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Engineering Mathematics

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Presentation on theme: "Engineering Mathematics"β€” Presentation transcript:

1 Engineering Mathematics
Kurdistan Technical Institute (KTI) Engineering Mathematics Lecture – 6 / Linear Equations with One Variable Fall Semester 2018 Instructor Twana. A, Hussein– MSc . Structural Engineering Civil Engineering Dept. Ishik University Department of Petroleum Refining

2 Ishik University Linear Equations Linear equations are first degree polynomial equations. A linear equation with one variable can be written in the form of 𝒂𝒙 + 𝒃 =𝒄. Where, a, b and c are all real numbers and π‘Žβ‰ 0 .

3 (to simplify or evaluate)
Ishik University Linear Equations Distinguish between expressions and equations. Equations and inequalities compare algebraic expressions. An equation is a statement that two algebraic expressions are equal. An equation always contains an equals symbol, while an expression does not. 3x – 7 3x – 7 = 2 Left side Right side Equation (to solve) Expression (to simplify or evaluate) .

4 Linear Equations with One Variable
Ishik University Linear Equations with One Variable So, can you tell these linear equations are linear equation with one variable or not? 𝒙 +πŸπ’š = 𝟏𝟎 NO, because there are 2 variables x and y 𝒙+ πŸ‘ 𝟐 =πŸ” = π‘₯ 2 +6π‘₯+3 NO, because π‘₯ 2 is not a linear equation πŸ“ πŸπ’™ +πŸ‘=π’™βˆ’πŸ“ NO, because there is variable at denominator .

5 Linear Equations with One Variable
Ishik University Linear Equations with One Variable Strategy for solving Linear Equations. Isolate variable terms on one side and constant terms on the other side. Problem-1. Solve the following Linear Equations. π‘₯βˆ’2=8 2π‘₯+4=0 8βˆ’2π‘₯=6 .

6 Solve Linear Equations with One Variable
Ishik University Solve Linear Equations with One Variable Example – 1: Solve the following Linear Equations. π’™βˆ’πŸ=πŸ– π‘₯=8+2 π‘₯=10 Check for x πŸπŸŽβˆ’πŸ=πŸ– πŸ–=πŸ– πŸπ’™+πŸ’=𝟎 2π‘₯=βˆ’4 π‘₯=βˆ’2 Check for x 𝟐 βˆ’πŸ +πŸ’=𝟎 βˆ’πŸ’+πŸ’=𝟎 .

7 Solve Linear Equations with One Variable
Ishik University Solve Linear Equations with One Variable Example – 1: Solve the following Linear Equations. 8βˆ’πŸπ’™=πŸ” βˆ’2π‘₯=6βˆ’8 βˆ’1 βˆ’2π‘₯=βˆ’2 2π‘₯=2 π‘₯=1 Check for x 8βˆ’πŸπ’™=πŸ” πŸ–βˆ’πŸ 𝟏 =πŸ” πŸ–βˆ’πŸ=πŸ” πŸ”=πŸ” .

8 Solve Linear Equations with One Variable
Ishik University Solve Linear Equations with One Variable To generate equivalents equations you can use the following operations. Simplify an expression by removing grouping symbols and combining like terms. Add/subtract the same real number/expression on both sides of the equation. Multiply/divide on both sides of the equation by the same nonzero quantity Interchange two sides of the equation. .

9 Solve Linear Equations with One Variable
Ishik University Solve Linear Equations with One Variable Example – 2: Solve the following linear equation. π’™βˆ’πŸ’=πŸ‘ π’™βˆ’πŸ“ π‘₯βˆ’4=3π‘₯βˆ’15 π‘₯βˆ’3π‘₯=4βˆ’15 βˆ’2π‘₯=βˆ’11 βˆ’1 βˆ’2π‘₯=βˆ’11 2π‘₯=11 π‘₯=5.5 Multiply number with inside parenthesis Put all variables at the left side and numbers at the right side. Add or subtract variables with same variables and number with number if there is any. Multiply both side by -1 Divide both side by 2 to get the value of x .

10 Solve Linear Equations with One Variable
Ishik University Solve Linear Equations with One Variable Example – 2: Solve the following linear equation. π’™βˆ’πŸ’=πŸ‘ π’™βˆ’πŸ“ Check for x π‘₯βˆ’4=3π‘₯βˆ’15 5.5βˆ’4=3 5.5 βˆ’15 1.5=16.5βˆ’15 1.5=1.5 .

11 Solve Linear Equations with One Variable
Ishik University Solve Linear Equations with One Variable Example – 3: Solve the following linear equation. π’™βˆ’πŸπŸŽ=πŸ“ πŸβˆ’πŸ“ πŸπ’™βˆ’πŸ π’™βˆ’πŸπŸŽ=πŸ“ πŸβˆ’πŸπŸŽπ’™+𝟏𝟎 π’™βˆ’πŸπŸŽ=πŸπŸŽβˆ’πŸ“πŸŽπ’™+πŸ“πŸŽ 𝒙+πŸ“πŸŽπ’™=𝟏𝟎+πŸ“πŸŽ+𝟏𝟎 πŸ“πŸπ’™=πŸ•πŸŽ 𝒙= πŸ•πŸŽ πŸ“πŸ .

12 Solve Linear Equations with One Variable
Ishik University Solve Linear Equations with One Variable Example – 4: Solve the following linear equation. πŸπ’™βˆ’πŸ=π’™βˆ’πŸ“ πŸπ’™βˆ’πŸ πŸ‘π’™βˆ’πŸ‘ πŸπ’™βˆ’πŸ=π’™βˆ’πŸ“ πŸπ’™βˆ’πŸ”π’™+πŸ” πŸπ’™βˆ’πŸ=π’™βˆ’πŸπŸŽπ’™βˆ’πŸ‘πŸŽπ’™+πŸ‘πŸŽ πŸπ’™βˆ’π’™+πŸπŸŽπ’™+πŸ‘πŸŽπ’™=πŸ‘πŸŽ+𝟏 πŸ’πŸπ’™=πŸ‘πŸ 𝒙= πŸ‘πŸ πŸ’πŸ .

13 Identification of Linear Equation
How to Determine the Solution Sets of Linear Equations

14 Identification of Linear Equation
Ishik University Identification of Linear Equation Identify conditional equations, contradictions, and identities. Type of Linear Equation Number of Solutions Indication when Solving Conditional One Final line is x = a number. Identity Infinite; solution set {all real numbers} Final line is true, such as 0 = 0. Contradiction None; solution set  Final line is false, such as 0 = –20 . .

15 Identification of Linear Equation
Ishik University Identification of Linear Equation Example – 5: Solve each equation, Decide whether it is conditional, identity, or contradiction. πŸ“π±βˆ’πŸ—=πŸ’ π±βˆ’πŸ‘ πŸ“π’™βˆ’πŸ—=πŸ’π’™βˆ’πŸπŸ πŸ“π’™βˆ’πŸ’π’™=πŸ—βˆ’πŸπŸ 𝒙=βˆ’πŸ‘ So, this equation is Conditional And solution set = {-3} .

16 Identification of Linear Equation
Ishik University Identification of Linear Equation Example – 5: Solve each equation, Decide whether it is conditional, identity, or contradiction. πŸ“π±βˆ’πŸπŸ“=πŸ“ π±βˆ’πŸ‘ πŸ“π’™βˆ’πŸπŸ“=πŸ“π’™βˆ’πŸπŸ“ πŸ“π’™βˆ’πŸ“π’™=πŸπŸ“βˆ’πŸπŸ“ 𝟎=𝟎 So, this equation is Identity And solution set = {real numbers} .

17 Identification of Linear Equation
Ishik University Identification of Linear Equation Example – 5: Solve each equation, Decide whether it is conditional, identity, or contradiction. πŸ“π±βˆ’πŸπŸ“=πŸ“ π±βˆ’πŸ’ πŸ“π’™βˆ’πŸπŸ“=πŸ“π’™βˆ’πŸπŸŽ πŸ“π’™βˆ’πŸ“π’™=πŸπŸ“βˆ’πŸπŸŽ 𝟎=βˆ’πŸ“ So, this equation is Contradiction And solution set = {} or no solution .

18 Identification of Linear Equation
Ishik University Identification of Linear Equation Example – 6: Solve each equation, Decide whether it is conditional, identity, or contradiction. 𝒙+𝟐 𝟐 βˆ’ 𝒙 πŸ‘ =πŸ“ πŸ” 𝒙+𝟐 𝟐 βˆ’ 𝒙 πŸ‘ =πŸ“ πŸ‘π’™+πŸ”βˆ’πŸπ’™=πŸ‘πŸŽ 𝒙=πŸπŸ’ So, this equation is Conditional And solution set = {24} .

19 Home Work πŸ”π± +πŸ’ πŸ‘ + 𝐱 πŸ‘ =𝟐𝐱+ 𝐱 πŸ‘ + πŸ’ πŸ‘ 𝐱 πŸ’ + πŸ’ πŸ” = 𝐱 𝟐
Ishik University Home Work Example – 7: Solve each equation, Decide whether it is conditional, identity, or contradiction. 3 𝟐𝐱+πŸ‘ βˆ’πŸ 𝟐𝐱+𝟐 =𝟐𝐱+πŸ’ πŸ”π± +πŸ’ πŸ‘ + 𝐱 πŸ‘ =𝟐𝐱+ 𝐱 πŸ‘ + πŸ’ πŸ‘ 𝐱 πŸ’ + πŸ’ πŸ” = 𝐱 𝟐 .


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