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- 1. CHAPTER 3 : SYSTEM OF EQUATION 3.1 3.2 3.3 3.4 LINEAR EQUATIONS QUADRATIC EQUATIONS SYSTEM OF EQUATIONS INEQUALITIES
- 2. 3.1 LINEAR EQUATION Introduction An equation is simply a statement that 2 quantities are the same. E.g. 45 = 9 + LHS = RHS LHS RHS A statement can be true or false is said to be a conditional (or open) equation Think of an equation as a pair of perfectly balanced as it is true for some values of the variable old – fashioned scales & not true for others. Equation = For example: 2x + 3 = 7 is true for x = 2 but is false for x = 8 or other values. Any number that makes the equation true is called a solution or root of the equation.
- 3. 3.1 LINEAR EQUATION Introduction A statement such as 2 x + 22 = 2( x + 11) is called an identity as it is true for all real numbers x. While solving equation, there may be no solution, one solution or there may be more than one solution. For example: Equation Solution Comment x + 9 = 12 x=3 Only one solution x2 = 81 x = 9 or – 9 Two solutions x=5+x No solution No solution x2 – 16 = (x – 4)(x + 4) All values of x Equation is true for all values of x.
- 4. 3.1 LINEAR EQUATION Solving Linear equation with 1 variable To solve an equation, we find the values of the unknown that satisfy it. How: Isolate the unknown / variable & solve its value. A linear equation is an equation that can be written in the form; ax + b = 0 where a & b are real numbers. Linear equations have only one root/ solution.
- 5. 3.1 LINEAR EQUATION Solving Linear equation with 1 variable x − 3 = 12 Add 3 to both sides x − 3 + 3 = 12 + 3 x = 15 x = 12 3 Multiply both sides by 3 x 3 = 3(12) 3 x = 36 x + 3 = 12 Subtract 3 from both sides x + 3 − 3 = 12 − 3 x=9 3 x = 12 Perform the same operation on both sides Divide both sides by 3 3 x 12 = 3 3 x=4 Always check in the original equation
- 6. Example 1 Solve the equation x − 7 = 3 x − ( 6 x − 8 ). Working Step x − 7 = 3x − 6x + 8 Parentheses removed x − 7 = −3 x + 8 4x −7 = 8 4x = 15 15 x = 4 x – terms combined 3x added to both sides 7 added to both sides Both sides divided by 4
- 7. PRACTICE 1 1. Solve the following equation : 2. Solve the given equation. a) 5( x − 3 ) = 10 a) 6 + 4L = 5 − 3L x 3−x b) 3( 4 − n ) = −n b) + = 12 2 4 c) 1.5 x − 0.3( x − 4 ) = 6 2 x 36 c) x + 10 = + d) 7 − 3(1 − 2 p ) = 4 + 2 p 3 5 5 3 − 5( 7 − 3V ) 2x − 3 2 e) 2V = d) = x +1 5 4
- 8. 3.2 QUADRATIC EQUATION Introduction A quadratic equation contains terms of the first and second degree of unknown is given by; 2 ax + bx + c = 0 Standard Standard form form where a, b, c are real numbers and a ≠ 0. A quadratic equation that contains terms of the 2nd degree only of the unknown is called a pure quadratic equation, can be expressed as ; ax 2 + c = 0 A value of the unknown that will satisfy the equation is called a solution or a root of the equation.
- 9. 3.2 QUADRATIC EQUATION Solving quadratic equation All quadratic equations have two roots . In pure quadratic equation , the roots are equal but of opposite sign. It can solve by extraction of the root. There are two methods of solving complete quadratic equation; By factorization By using quadratic formula
- 10. Example 2 Solve the equation : a) x 2 = 9 x = 9 = ±3 b) x 2 − 16 = 0 x 2 = 16 x = 16 = ±4 c) 5R 2 − 89 = 0 5R 2 = 89 89 R2 = 5 R= 89 = ±4.219 5 d) x 2 + 36 = 0 x 2 = −36 x = − 36 x = − 1 36 x = ±i 6 e) ( 2 x − 5 ) + 5 = 3 2 Where Where −1 = i
- 11. 3.2 QUADRATIC EQUATION Solving quadratic equation Factorization The expression ax 2 + bx + c is written as the product of two factors. If the product of 2 factors is zero, then one of those factors must be zero. Thus if mn = 0 then either m = 0 or n = 0. Only be used if the quadratic expression can be factorized completely.
- 12. Example 3 Solve the quadratic equation : a) x 2 − 6 x = 0 x ( x − 6) = 0 x = 0 or x − 6 = 0 x=6 b) 5 x 2 − 2 x = 0 x ( 5 x − 2) = 0 x = 0 or 5 x − 2 = 0 x= 2 5 c) x 2 + x − 12 = 0 (x+4) x 4 = 4x x -3 = -3x -12 =x ( x + 4)( x − 3 ) = 0 x + 4 = 0 or x − 3 = 0 x = −4 x =3 (x-3) x2
- 13. Example 3 Solve the quadratic equation : 2 d) x + 7 x + 12 = 0 ( x + 4)( x + 3) = 0 x + 4 = 0 or x + 3 = 0 x = −4 x = −3 e) x 2 + 6 x − 7 = 0 ( x − 1)( x + 7) = 0 x − 1 = 0 or x + 7 = 0 x =1 x = −7 f) x ( x − 1) = 6 x2 − x − 6 = 0 ( x + 2)( x − 3) = 0 x + 2 = 0 or x − 3 = 0 x = −2 x =3 (x+4) x 4 = 4x x 3 = 3x 12 = 7x (x+3) x2
- 14. PRACTICE 2 Solve the quadratic equation : a) x 2 + 5 x − 14 = 0 b) 3 x 2 + 10 x = 8 c) 3 x 2 − 5 x − 2 = 0 d) 6 x 2 − 11x − 35 = 0 e) ( x − 2)( x + 4 ) = 16
- 15. 3.2 QUADRATIC EQUATION Solving quadratic equation – Quadratic Formula For any quadratic equation ax2 + bx + c = 0, the solutions for x can be found by using the quadratic formula: − b ± b 2 − 4ac x= 2a The quantity b2 – 4ac is called discriminant of the quadratic equation that characterize the solution of the quadratic equations: If b2 – 4ac = 0 one real solution If b2 – 4ac > 0 2 real and unequal solutions If b2 – 4ac < 0 no real but 2 imaginary solutions (complex number)
- 16. Example 4 1. Solve the equation 2k 2 + 8k − 5 = 0 by using the quadratic formula Step 1 Identify a, b & c, by comparing with ax2 + bx + c = 0. a = 2, b = 8, c = -5 Step 2 − b ± b 2 − 4ac Substitute into the formula: k = 2a − 8 ± 8 2 − 4( 2)( − 5 ) = 2( 2) = Step 3 The solutions are: − 8 + 104 4 = 0.5495 k1 = − 8 ± 104 4 − 8 − 104 4 = −4.5495 k2 = b22--4ac > 0, b 4ac > 0, 2 real & 2 real & unequal unequal roots roots
- 17. Example 4 2. Solve the equation 4 p 2 − 12 p + 9 = 0 by using the quadratic formula Step 1 Identify a, b & c, by comparing with ax2 + bx + c = 0. a = 4, b = -12, c = 9 Step 2 − b ± b 2 − 4ac Substitute into the formula: p = 2a = = Step 3 The solution is: 12 3 p= = 8 2 − ( − 12) ± 12 ± 0 8 ( − 12) 2 − 4( 4)( 9 ) 2( 4 ) b22--4ac = 0, b 4ac = 0, 1 real root 1 real root
- 18. Example 4 3. Solve the equation 3k 2 − 5k + 4 = 0 by using the quadratic formula Step 1 Identify a, b & c, by comparing with ax2 + bx + c = 0. a = 3, b = -5, c = 4 Step 2 − b ± b 2 − 4ac Substitute into the formula: k = 2a b22--4ac < 0, b 4ac < 0, no real, 2 no real, 2 imaginary roots imaginary roots Step 3 = 5 + i 23 6 = 0.83 + i 0.8 ( − 5 ) 2 − 4( 3)( 4) 2( 3 ) = 5 ± − 23 6 = The solutions are: k1 = − ( − 5) ± 5 ± − 1 23 5 ± i 23 = 6 6 5 − i 23 6 = 0.83 − i 0.8 k2 =
- 19. PRACTICE 3 Solve the quadratic equation below by using quadratic formula : a) x 2 + 7 x − 8 = 0 b) 9 x 2 + 49 = 42 x c) 3 x 2 + 7 x + 3 = 0 3 d) = 2R 5−R e) s 2 = 9 + s (1 − 2s )
- 20. 3.3 SYSTEMS OF EQUATION Introduction A system of equations contains 2 or more equations. Each equation contains 1 or more variables. a1x + b1y = c1 a2 x + b2 y = c 2 Where a & b are Where a & b are coefficients of xx& y, coefficients of & y, ccis a constant. is a constant. A solution of the system is any pair of values (x , y) that satisfies the both equations. 2 methods that are usually used: Substitution method Elimination method
- 21. 3.3 SYSTEMS OF EQUATION Substitution Method Step for solving by substitution: Pick 1 of the equation & solve for 1 of the variables in terms of the remaining variables. Substitute the result in the remaining equations. If 1 equation in 1 variable results, solve the equation. Otherwise, repeat Step 1 – 3 again. Find the values of the remaining variables by back – substitution. Check the solution found.
- 22. Example 5 a Solve the following system of equations x − 3 y = 6 → Eq 1 Solution: 2 x + 3 y = 3 → Eq 2 Step 1 From Eq 1, make x as subject: x = 6 + 3 y Step 2 Substitute x in Eq 2 : 2( 6 + 3 y ) + 3 y = 3 Step 3 Solve for y : 12 + 6 y + 3 y = 3 9 y = −9 y = −1 Step 4 Substitute y in Eq 1, to solve for x: x = 6 + 3( −1) = 3 Step 5 Check the solution : Eq 1, 3 – 3(-1) = 6 Eq 2, 2(3) + 3(-1) = 3
- 23. Example 5 b Solve the following system of equations x + 2y = 8 → Eq 1 The solution is The solution is Solution: the intersection the intersection 3 y − 4 x = 1 → Eq 2 Step 1 From Eq 1, make x as subject: x = 8 − 2y Step 2 Substitute x in Eq 2 : 3 y − 4( 8 − 2 y ) = 1 Step 3 Solve for y : 3 y − 32 + 8 y = 1 11y = 33 y =3 Step 4 Substitute y in Eq 1, to solve for x: x = 8 − 2( 3 ) = 2 Step 5 Check the solution : Eq 1; 2 + 2(3) = 8 Eq 2; 3(3) – 4(2) = 1 point (2,3) point (2,3)
- 24. Example 5 c Solve the following system of equations The lines are 4 x + 2y = 8 → Eq 1 The lines are Solution: parallel & no parallel & no 2 x + y = 5 → Eq 2 Step 1 From Eq 2, make y as subject: y = 5 − 2 x Step 2 Substitute y in Eq 1 : 4 x + 2( 5 − 2 x ) = 8 4 x + 10 − 4 x = 8 10 = 8 Step 3 The result is a false statement. Therefore the system is inconsistent. It means that there is no solutions for this system of equations intersection intersection
- 25. Example 5 d Solve the following system of equations Solution: 2 x + y = 4 → Eq 1 −6 x − 3 y = −12 → Eq 2 The lines is The lines is coincide, the coincide, the systems has systems has infinitely many infinitely many solution solution Step 1 From Eq 1, make y as subject: y = 4 − 2 x Step 2 Substitute y in Eq 2 : − 6 x − 3( 4 − 2 x ) = − 12 − 6 x − 12 + 6 x = − 12 − 12 = − 12 Step 3 The result is a true statement. Therefore the system is dependent. It means that there is no unique solutions can be determined.
- 26. 3.3 SYSTEMS OF EQUATION Elimination Method Step for solving by elimination: Write x’s, y’s & numbers in the same order in both equations. Compare x’s & y’s in the 2 equations & decide which unknown is easier to eliminate. Make sure the number of this unknown is the same in both equations. Eliminate the equal terms by adding or subtracting the 2 equations Solve the resulting equation & substitute in the simpler original equation in order to find the value of the other unknown. Check the solution found.
- 27. Example 6 a Solution: Solve the following system of equations 7 y = 6 − 4 x → Eq1 3 x = 2y + 19 → Eq 2 Multiply by a Multiply by a constants to constants to get same get same coefficient of x coefficient of x Step 1 Rearrange the eq: 4 x + 7 y = 6 →Eq1 3 x − 2 y = 19 →Eq 2 Step 2 To eliminate x, Eq1 multiply by 3, Eq 2 multiply by 4 12 x + 21y = 18 → Eq 3 12 x − 8 y = 76 → Eq 4 Step 3 Eq 3 – Eq 4, solve for y : 21y − ( − 8 y ) = 18 − 76 29 y = −58 y = −2 Step 4 Substitute y in Eq 1, to solve for x: 4 x = 6 − 7( − 2) 20 x= =5 4 Step 5 Check the solution :
- 28. Example 6 b Solve the following system of equations 3 x − 2y = 4 → Eq1 x + 3 y = 2 → Eq 2 Step 1 To eliminate x, Eq2 multiply by 3; 3 x − 2y = 4 → Eq1 3 x + 9 y = 6 → Eq 3 Solution: Step 2 Eq 1 – Eq 3, solve for y : − 2 y − 9 y = 4 − 6 −11y = −2 2 y = 11 2 Step 3 Substitute y in Eq 2, to solve for x: x + 3 = 2 11 11x + 6 = 22 Step 4 Check the solution : x= 16 11
- 29. PRACTICE 4 Solve the following system by using A. Substitution method B. Elimination method 1) 2 x − y = 5 1) − 2 x + 2y = 5 x + 6y = 1 6 x + 2y = −5 2) 9m − 6n = 12 2) 2m + 6n = −3 2m + 6n = 4 − 6m − 18n = 5 3) 2R + S = 4 3) − 5K + 2L = −4 − 6R − 3S = −12 10K + 6L = 3 4) x 2 + y 2 = 4 x+y =2 4) y = x 2 + 2 x + 2 y = 3x + 8
- 30. 3.4 INEQUALITY Introduction An inequality is a statement involving 2 expressions separated by any of the inequality symbols below. Inequality Meaning a>b a is greater than b a<b a is less than b a≥b a is greater than or equal to b a≤b a is less than or equal to b A real number is a solution of an inequality involving a variable if a true statement is obtained when the variable is replaced by the number. The set of all real numbers satisfy the inequality is called the solution set.
- 31. 3.4 INEQUALITY Inequality, Interval Notation & Line Graph Inequality Notation Interval Notation a≤x≤b a<x≤b a≤x<b a<x<b x≥b x>b x≤a x<a [ a, b ] ( a, b ] [ a, b ) ( a, b ) [ b, ∞ ) ( b, ∞ ) ( -∞, a ] ( -∞, a ) a Graph b
- 32. 3.4 INEQUALITY Properties of Inequality Properties Example Let a, b & c be any real number If a < b and b < c, then a < c If 2 < 3 & 3 < 8 so 2 < 8 If a < b, then a ± c < b ± c 5 < 7, 5 + 3 < 7 + 3, 8 < 10 5 < 7, 5 - 3 < 7 - 3, 2 < 4 If a < b, c > 0 then ac < bc 5 < 7, 5(2) < 7(2), 10 < 14 If a < b, c < 0 then ac > bc 5 < 7, 5(-2) > 7(-2), -10 > -14 If a < b, c > 0 then a/c < b/c 5 < 7, 5/2 < 7/2, 2.5 < 3.5 If a < b, c < 0 then a/c > b/c 5 < 7, 5/(-2) > 7/(-2), -2.5 > -3.5 Similar properties hold if < between a and b is replaced by ≥ , > , or ≤ .
- 33. Example 7 Solve each of the following inequalities : a) 3 x + 10 < 4 x − 7 b) 4 x + 5 ≥ 2 x + 9 3 x − 4 x < −7 − 10 − x < −17 Divide by Divide by –ve no , , ∴ x > 17 –ve no reverse reverse sign sign c) 3 − 7 x < 2 x + 21 d) 6 x + 4 2x − 1 > 5 3
- 34. Example 8 Solve each of the following inequalities : a) − 11 < 3 x − 2 < 4 − 11 < 3 x − 2 and 3x − 2 < 4 − 9 < 3x 3x < 6 −3< x x<2 Therefore − 3 < x < 2 or x = { − 2, − 1, 0,1} b) 10 − 3 x ≤ 2 x − 7 < x − 13 10 − 3 x ≤ 2 x − 7 and − 5 x ≤ −17 Solution Solution set set 2 x − 7 < x − 13 x < −6 17 x≥ 5 Therefore there is no solution. Impossible to have x that Impossible to have x that satisfies both inequalities satisfies both inequalities
- 35. PRACTICE 5 1. If x is an integer, find the solution set for 3 ≤ x + 1 ≤ 5. 1 2. Determine the range of values of x satisfying < −4. x 3. Solve the inequality : a) 3( 2 x + 1) − 5( x − 1) ≥ 7 x b) − 1 ≤ 2 x − 5 < 7 4. Find the minimum cost C (in dollars) given that 5(C − 25 ) ≥ 1.75 + 2.5C 5. Find the maximum profit P (in dollars) given that 6( P − 2500 ) ≤ 4( P + 2400 )

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