Solving Linear and Quadratic Equations

Grade 10 CAPS Mathematics Series Outcomes for this Topic

In this Topic we will solve: In this DVD you will:

Linear Equations (Unit 1) • Revise factorization. Unit 1. Quadratic Equations (Unit 2) • Revise simplification of algebraic fractions. Unit 2. Two Linear Equations Simultaneously (Unit 3) • Discuss when trinomials can be factorized. Unit 3.

2 Unit 1

Solving Linear Equations

Grade 10 CAPS Mathematics Series Outcomes for Unit 1

In this Unit we will focus on:

• Examples of linear equations

• General steps to solve linear equations

• How to verify the correctness of a solution

• How to verify the solution (root) graphically Examples of Linear Equations

1. 3x  4 5 General remarks : • Simplest equation to solve is a linear equation 3 x • A linear equation is an equation where the power 2.  4 31x  of the variable is one • Also known as an equation of the first degree 4x • We will learn how to find what value of x makes 3.  6  7x  4 3 both sides of the given linear equation true • There is at most one solution or root for a linear equation Consider the following :

1. From solution to equation: 5 3x  4 25 x 33 3 x  9  4 3 x  4 5  5 5 3 x  4  25   2  22 2. From equation to solution (Reversing process above): 5 3x  4 25  25 5 3x  4  25 3x  4  5 43 3 x  9  x  3 22 General steps to solve linear equations

General steps : 1. If there are any fractions multiply all the terms by the LCM of the denominators. 2. Remove brackets where possible. 3. Transfer terms containing the variable to the LHS of the equation (Transpose, change signs). 4. Transfer the other terms to the RHS of the equations (Transpose, change sign). 5. Simplify both sides where possible. 6. Divide by the coefficient of the unknown. 7. Test the correctness of the solution. Apply general steps to solve a linear equation

x3 x 2 x  3 Solve for x if 3   2  4 2 3 6 Step 1: 33 x 18  4 x  2422  x  3 Multiply each term by LCM 12 Step 2 : 9x 54  4 x  24  4 x  6 Remove brackest  Steps 3 & 4 : 9x 4 x  4 x  24  54  6 Transpose, change signs Step 5 : 9x  24 Simplify both si des 24 8 2 Step 6 : x      2 Divide by coefficient of x 9 3 3 Step 7 : Test correctness of solution Substitute solution into both sides of original equation:

8 16 2  3  3 81381   985861   1692561  LHS 3 2 and RHS  233  2  2  2 34233   292918  6 6 18 1818  LHS RHS Checking the solution graphically

x3 x 2 x  3 In the linear equation 3   2  we showed and 4 2 3 6 82 tested that x     2 is the solution or root for this equation. 33

• In the solution process the linear equation is reduced to the form x k , k . • A linear equation of the form x  k , k  is a straight line to the y  axis. • k is the root for this linear equation. • k;0 is the point of intersection between x k and y  0  x  axis . Solve a more complex linear equation

1 1 1 Solve for x if  6x2 x  15 9  4 x 2 6 x 2  19 x  15 Factorize denominators: 1 1 1  3523x x   3232  x  x  3523 x  x   Multiply by LCM 3xx  5 2  3 2x  3 : 53  2xx 3  3 5  2x 3 x   and x   32 2x  3 x  2 x   3  3  5 3x   11 11 x   is the solution to given equation. 3 Use your calculator to test the correctness of solution. Tutorial 1: Solving Linear Equations

Apply some or all of the general steps, where applicable, to solve the following linear equations. Test in each case whether the solution is correct: PAUSE Topic

1. 10xx 1  3  2  3 • Do Tutorial 1     • Then View Solutions 3 2xx 4 3 2.  x  3   2  2 3 6 12yy5 3.  y  5    0 4 3 3 2 6 1 1 1 4.  24x2 2 x  15 9  16 x 2 24 x 2  38 x  15 Tutorial 1: Problem 1: Suggested Solution

1. Solve for x and test solution : 10 x 1  3  2 x  3 10xx  10  3  2  3  Remove brackets 10xx  2  3  3  10  Transpose and change signs 12x   4  Simplify both sides 1 x    Divide by coefficient of unknown 3 TEST 1 10 xx 1  3  2  3 1 2 20 2 LHS 10   1  10    6  6,6 3 3 3 3 1   2  2 2 RHS 3 2  3  3  3  3 3  6 3   3  3 3 LHS RHS Tutorial 1: Problem 2: Suggested Solution

3 2xx 4 3 2. Solve the equation and test the solution:  x  3   2  2 3 6

9 x  3  4 x  12  4 x  3 Multiply both sides by LCD 6 9x  27  4 x  12  4 x  3 Remove brackets 9x  4 x  4 x  12  3  27 Transpose and change sides 9x   12 Simplify both sides 4 x    Divide by coefficient of unknown 3

TEST (Use Calculator) 16 3 3 4 8 613 61 LHS   3   and RHS  2    LHS  RHS 2 3 9 18 6 18 Tutorial 1: Problem 3: Suggested Solution

1 2yy 5 3. Solve equation and test solution:  y 50     4 3 3 2 6

y5 2 y 5 y       0 Often better to first remove brackets 4 4 9 3 6 9y  45  8 y  60  6 y  0  Multiply by LCD 36 9y  8 y  6 y  45 60 Transpose and change signs 7y 105 Simplify both sides y 15 Divide by coefficient of unkn own

TEST (Use Calculator) 1 2 5 15 LHS 15  5  5    0  RHS  LHS  RHS 4 3 2 6 Tutorial 1: Problem 4: Suggested Solution

1 1 1 4. Solve equation and test solution:  24x2 2 x  15 9  16 x 2 24 x 2  38 x  15 1 1 1    Factorize denominators 4365x x   4343 x  x   4365 x  x   4x 3  6 x  5  4 x  3 Multiply by LCD 4x 3 4 x 3 6 x  5 4xxx  6  4  3  3  5 Transpose and change signs 11 6xx   11    Simplify and Divide by coefficient of unknown 6 TEST (Use Calculator) : 1 1 1 LHS 22   11   11   11  26 2    2      15 9  16     6   6   6  11 RHS2   LHS  RHS 11   11  26 24    38      15 66    Unit 2

Solving Quadratic Equations Grade 10 CAPS Mathematics Series Outcomes for Unit 2

In this Unit we will focus on:

• Examples of quadratic equations

• The standard form of a quadratic equation

• Zero products

• General steps to solve quadratic equations

• Different formats of quadratic equations

• Graphic interpretation of quadratic equations Examples of Quadratic Equations

General remarks : • A quadratic equation is an equation where the power of the variable is at most two • Such an equation is also called an equation of the second degree • We will ony solve quadratic equations by factorization • Quadratic equations has at most two solutions or roots Examples of quadratic equations :

1. 3xx2  4 5 31 x 2.  4x  3  x  4 x 3 x  1 with x   3x  1 3 4x 3.  6  7x2  4 3 Standard form of a Quadratic Equation

The standard form of a quadratic equation is ax2  bx  c  0 where: • x is a variable • ax is the coefficient of 2 • bx is the coefficient of • cx is independent of also known as the constant Zero Products

For all real numbers ab and it follows that: If a b  0 then a  0 or b  0 So we have that: • If ab 0 then may have any real value • If ab  0 and a  0 then b  0 • If ab  0 and b  0 then a  0 In general : If the product of two or more factors is equal to zero, at least one of the factors is zero. Example : Given 2 xx 3 2  1  0 then because 2 0 either xx  3  0 or 2  1  0 Apply general steps to solve a quadratic equation

Solve for x if 6 x2  6 13 x Step 1: 6xx2  13  6  0 Write equation in standard form Step 2 :  3xx 2 2  3  0 Factorize the LHS of the equation 23 Step 3 : 3x 2  0  x  or 2 x  3  0  x  32 23 xx  or  Determine the two possible solutions 32 Step 4 : Test correctness of solution(s): Use calculator where needed 2 2 2 2 • x  is a solution because 6   13   6  0 and 3 3 3 2 3 3 3 • x  is a solution because 6   13   6  0 2 2 2 Quadratic Equations in Disguise

NOTE : Sometimes an equation might not look quadratic at first glance!!

c Case 1: ax b Multiply both sides by x and write in standard form x ax22  bx  c  ax  bx  c  0 where x  0 1 Case 2 :  c Cross Multiply ax2  bx b c ax22 bx 1  ac x  bc x 1 0 where x ax  b 0 x 0 and x  a

Case 3 : ax b cx Square both sides ax b  c2 x 2  c 2 x 2  ax  b 0 where ax  b  0 Example 1: Solving Quadratic Equations

Example 1: Solve for x if 2 x2  5 x 12 Solution : 2x2  5 x  12  0  2 x  3 x  4  0 3 2x  3  0 or x  4  0  x   or x  4 2 3 xx   or  4 are both solutions 2

2 33    Check : 2    5      12  0 22    2 and 2 4  5  4  12  0 Example 1: Graphical Solution

3 We showed algebraically that for 2x2  5 x  12, both x   and x  4 are solutions. 2

Consider the graphs of the parabola and line defined by y 2 x2  5 x and y  12 respectively.

Then  x; y / y 2 x2  5 x  x ; y / y  12 3 ;12 ; 4;12 the solution set 2

y coordinates of parabola and straight line 3 are equal when xx  or  4. 2 Example 2: Solving Quadratic Equations

Grade 11 Level Example 2: Solve for x if 14 5 x x

Positive square root Solution : 14 5x  x22  x  5 x  14  0  x  7 x  2  0 x 7  0 or x  2  0  x   7 or x  2 x 2 is the only solution

Check : 14 5   7  49   7  and 14 5  2  4  2 Example 2: Graphical Solution

Grade 11 Level We showed algebraically that 14 5xx xx 2 is a solution but   7 is not a solution. Consider the graphs of one branch of a parabola defined by y 14  5 x and the straight line defined by y  x. Note : The branch defined by yx  14  5 is excluded. From the graphs it should be clear why x  2 is a solution but x  7 is not a solution. Example 3: Solving Quadratic Equations

28 Example 3 : Solve for xx if  3 x Solution : x x3  28  x2  3 x  28  0  x  7 x  4  0 x 7  0 or x  4  0  x  7 or x   4 xx 7 or   4 are both solutions 28 Check : If x  7 then 7  3  4  7  28 and if x   4 then  4  3   7   4  Example 3: Graphical Solution

28 Algebraically we showed that x 3   x  7 or x   4 x

Consider the graphs of the hyperbola defined by 28 y and the straight line defined by y  x  3 x

From the graphs it should be clear why xx7 and   4 are both solutions. Tutorial 2: Solving Quadratic Equations

Solve the following quadratic equations and test the correctness of the solutions: 1. 14xx2  17 6 Grade 11 Level 2. xx 2  2  PAUSE Topic 15 3.  x • Do Tutorial 2 x  2 • Then View Solutions Tutorial 2: Problem 1: Suggested Solution

1. Solve 14xx2  17 6 and test the correctness of the solutions.

14x22 17 x  614  x  17 x  60   7 x  22 x  3  0 23 7x  2  0 or 2 x  3  0  x  or x   72 Check (Use Calculator) : 2 22 • 14   17   6 and 77 2 33    • 14     17      6 22    Tutorial 2: Problem 2: Suggested Solution

Grade 11 Level 2. Solve xx 2  2  and test the correctness of the solutions.

x2 2 x x 2 2 x2  x 2 4 4 x  x2

x2 560 x x 2 x  30 x 2 or x  3

x 2 is the only solution

Check : • If x  2 then x  2  0 and 2  x  0  x  2 is a solution • If x  3 then x  2  1 and 2  x   1  x  3 is not a solution Tutorial 2: Problem 3: Suggested Solution

15 3. Solve  x and test the correctness of the solutions. x  2 15 x  x x 2  15  x2  2 x  15  0 x  2 x5 x  3 0 x 5 0 or x  3 0 x 5 or x  3 xx  5 and  3 are both solutions Check : 15 15 • x 5   5 x x 5 is a solution x 23 15 15 • x  3    3  x  x  3 is a solution x  25 Unit 3

Solving Simultaneous Linear Equations

Grade 10 CAPS Mathematics Series Outcomes for Unit 3

In this Unit we will focus on:

• The meaning of solving a system of linear equations in two unknowns.

• Solving such a system:

• graphically,

• by means of the substitution method,

• by means of the elimination method. System of Linear Equations in two unknowns

General remarks : • Thus far we only solved equations in one unknown variable. • When two unknown variables need to be solved for, two equations are required and these equations are known as simultaneous equations. • The solutions to the system of simultaneous equations are the values of the unknown variables which satisfy the system of equations simultaneously, that means all equations at the same time. • In general: If there are nn unknown variables, then equations are required to (possibly) obtain a solution for each of the n variables. Examples of sytems of simultaneous linear equations : Focus in this lesson will be on : 2 unknowns 3 unknowns 4 unknowns Two variables 2x 3 y  5 2 x  3 y  4 z  7 2 a  2 b  3 c  4 d  12 Two linear equations 3x 5 y  7 3 x  5 y  8 z  12 3 a  4 b  2 c  7 d  11 5x 6 y  10 z  15 5 a  3 b  2 c  4 d  19 Solving this 2 2 system of 6a 4 b  7 c  8 d  16 linear equations simultaneously Method 1: Graphic Solution of a 2x2 system of linear equations

Solve the following system Method : of linear equations graphically: Step 1: Sketch the two lines Step 2: Find the point of intersection xy5 1 Step 3: Write down the solution for the system  Step 4: Test the correctness of the solution 3xy3 2 Assume:

1  x; y / x  y  5 and

2  x; y / 3 x  y  3

12   2;3 or xy 2 and 3 is a solution for the system or  2;3 is a point on both lines 12 and Check: x y  2  3  5 and 3 x  y  3  2  3  3 Method 2: Solving a 2x2 system of linear equations by means of substitution

Solve the following system of linear equations by means of the substitution method: 2x y  7 12 and 3 x 2 y  7

Method :  Step 1: yx 7 2 3  Use 1 and make yx  or  the subject of the formula  Step 2 : 3xx 2 7  2  7 Substitute yx into 2 to obtain a linear equation in  Step 3 : 3x 14 4 x  7  7 x  21  x  3 Solve linear equation in x  Step 4 : y  7  2  3  1 Back-substitute xy 3 into 3 to determine corresponding Step 5 : x  3 and y  1 is the solution for the system Write down the solution Step 6 : 2 3  1  7  solution satisfies 1 3 3  2  1  7  solution satisfies 2 Check correctness of sol u tion Method 3: Solving a 2x2 system of linear equations by means of elimination

Solve by means of the elimination method the following system of linear equations: 2x3 y  5 1 and 3 x  2 y  6 2 • Decide to eliminate Makexx coefficients o f in equations identical 1 3: 6915xy 3 Check or test correctness of solution: 8 3 16 9 2  3     5  Solution satisfies 1 22 : 6412xy 4 5 5 5 5 8 3 24 6 3  2     6  Solution satisfies 2 3 5 5 5 5 43 : 53  5yy 5

9168 Substitute 5 into 1 : 252xxx  555 83  xy  and  is the solution of the system. 55 Tutorial 3: Solving Simultaneous Linear Equations

2xy 3   8 1 Given the 2 2 system of linear equations  3xy 4 39 2 Solve the given system by using the: 1. Graphical method 2. Subtitution method PAUSE Topic 3. Elimination method • Do Tutorial 3 • Then View Solutions Tutorial 3: Problem 1: Suggested Solution

1. Solve graphically the 2 2 system of linear equations: 2xy 3   8 1  3xy 4 39 2 Use any method from Grade 9 to sketch the two lines. xy 5 and 6 satisfies both equations 1 and 2  Left as an exercise Tutorial 3: Problem 2: Suggested Solution

2. Solve the 2 2 system of linear equations by means of substitution 2xy 3   8 1  3xy 4 39 2 3y From 1 it follows that x  4 3 Back-substitute 5 into 3 : 2 36 x  4  9  4  5 Substitute 3 into 2 : 2 9y  12  4y  39 4 2 Show that xy 5 and 6 Solve equation 4 : is a solution of the system 9y 24  8 y  78  17 y  102  y  6 5 (Left as an exercise) Tutorial 3: Problem 3: Suggested Solution

3. Solve by means of elimination the 2 2 system of linear equations: 2xy 3   8 1  3xy 4 39 2 Decide to eliminate y : 1 4 : 8xy  12   32 3 Check (Test) that 2 3: 9xy  12  117 4 xy5 and 6 3 4 : 17xx  85   5 5 is the solution of the system (Exercises!) Substitute 5 into 4 : 45 12y  117  12 y  72  y  6 End of Topic Slides on Solving Linear and Quadratic Equations REMEMBER! •Consult text-books and past papers and memos for additional examples. •Attempt as many as possible other similar examples on your own. •Compare your methods with those that were discussed in these Topic slides. •Repeat this procedure until you are confident. •Do not forget: Practice makes perfect!