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Linear Algebra

Linear Algebra. Achievement Standard 1.4. SOLVING EQUATIONS. - Remember that addition/subtraction undo each other as do multiplication/division. - Terms containing the variable ( x ) should be placed on one side (often left). e.g. Solve. a) 5 x = 3 x + 6. b) -6 x = - 2 x + 12. -3 x.

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Linear Algebra

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  1. Linear Algebra Achievement Standard 1.4

  2. SOLVING EQUATIONS - Remember that addition/subtraction undo each other as do multiplication/division - Terms containing the variable (x) should be placed on one side (often left) e.g. Solve a) 5x =3x + 6 b) -6x = -2x + 12 -3x -3x +2x +2x 2x = 6 -4x = 12 Don’t forget the integer rules! ÷2 ÷2 ÷-4 ÷-4 x = 3 x = -3 Always line up equals signs and each line should contain the variable and one equals sign You should always check your answer by substituting into original equation Always look at the sign in front of the term/number to decide operation - Numbers should be placed on the side opposite to the variables (often right) e.g. Solve a) 6x – 5 =13 b) -3x + 10= 31 +5 +5 -10 -10 6x = 18 -3x = 21 ÷6 ÷6 ÷-3 ÷-3 x = 3 x = -7

  3. - Same rules apply for combined equations e.g. Solve a) 5x + 8 =2x + 20 b) 4x - 12= -2x + 24 -2x -2x +2x +2x 3x + 8= 20 6x - 12= 24 -8 -8 +12 +12 3x = 12 6x = 36 ÷3 ÷3 ÷6 ÷6 x = 4 x = 6 - Answers can also be negatives and/or fractions e.g. Solve a) 8x + 3 = -12x - 17 b) 5x + 2= 3x + 1 +12x +12x -3x -3x 20x + 3= -17 2x + 2= 1 -3 -3 -2 -2 20x = -20 2x = -1 ÷20 ÷20 ÷2 ÷2 x = -1 x = -1 2 Make sure you don’t forget to leave the sign too! Answer can be written as a decimal but easiest to leave as a fraction

  4. - Expand any brackets first e.g. Solve a) 3(x + 1) = 6 b) 2(3x – 1) = x + 8 3x + 3 = 6 6x - 2 = x + 8 -3 -3 -x -x 3x = 3 5x - 2= 8 ÷3 ÷3 +2 +2 x = 1 5x = 10 ÷5 ÷5 x = 2 - For fractions, cross multiply, then solve e.g. Solve a) x = 9 4 2 b) 3x - 1 = x + 3 5 2 2x = 36 2(3x - 1) = 5(x + 3) ÷2 ÷2 x = 18 6x - 2 = 5x + 15 -5x -5x x - 2= 15 +2 +2 x = 17

  5. - For two or more fractions, find a common denominator, multiply it by each term, then solve e.g. Solve 4x - 2x = 10 5 3 ×15 ×15 ×15 5 × 3 = 15 60x 5 - 30x 3 = 150 Simplify terms by dividing numerator by denominator 12x - 10x = 150 2x = 150 ÷2 ÷2 x = 75 e.g. Solve 5x - (x + 1) = 2x 6 4 ×24 ×24 ×24 120x 6 - (24x + 24) 4 = 48x 20x - 6x – 6 = 48x 14x – 6 = 48x -48x -48x -34x – 6 = 0 + 6 + 6 -34x = 6 ÷ -34 ÷ -34 x = -6 34

  6. WRITING AND SOLVING e.g. Write an equation for the following information and solve a) A rectangular pool has a length 5m longer than its width. The perimeter of the pool is 58m. Find its width x + 5 x + 5 + x + x + 5 + x = 58 4x + 10 = 58 Draw a diagram -10 -10 x x 4x = 48 Let x = width ÷4 ÷4 x + 5 x = 12 Therefore width is 12 m b) I think of a number and multiply it by 7. The result is the same as if I multiply this number by 4 and add 15. What is this number? Let n = a number 7 n = n 4 + 15 -4n -4n 3n = 15 ÷3 ÷3 n = 5 Therefore the number is 5

  7. SOLVING INEQUATIONS - Inequations contain one of four inequality signs: < > ≤ ≥ - To solve follow the same rules as when solving equations - Except: Reverse the direction of the sign when dividing by a negative e.g. Solve a) 3x + 8 >24 b) -2x -5 ≤ 13 -8 -8 +5 +5 3x >16 -2x ≤ 18 As answer not a whole number, leave as a fraction ÷3 ÷3 ÷-2 ÷-2 Sign reverses as dividing by a negative x >16 3 x ≥ -9 e.g. At an upcoming tournament, Jake has got $80 to spend. Jake wants to have at least $30 left by the time he returns home so he can buy a CD. At the tournament he is only allowed to spend his money on ‘V’ drinks which each cost $3.50. Form and solve an inequation to find out the maximum number of ‘V’ drinks he can buy at the tournament. 3.5x + 30 <80 Let x = number of V drinks -30 -30 3.5x <50 ÷3.5 ÷3.5 x <14.3 14 Therefore the maximum number of V drinks is:

  8. SUBSTITUTION - Involves replacing variables with numbers and calculating the answer - Remember the BEDMAS rules e.g. If m = 5, calculate m2 – 4m - 3 = 52 – 4×5 - 3 = 25 – 4×5 - 3 = 25 – 20 - 3 = 2 - Formulas can also have more than one variable e.g. If x = 4 and y = 6, calculate 3x – 2y = 3×4 - 2×6 = 12 - 12 = 0 e.g. If a = 2, and b = 5, calculate 2b – a 4 = (2 × 5 – 2) 4 = (10 – 2) 4 Because the top needs to be calculated first, brackets are implied = 8 4 = 2

  9. Plotting Points - To draw straight line graphs we can use a rule to find and plot co-ordinates e.g. Complete the tables below to find co-ordinates in order to plot the following straight lines: a) y = 2x b) y = ½x – 1 c) y = -3x + 2 2 x -2 -4 ½ x -2 – 1 -2 2 x -1 -2 ½ x -1 – 1 -1 ½ 0 -1 2 -½ 4 0 -3 x -2 + 2 8 -3 x -1 + 2 5 2 -1 -4

  10. Gradients of Lines - The gradient is a number that tells us how steep a line is. - The formula for gradient is: Gradient = rise run When calculating gradients it is best to write as simplest fraction  e.g. Write the gradients of lines A and B  e.g. Draw lines with the following gradients a) 1 b) 3 c) 2 25 = 3 1 4 6 8 4 4 = 1 8 2 6 = 3 4 2 A = B = To draw, write gradients as fractions

  11. y = mx • This is a rule for a straight line, where the gradient (m) is the number • directly in front of the x • When drawing graphs of the form y = mx, the line always goes through the • origin i.e. (0,0) e.g. Draw the following lines: a) y = 2x b) y = 4x c) y = 3x 54 = 4x 1 gradient 1. Step off the gradient from the origin (0,0) 2. Join the plotted point back to the origin To draw, always write gradients as fractions

  12. Negative Gradients  e.g. Write the gradients of lines A and B -3 -5 2 10 A = -5 = -1 10 2 B = -3 2 When calculating gradients it is best to write as simplest fraction

  13. e.g. Draw the following lines: a) y = -2x b) y = -4x c) y = -3x 54 = -4x 1 gradient 1. Step off the gradient from the origin (0,0) 2. Join the plotted point back to the origin To draw, always write gradients as fractions

  14. Intercepts - Is a number telling us where a line crosses the y-axis (vertical axis) i.e. The line y = mx + chas m as the gradient and c as the intercept  e.g. Write the intercepts of the lines A, B and C A = 8 B = 4 C = -3

  15. Drawing Lines: Gradient and Intercept Method - A straight line can be expressed using the rule y = mx + c e.g. Draw the following lines: a) y = 1x + 2 b) y = -3x – 2 c) y = -4x + 8 2 7 = -3x – 2 1 To draw: 1. Mark in intercept 2. Step off gradient 3. Join up points Note: Any rule with no number in front of x has a gradient of 1 1 e.g. y = x – 1

  16. Writing Equations of Lines - A straight line can be expressed using the rule y = mx + c e.g. Write equations for the following lines 3 4 -2 3 4 1 A: m = c = -6 B: m = c = +1 C: m = c = +4 y = -2x + 1 3 y = 4x + 4 1 y = 3x – 6 4

  17. Horizontal and Vertical Lines - Horizontal lines have a gradient of: 0 Rule: y = c (c is the y-axis intercept) - Vertical lines have a gradient that is: undefined Rule: x = c (c is the x-axis intercept) e.g. Draw or write equations for the following lines: a) y = 2 b) c) x = 4 d) y = -3 x = -1 b) d)

  18. Writing Equations Cont. When you are given two points and are expected to write an equation: - One method is set up a set of axes and plot the two points. - Or, substitute the gradient and a point into y = mx + c to find ‘c’, the intercept e.g. Write an equation for the line joining the points A=(1, 3) and B = (3, -1) -2 1 m = c = 5 y = -2x + 5 -2 1 m = using point (1, 3) y = mx + c 3 = -2 x 1 + c 3 = -2 + c +2 +2 5 = c y = -2x + 5 Sometimes when plotting the points, you may need to extend the axes to find the intercept.

  19. Equations in the Form ‘ax + by = c’ - Can use the cover up rule to find the two intercepts: e.g. Draw the following lines: a) 2y – x = 4 b) 4x – 3y =12 - x = 4 2y = 4 4x = 12 -3y = 12 ÷ -1 ÷ -1 ÷ 2 ÷ 2 ÷ 4 ÷ 4 ÷ -3 ÷ -3 x = -4 y = 2 x = 3 y = -4 1. Cover up ‘y’ term to find x intercept 2. Cover up ‘x’ term to find y intercept 3. Join up intercepts with a straight line It is also possible to rearrange equations into the form y = mx + c e.g. Rearrange 2x – y = 6 -2x -2x - y = 6 – 2x ÷ -1 ÷ -1 y = -6 + 2x y = 2x – 6

  20. Applications e.g. A pizzeria specializes in selling large size pizzas. The relationship between x, number of pizzas sold daily, and y, the daily costs is given by the equation, y = 10x + 50 1. Draw a graph of the equation 2. What are the costs if they sell 8 pizzas? $130 3a. What is the cost per pizza? $10 3b. How is this shown by the graph? The gradient of the line 4a. What are the costs if they sell no pizzas? $50 4b. How is this shown by the graph? Where the line crosses the y-axis

  21. SIMULTANEOUS EQUATIONS - Are pairs of equations with two unknowns To solve we can use one of three methods: 1. ELIMINATION METHOD - Line up equations and either add or subtract so one variable disappears e.g. Solve To remove the ‘y’ variable we add as the signs are opposite. To remove the ‘y’ variable we subtract as the signs are the same. • 2x + y = 20 • x – y = 4 b) 2x + y = 7 x + y = 4 + ( ) - ( ) 3x = 24 x = 3 ÷3 ÷3 Now we substitute x-value into either equation to find ‘y’ 2 × 3 + y = 7 Now we substitute x-value into either equation to find ‘y’ x = 8 6 + y = 7 2 × 8 + y = 20 -6 -6 16 + y = 20 y = 1 -16 -16 y = 4 Check values in either equation Check values in either equation 8 – 4 = 4 3 + 1 = 4

  22. - You may need to multiply an equation by a number to be able to eliminate a variable e.g. Solve • 2x – y = 0 • x + 2y = 5 × 2 × 1 4x – 2y = 0 x + 2y = 5 b) 4x – 2y = 28 3x + 3y = 12 × 3 × 4 12x – 6y = 84 12x + 12y = 48 + ( ) - ( ) Multiply the 1st equation by ‘2’ then add to eliminate the y 5x = 5 Multiply the 1st equation by ‘3’ and the 2nd by ‘4’ then subtract to eliminate the x -18y = 36 ÷5 ÷5 ÷-18 ÷-18 x = 1 y = -2 2 × 1 – y = 0 4x – 2 ×-2 = 28 2 – y = 0 4x + 4 = 28 Now we substitute x-value into either equation to find ‘y’ Now we substitute y-value into either equation to find ‘x’ -2 -2 -4 -4 – y = -2 4x = 24 ÷-1 ÷-1 ÷4 ÷4 y = 2 x = 6 Check values in either equation Check values in either equation 1 + 2 ×2 = 5 3 × 6 + 3 ×-2 = 12 Note that it was possible to eliminate the ‘x’ variable by multiplying second equation by 2 and then subtracting Note that it was possible to eliminate the ‘y’ variable by multiplying the 1st equation by 3 and the 2nd by 2 and then adding

  23. 2. SUBSTITUTION METHOD Substitute what the subject equals in for that variable in the other equation - Make x or y the subject of one of the equations - Substitute this equation into the second e.g. Solve a) y = 3x+ 1 9x – 2 y = 4 b) x – y = 2 y = 2x + 3 As we are subtracting more than one term, place in brackets and put a one out in front. 9x – 2(3x + 1) = 4 x – (2x + 3) = 2 1 9x – 6x – 2 = 4 x – 2x – 3 = 2 3x – 2 = 4 -x – 3 = 2 +2 +2 +3 +3 3x = 6 -x = 5 ÷3 ÷3 ÷-1 ÷-1 x = 2 x = -5 Now we substitute x-value into first equation to find ‘y’ y = 3×2 + 1 Now we substitute x-value into second equation to find ‘y’ y = 2×-5 + 3 y = 7 y = -7 To check values you can substitute both values into second equation To check values you can substitute both values into second equation

  24. GRAPHING LINEAR INEQUALITIES - For < and > we used a dotted line ------------------- - For ≤ and ≥ we used a solid line - Shading in includes the region that contains the solution e.g. Solve by Shading in, and show the solution to the following system of inequalities 1. y < 2 2. x ≥ 3 3. y < 3x - 2 Test which region you will shade by substituting (0,0) into the equation. 1. Is 0 < 2 Yes, so shade below the line 2. Is 0 ≥ 3 No, so shade to the right of the line (excluding the point (0,0) 2. Is 0 < 3x0 - 2 No, so shade to the right of the line (excluding the point (0,0)

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