): $$x\ge -1$$
3. The solution is $x\in (-\infty ,-1]$, represented by a closed circle at -1 and an arrow extending to the left.

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2. Roots of Quadratic Inequalities

A quadratic inequality is of the form:

ax2+bx+c<0,ax2+bx+c>0,ax2+bx+c≤0,ax2+bx+c≥0ax^2 + bx + c < 0, \quad ax^2 + bx + c > 0, \quad ax^2 + bx + c \leq 0, \quad ax^2 + bx + c \geq 0ax2+bx+c<0,ax2+bx+c>0,ax2+bx+c≤0,ax2+bx+c≥0

where $a$, $b$, and $c$ are constants.

Steps to Solve Quadratic Inequalities

1. Find the roots (or critical points) by solving the corresponding quadratic equation ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0. These roots divide the number line into intervals.
2. Test intervals between the roots to determine where the inequality holds true (positive or negative intervals).
3. Write the solution as an interval or union of intervals.

Example: Solve x2−3x−10<0x^2 – 3x – 10 < 0x2−3x−10<0.

1. Solve the quadratic equation x2−3x−10=0x^2 – 3x – 10 = 0x2−3x−10=0 by factoring:

   $$(x-5)(x+2)=0$$The roots are $x=5$ and $x=-2$.

2. Divide the number line into intervals based on the roots: $(-\infty ,-2)$, $(-2,5)$, and $(5,\infty )$.
3. Test points within each interval to check whether the inequality is satisfied:
   • For $x=-3$ (in $(-\infty ,-2)$):

     (−3)2−3(−3)−10=9+9−10=8(positive, so inequality does not hold)(-3)^2 – 3(-3) – 10 = 9 + 9 – 10 = 8 \quad \text{(positive, so inequality does not hold)}(−3)2−3(−3)−10=9+9−10=8(positive, so inequality does not hold)

   • For $x=0$ (in $(-2,5)$):

     02−3(0)−10=−10(negative, inequality holds here)0^2 – 3(0) – 10 = -10 \quad \text{(negative, inequality holds here)}02−3(0)−10=−10(negative, inequality holds here)

   • For $x=6$ (in $(5,\infty )$):

     62−3(6)−10=36−18−10=8(positive, so inequality does not hold)6^2 – 3(6) – 10 = 36 – 18 – 10 = 8 \quad \text{(positive, so inequality does not hold)}62−3(6)−10=36−18−10=8(positive, so inequality does not hold)

4. Conclusion: The inequality holds in the interval $(-2,5)$.

Thus, the solution is:

$$x\in (-2,5)$$

Graphically, this is represented as an open interval between -2 and 5 on the number line.

Example 2: Solve x2−4x+3≥0x^2 – 4x + 3 \geq 0x2−4x+3≥0.

1. Solve the quadratic equation x2−4x+3=0x^2 – 4x + 3 = 0x2−4x+3=0:

   $$(x-3)(x-1)=0$$The roots are $x=3$ and $x=1$.

2. Divide the number line into intervals: $(-\infty ,1)$, $(1,3)$, and $(3,\infty )$.
3. Test points:
   • For $x=0$ (in $(-\infty ,1)$): 02−4(0)+3=3(positive, inequality holds)0^2 – 4(0) + 3 = 3 \quad \text{(positive, inequality holds)}02−4(0)+3=3(positive, inequality holds)
   • For $x=2$ (in $(1,3)$): 22−4(2)+3=4−8+3=−1(negative, inequality does not hold)2^2 – 4(2) + 3 = 4 – 8 + 3 = -1 \quad \text{(negative, inequality does not hold)}22−4(2)+3=4−8+3=−1(negative, inequality does not hold)
   • For $x=4$ (in $(3,\infty )$): 42−4(4)+3=16−16+3=3(positive, inequality holds)4^2 – 4(4) + 3 = 16 – 16 + 3 = 3 \quad \text{(positive, inequality holds)}42−4(4)+3=16−16+3=3(positive, inequality holds)
4. Conclusion: The inequality holds in the intervals $(-\infty ,1]\cup [3,\infty )$.

Thus, the solution is:

$$x\in (-\infty ,1]\cup [3,\infty )$$

Graphically, this is represented with a closed circle at 1 and 3, with the regions extending left of 1 and right of 3.

 

5.4 System of first-degree equations of two variables

A system of first-degree equations (also called a system of linear equations) consists of two or more linear equations involving two variables. These systems can have one solution, infinitely many solutions, or no solution depending on the nature of the lines represented by the equations.

A system of linear equations in two variables can be written as:

a1x+b1y=c1(Equation 1)a2x+b2y=c2(Equation 2)\begin{aligned} a_1x + b_1y &= c_1 \quad \text{(Equation 1)} \\ a_2x + b_2y &= c_2 \quad \text{(Equation 2)} \end{aligned}a1​x+b1​ya2​x+b2​y​=c1​(Equation 1)=c2​(Equation 2)​

where $x$ and $y$ are variables, and a1,a2,b1,b2,c1,c2a_1, a_2, b_1, b_2, c_1, c_2a1​,a2​,b1​,b2​,c1​,c2​ are constants.

Solutions of a System of Equations

There are three possible outcomes when solving a system of linear equations:

1. One Solution (Consistent and Independent System): The lines intersect at exactly one point. The coordinates of this point give the solution to the system.
2. No Solution (Inconsistent System): The lines are parallel and never intersect. There is no solution in this case.
3. Infinitely Many Solutions (Consistent and Dependent System): The lines are coincident (overlap entirely). Every point on the line is a solution.

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Methods for Solving Systems of First-Degree Equations

1. Substitution Method

In the substitution method, one equation is solved for one variable, and then this expression is substituted into the other equation.

Example:

Solve the system:

x+2y=10(1)3x−y=5(2)\begin{aligned} x + 2y &= 10 \quad \text{(1)} \\ 3x – y &= 5 \quad \text{(2)} \end{aligned}x+2y3x−y​=10(1)=5(2)​

1. Solve Equation (1) for $x$:

   $$x=10-2y$$

2. Substitute this expression for $x$ into Equation (2):

   $$3(10-2y)-y=5$$Simplify:

   $$30-6y-y=5$$ $$30-7y=5$$ $$-7y=-25$$ y=257y = \frac{25}{7}y=725​

3. Substitute y=257y = \frac{25}{7}y=725​ back into $x=10-2y$:

   x=10−2(257)=10−507=707−507=207x = 10 – 2\left(\frac{25}{7}\right) = 10 – \frac{50}{7} = \frac{70}{7} – \frac{50}{7} = \frac{20}{7}x=10−2(725​)=10−750​=770​−750​=720​

Thus, the solution is x=207x = \frac{20}{7}x=720​ and y=257y = \frac{25}{7}y=725​.

2. Elimination (or Addition) Method

The elimination method involves adding or subtracting the two equations to eliminate one of the variables, then solving for the remaining variable.

Example:

Solve the system:

2x+3y=9(1)4x−3y=1(2)\begin{aligned} 2x + 3y &= 9 \quad \text{(1)} \\ 4x – 3y &= 1 \quad \text{(2)} \end{aligned}2x+3y4x−3y​=9(1)=1(2)​

1. Add the two equations together to eliminate $y$:

   $$(2x+3y)+(4x-3y)=9+1$$ $$6x=10$$ x=106=53x = \frac{10}{6} = \frac{5}{3}x=610​=35​

2. Substitute x=53x = \frac{5}{3}x=35​ into one of the original equations, say Equation (1):

   2(53)+3y=92\left(\frac{5}{3}\right) + 3y = 92(35​)+3y=9 103+3y=9\frac{10}{3} + 3y = 9310​+3y=9Multiply everything by 3 to clear the fraction:

   $$10+9y=27$$ $$9y=17$$ y=179y = \frac{17}{9}y=917​

Thus, the solution is x=53x = \frac{5}{3}x=35​ and y=179y = \frac{17}{9}y=917​.

3. Graphical Method

The graphical method involves graphing both equations on the same coordinate plane and identifying the point of intersection.

Example:

Solve the system:

x+y=52x−y=1\begin{aligned} x + y &= 5 \\ 2x – y &= 1 \end{aligned}x+y2x−y​=5=1​

1. Graph Equation 1: $x+y=5$ can be rewritten as $y=5-x$, a line with slope -1 and y-intercept 5.
2. Graph Equation 2: $2x-y=1$ can be rewritten as $y=2x-1$, a line with slope 2 and y-intercept -1.

By graphing these equations, the point where they intersect is the solution. In this case, the intersection occurs at $x=2$ and $y=3$.

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Types of Solutions: Analyzing the System

• One Solution: The two lines have different slopes, meaning they intersect at one unique point.
• No Solution: The lines are parallel, meaning their slopes are equal but they have different y-intercepts.
• Infinitely Many Solutions: The lines are identical, meaning they have the same slope and the same y-intercept.

Example of No Solution:

2x+3y=64x+6y=15\begin{aligned} 2x + 3y &= 6 \\ 4x + 6y &= 15 \end{aligned}2x+3y4x+6y​=6=15​

The second equation is simply a multiple of the first, but the right-hand side is different, meaning the lines are parallel but never intersect. Therefore, the system has no solution.

Example of Infinitely Many Solutions:

x−y=32x−2y=6\begin{aligned} x – y &= 3 \\ 2x – 2y &= 6 \end{aligned}x−y2x−2y​=3=6​

The second equation is simply a multiple of the first, so the lines are coincident. Therefore, there are infinitely many solutions, and every point on the line $x-y=3$ is a solution.

 

5.5 Inequalities and their properties

An inequality is a mathematical statement that expresses the relative size or order of two values. Unlike equations, which state that two expressions are equal, inequalities show that one expression is greater than, less than, greater than or equal to, or less than or equal to another.

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Basic Types of Inequalities

1. Strict Inequality:
   • $a<b$ (a is less than b)
   • $a>b$ (a is greater than b)
2. Inclusive Inequality:
   • $a\le b$ (a is less than or equal to b)
   • $a\ge b$ (a is greater than or equal to b)

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Properties of Inequalities

The following properties of inequalities help in manipulating and solving them.

1. Transitive Property

• If $a<b$ and $b<c$, then $a<c$.
• Similarly, if $a\le b$ and $b\le c$, then $a\le c$.

Example:

If $3<5$ and $5<7$, then $3<7$.

2. Addition and Subtraction Property

• You can add or subtract the same number from both sides of an inequality without changing the inequality.
  • If $a<b$, then $a+c<b+c$ and $a-c<b-c$.

Example:

If $4<6$, adding 3 to both sides gives:

$$4+3<6+3\Rightarrow 7<9$$

Similarly, subtracting 2 from both sides:

$$4-2<6-2\Rightarrow 2<4$$

3. Multiplication and Division Property

• You can multiply or divide both sides of an inequality by the same positive number without changing the inequality.
  • If $a<b$, then $a\times c<b\times c$ and ac<bc\frac{a}{c} < \frac{b}{c}ca​<cb​ for $c>0$.
• Important: If you multiply or divide both sides by a negative number, you must reverse the inequality.
  • If $a<b$ and $c<0$, then $a\times c>b\times c$.

Example:

If $3<5$, multiplying both sides by 2 gives:

$$3\times 2<5\times 2\Rightarrow 6<10$$

If $3<5$, multiplying both sides by -2 (and reversing the inequality) gives:

$$3\times (-2)>5\times (-2)\Rightarrow -6>-10$$

4. Substitution Property

• If $a=b$, you can replace $a$ with $b$ in any inequality where $a$ occurs.
  • If $a=b$ and $b<c$, then $a<c$.

5. Distributive Property

• The distributive property of multiplication over addition or subtraction holds for inequalities as well.
  • If $a(b+c)\le d$, you can expand this to $ab+ac\le d$.

Example:

If $2(3+4)<15$, expanding gives:

$$2(7)<15\Rightarrow 14<15$$

6. Properties of Absolute Value Inequalities

An absolute value inequality involves expressions of the form $|x|$, which measures the distance of $x$ from 0.

• For $|x|<a$ where $a>0$: $$-a<x<a$$
• For $|x|\le a$: $$-a\le x\le a$$
• For $|x|>a$: $$x<-a\text{or}x>a$$
• For $|x|\ge a$: $$x\le -a\text{or}x\ge a$$

Example:

Solve $|x-2|<5$:

$$-5<x-2<5$$

Add 2 to all parts:

$$-3<x<7$$

Thus, the solution is $x\in (-3,7)$.

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Inequalities and Intervals

The solution to an inequality is often an interval of values that satisfy the inequality. Intervals can be expressed in different ways:

1. Open Interval: $(a,b)$ means $a<x<b$ (does not include $a$ and $b$).
2. Closed Interval: $[a,b]$ means $a\le x\le b$ (includes $a$ and $b$).
3. Half-Open Interval: $[a,b)$ or $(a,b]$ means one end is included and the other is not.
4. Infinite Interval: $(-\infty ,a)$, $(a,\infty )$, or $(-\infty ,\infty )$.

Example:

Solve $2x-3\ge 5$.

1. Add 3 to both sides: $$2x\ge 8$$
2. Divide by 2: $$x\ge 4$$

Thus, the solution is $x\in [4,\infty )$.

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Compound Inequalities

A compound inequality involves two or more inequalities combined by “and” or “or.”

• And (Intersection): The solution must satisfy both inequalities.

  $$a<x<b$$This can be written as $a<x\wedge x<b$.

• Or (Union): The solution satisfies at least one inequality.

  $$x<a\text{or}x>b$$

Example of “And”:

Solve $2<3x+1\le 8$.

1. Subtract 1 from all parts: $$1<3x\le 7$$
2. Divide by 3: 13<x≤73\frac{1}{3} < x \leq \frac{7}{3}31​<x≤37​

The solution is x∈(13,73]x \in \left(\frac{1}{3}, \frac{7}{3}\right]x∈(31​,37​].

Example of “Or”:

Solve $x+2<1\text{or}3x-1>8$.

1. Solve each inequality: $$x<-1\text{or}x>3$$

Thus, the solution is $x\in (-\infty ,-1)\cup (3,\infty )$.

 

5.6 Graph of inequalities of one and two variables and their solution set

Graphing inequalities is a visual method for representing solutions to inequalities in one or two variables. This approach provides a clear understanding of the solution set.

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Graphing Inequalities in One Variable

An inequality in one variable has the general form:

$$ax+b<0,ax+b>0,ax+b\le 0,ax+b\ge 0$$

where $x$ is the variable, and $a$ and $b$ are constants.

Steps to Graph Inequalities in One Variable:

1. Solve the Inequality: Solve the inequality as you would solve a linear equation.
2. Plot the Solution on a Number Line:
   • Use a solid dot to indicate that the endpoint is included (for inequalities like $\le$ or $\ge$).
   • Use an open dot to indicate that the endpoint is not included (for inequalities like $<$ or $>$).
3. Shade the Region Representing the Solution:
   • Shade to the left of the point for $x<$ or $x\le$.
   • Shade to the right of the point for $x>$ or $x\ge$.

Example 1 (One Variable):

Graph the inequality $x-3\le 4$.

Steps:

1. Solve the inequality:

   $$x-3\le 4\Rightarrow x\le 7$$

2. Plot the point $x=7$ on the number line and use a solid dot because the inequality includes the equals sign (≤).
3. Shade the region to the left of 7 to represent $x\le 7$.

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Graphing Inequalities in Two Variables

An inequality in two variables has the general form:

$$ax+by<c,ax+by>c,ax+by\le c,ax+by\ge c$$

where $x$ and $y$ are variables, and $a$, $b$, and $c$ are constants.

Steps to Graph Inequalities in Two Variables:

1. Rewrite the Inequality as an Equation:
   • Convert the inequality to an equation by replacing the inequality symbol with an equals sign (i.e., graph the line $ax+by=c$).
2. Graph the Boundary Line:
   • If the inequality is strict ($<$ or $>$), draw a dashed line to indicate that points on the line are not part of the solution.
   • If the inequality is inclusive ($\le$ or $\ge$), draw a solid line to indicate that points on the line are included in the solution.
3. Test a Point:
   • Choose a test point that is not on the boundary line (commonly $(0,0)$ if it is not on the line) to check if it satisfies the inequality.
   • If the test point satisfies the inequality, shade the region that includes the test point. If not, shade the opposite region.
4. Shade the Solution Region: The shaded region represents all the points that satisfy the inequality.

Example 2 (Two Variables):

Graph the inequality $2x-y\ge 3$.

Steps:

1. Rewrite the inequality as an equation:

   $$2x-y=3$$

2. Graph the boundary line $2x-y=3$:
   • Solve for $y$: $y=2x-3$.
   • This is a straight line with slope 2 and y-intercept -3.
   • Since the inequality is $\ge$, the line will be solid because points on the line are part of the solution.
3. Test a point:
   • Choose $(0,0)$. Substituting into the inequality:

   $$2(0)-(0)\ge 3\Rightarrow 0\ge 3\text{(false)}$$

   • The point $(0,0)$ does not satisfy the inequality, so shade the region above the line.

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Solution Set of Inequalities

The solution set of an inequality consists of all points or values that satisfy the inequality.

• For one variable: The solution set is a range of values on the number line.
• For two variables: The solution set is a region in the coordinate plane.

Intersection of Solution Sets (Systems of Inequalities):

When solving a system of inequalities, the solution set is the overlapping region that satisfies all inequalities in the system.

Example 3 (System of Inequalities):

Solve the system of inequalities and graph the solution set:

1. $x+y\ge 2$
2. $y<3x+1$

Steps:

1. Graph the first inequality $x+y\ge 2$:
   • Convert to the equation $x+y=2$, which is a line with slope -1 and y-intercept 2.
   • Draw a solid line because the inequality is $\ge$.
   • Choose a test point, such as $(0,0)$. Since $0+0\ge 2$ is false, shade the region above the line.
2. Graph the second inequality $y<3x+1$:
   • Convert to the equation $y=3x+1$, which is a line with slope 3 and y-intercept 1.
   • Draw a dashed line because the inequality is $<$.
   • Test the point $(0,0)$. Since $0<1$, shade the region below the line.
3. The solution set is the region where the two shaded areas overlap. This is the region that satisfies both inequalities.

 

5.7 Graph of quadratic function

A quadratic function is a polynomial function of degree 2 and has the general form:

f(x)=ax2+bx+cf(x) = ax^2 + bx + cf(x)=ax2+bx+c

where $a$, $b$, and $c$ are constants, and a≠0a \neq 0a=0. The graph of a quadratic function is a parabola, which can either open upwards or downwards depending on the sign of $a$.

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Key Features of the Graph of a Quadratic Function

1. Vertex: The highest or lowest point on the graph.
   • The vertex of the parabola occurs at x=−b2ax = -\frac{b}{2a}x=−2ab​.
   • To find the y-coordinate of the vertex, substitute x=−b2ax = -\frac{b}{2a}x=−2ab​ into the quadratic equation to get y=f(−b2a)y = f\left(-\frac{b}{2a}\right)y=f(−2ab​).
2. Axis of Symmetry: A vertical line that passes through the vertex, dividing the parabola into two symmetric halves. The equation of the axis of symmetry is:

   x=−b2ax = -\frac{b}{2a}x=−2ab​

3. Direction of Opening:
   • If $a>0$, the parabola opens upwards.
   • If $a<0$, the parabola opens downwards.
4. Intercepts:
   • y-intercept: This is the point where the graph crosses the y-axis. To find the y-intercept, substitute $x=0$ into the quadratic equation:

     $$y=c$$The y-intercept is $(0,c)$.

   • x-intercepts (Roots): The points where the graph crosses the x-axis. These are the solutions to the equation ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0. You can find the x-intercepts using the quadratic formula:

     x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}x=2a−b±b2−4ac​​If the discriminant Δ=b2−4ac\Delta = b^2 – 4acΔ=b2−4ac is:

     • Positive, the quadratic has two real roots (two x-intercepts).
     • Zero, the quadratic has one real root (the vertex lies on the x-axis).
     • Negative, the quadratic has no real roots (the parabola does not cross the x-axis).

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Steps to Graph a Quadratic Function

1. Identify the Coefficients: Start by identifying $a$, $b$, and $c$ in the quadratic equation f(x)=ax2+bx+cf(x) = ax^2 + bx + cf(x)=ax2+bx+c.
2. Find the Vertex: Use the formula x=−b2ax = -\frac{b}{2a}x=−2ab​ to find the x-coordinate of the vertex, and then calculate the y-coordinate by plugging the x-value into the function.
3. Determine the Axis of Symmetry: The axis of symmetry is the vertical line x=−b2ax = -\frac{b}{2a}x=−2ab​.
4. Find the y-intercept: The y-intercept is $(0,c)$.
5. Find the x-intercepts (if they exist): Use the quadratic formula or factor the quadratic if possible.
6. Plot Additional Points: Choose additional values of $x$ around the vertex and calculate corresponding $y$-values to get more points for accuracy.
7. Sketch the Parabola: Use the points to sketch the parabola, ensuring it is symmetric about the axis of symmetry and either opens upwards or downwards depending on the value of $a$.

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Example 1: Graph the Quadratic Function

Given the quadratic function:

f(x)=2×2−4x+1f(x) = 2x^2 – 4x + 1f(x)=2x2−4x+1

Step 1: Identify the coefficients:
$a=2$, $b=-4$, $c=1$.

Step 2: Find the vertex:
The x-coordinate of the vertex is:

x=−−42(2)=44=1x = -\frac{-4}{2(2)} = \frac{4}{4} = 1x=−2(2)−4​=44​=1

To find the y-coordinate, substitute $x=1$ into the function:

f(1)=2(1)2−4(1)+1=2−4+1=−1f(1) = 2(1)^2 – 4(1) + 1 = 2 – 4 + 1 = -1f(1)=2(1)2−4(1)+1=2−4+1=−1

So, the vertex is $(1,-1)$.

Step 3: Axis of symmetry:
The axis of symmetry is the vertical line $x=1$.

Step 4: y-intercept:
The y-intercept is f(0)=2(0)2−4(0)+1=1f(0) = 2(0)^2 – 4(0) + 1 = 1f(0)=2(0)2−4(0)+1=1, so the y-intercept is $(0,1)$.

Step 5: Find the x-intercepts (if they exist):
Use the quadratic formula to find the x-intercepts:

x=−(−4)±(−4)2−4(2)(1)2(2)=4±16−84=4±84=4±224x = \frac{-(-4) \pm \sqrt{(-4)^2 – 4(2)(1)}}{2(2)} = \frac{4 \pm \sqrt{16 – 8}}{4} = \frac{4 \pm \sqrt{8}}{4} = \frac{4 \pm 2\sqrt{2}}{4}x=2(2)−(−4)±(−4)2−4(2)(1)​​=44±16−8​​=44±8​​=44±22​​

Thus, the x-intercepts are:

x=4+224andx=4−224x = \frac{4 + 2\sqrt{2}}{4} \quad \text{and} \quad x = \frac{4 – 2\sqrt{2}}{4}x=44+22​​andx=44−22​​

Step 6: Plot additional points:
To get more points, plug in values of $x$ around the vertex, such as $x=0$ and $x=2$, and calculate the corresponding y-values.

Step 7: Sketch the parabola:
Plot the vertex, intercepts, and additional points, then sketch a parabola that opens upwards (since $a=2>0$).

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Example 2: Graph the Quadratic Function

Given the quadratic function:

f(x)=−x2+6x−8f(x) = -x^2 + 6x – 8f(x)=−x2+6x−8

Step 1: Identify the coefficients:
$a=-1$, $b=6$, $c=-8$.

Step 2: Find the vertex:
The x-coordinate of the vertex is:

x=−62(−1)=62=3x = -\frac{6}{2(-1)} = \frac{6}{2} = 3x=−2(−1)6​=26​=3

To find the y-coordinate, substitute $x=3$ into the function:

f(3)=−(3)2+6(3)−8=−9+18−8=1f(3) = -(3)^2 + 6(3) – 8 = -9 + 18 – 8 = 1f(3)=−(3)2+6(3)−8=−9+18−8=1

So, the vertex is $(3,1)$.

Step 3: Axis of symmetry:
The axis of symmetry is $x=3$.

Step 4: y-intercept:
The y-intercept is f(0)=−(0)2+6(0)−8=−8f(0) = -(0)^2 + 6(0) – 8 = -8f(0)=−(0)2+6(0)−8=−8, so the y-intercept is $(0,-8)$.

Step 5: Find the x-intercepts:
Use the quadratic formula:

x=−6±62−4(−1)(−8)2(−1)=−6±36−32−2=−6±4−2x = \frac{-6 \pm \sqrt{6^2 – 4(-1)(-8)}}{2(-1)} = \frac{-6 \pm \sqrt{36 – 32}}{-2} = \frac{-6 \pm \sqrt{4}}{-2}x=2(−1)−6±62−4(−1)(−8)​​=−2−6±36−32​​=−2−6±4​​ x=−6±2−2x = \frac{-6 \pm 2}{-2}x=−2−6±2​

So, the x-intercepts are $x=2$ and $x=4$.

Step 6: Plot additional points:
To get more points, plug in values of $x$ around the vertex, such as $x=1$ and $x=5$, and calculate the corresponding y-values.

Step 7: Sketch the parabola:
Plot the vertex, intercepts, and additional points, then sketch a parabola that opens downwards (since $a=-1<0$).

 

5.8 Solution of linear programming problems by graphical method

Linear programming (LP) is a mathematical technique used to find the optimal solution to a problem with constraints, typically involving maximizing or minimizing a linear objective function. The graphical method is a visual approach used to solve LP problems when there are only two variables involved.

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Steps to Solve Linear Programming Problems by the Graphical Method

1. Define the Objective Function:
   The objective function is the function you want to maximize or minimize. It has the general form:

   $$Z=ax+by$$where $x$ and $y$ are the decision variables, and $a$ and $b$ are constants.

2. Identify the Constraints:
   Constraints are inequalities that limit the values of the decision variables. They are usually linear inequalities of the form:

   $$Ax+By\le C\text{or}Ax+By\ge C$$These constraints define the feasible region, which contains all possible solutions that satisfy the constraints.

3. Graph the Constraints:
   • Convert each inequality constraint into an equation by replacing the inequality symbol with an equals sign.
   • Graph each line on the coordinate plane.
   • Determine which side of the line satisfies the inequality by testing a point (typically $(0,0)$, if it is not on the line).
   • Shade the region that satisfies each constraint.
4. Find the Feasible Region:
   The feasible region is the area where all constraints overlap. This region represents all possible solutions that satisfy all constraints simultaneously.
5. Determine the Corner Points:
   The optimal solution (maximum or minimum) of a linear programming problem occurs at one of the corner points (vertices) of the feasible region. Find these corner points by solving the system of equations at each intersection of the constraint lines.
6. Evaluate the Objective Function at Each Corner Point:
   Substitute the coordinates of each corner point into the objective function to calculate the value of the objective function at each point.
7. Identify the Optimal Solution:
   • For a maximization problem, choose the corner point that gives the highest value of the objective function.
   • For a minimization problem, choose the corner point that gives the lowest value of the objective function.

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Example of Linear Programming Problem (Graphical Method)

Problem Statement:

Maximize the objective function:

$$Z=3x+2y$$

subject to the following constraints:

$$x+y\le 4(1)$$ $$x\le 2(2)$$ $$y\le 3(3)$$ $$x,y\ge 0(non-negativityconstraints)$$

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Step 1: Define the Objective Function:

The objective function is $Z=3x+2y$, and we want to maximize it.

Step 2: Identify the Constraints:

The constraints are:

1. $x+y\le 4$
2. $x\le 2$
3. $y\le 3$
4. $x,y\ge 0$

Step 3: Graph the Constraints:

• For the first constraint $x+y\le 4$, rewrite it as an equation $x+y=4$. Plot the line and shade the region below the line (since it’s a “less than or equal to” inequality).
• For the second constraint $x\le 2$, draw a vertical line at $x=2$ and shade to the left.
• For the third constraint $y\le 3$, draw a horizontal line at $y=3$ and shade below.
• The non-negativity constraints $x\ge 0$ and $y\ge 0$ confine the feasible region to the first quadrant.

Step 4: Find the Feasible Region:

The feasible region is the intersection of all shaded regions, forming a polygon bounded by the lines.

Step 5: Determine the Corner Points:

The corner points of the feasible region can be found at the intersection of the boundary lines:

• Intersection of $x=2$ and $y=3$ is $(2,3)$.
• Intersection of $x+y=4$ and $x=2$ is $(2,2)$.
• Intersection of $x+y=4$ and $y=0$ is $(4,0)$.
• Intersection of $x=0$ and $y=0$ is $(0,0)$.

So, the corner points are: $(2,3)$, $(2,2)$, $(4,0)$, and $(0,0)$.

Step 6: Evaluate the Objective Function at Each Corner Point:

Substitute the coordinates of each corner point into the objective function $Z=3x+2y$:

• At $(2,3)$: $$Z=3(2)+2(3)=6+6=12$$
• At $(2,2)$: $$Z=3(2)+2(2)=6+4=10$$
• At $(4,0)$: $$Z=3(4)+2(0)=12+0=12$$
• At $(0,0)$: $$Z=3(0)+2(0)=0$$

Step 7: Identify the Optimal Solution:

The maximum value of $Z$ is 12, which occurs at both points $(2,3)$ and $(4,0)$.

Thus, the optimal solution is either $(2,3)$ or $(4,0)$, with $Z=12$.

 

 

 

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