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Solving Linear Equations, Inequalities, and Parabola Analysis

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This guide walks you through various mathematical problems including determining the equation of a straight line, solving inequalities, analyzing equations of quadratic functions, and evaluating systems of equations. It covers practical applications such as budgeting for paint purchases and calculating dimensions of geometric shapes. Key topics include line equations with gradients, solving for variable values in systems, and understanding the properties of parabolas including symmetry and turning points. Enhance your algebra skills with practical examples.

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Solving Linear Equations, Inequalities, and Parabola Analysis

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  1. Relationships

  2. 1.1 m (a , b) 1 A straight line with gradient –3 passes through the point (– 2, 4).Determine the equation of this straight line. y – b = m(x – a) y – 4 = -3(x + 2) y – 4 = -3x – 6 y = -3x – 2

  3. 1.1 1.1 1.1 2 Solve the inequation 4p – 18 < p – 3. 3p – 18 < -3 3p < 15 p < 5 3 Gertrude buys 3 small tins and 5 medium tins of paint. Total cost is £44.92 Write an equation to represent this information. 3s + 5m = 4492 (-3 + 18) # 2.1

  4. 1.1 1.1 1.1 5a + 2b = 24 X2 → 2a – 2b = 4 4 Solve the following system of equations algebraically: 5a + 2b = 24 a – b = 2 Add 7a = 28 a = 4 4 – b = 2 b = 2 a = 4 , b = 2

  5. 1.1 1.1 1.1 5T – P = H r 5 Change the subject of the formula to T 5T = H + P r = r(H + P) 5T = r(H + P) T 5 1.1 Out of 10

  6. 1.2 y 18 6 The diagram shows the parabola with equation y = kx2What is the value of k? y = kx2 → 12 = k X 22 12 = 4k k = 3 16 14 (2 , 12) x y 12 10 8 6 4 2 x -3 -2 -1 0 1 2 3

  7. y 1.2 1.2 8 6 4 2 x -2 -1 0 1 2 -4 -3 7 The equation of the quadratic function whose graph is shown below is of the form y = (x + a)2 + b, where a and b are integers. Write down the values of a and b. y = (x – 1)2 + 3 a = -1 , b = 3 3 1

  8. 1.2 1.2 1.2 8 Sketch the graph y = (x + 6)(x – 2) on plain paper.Mark clearly where the graph crosses the axes and state the coordinates of the turning point. For roots, y = 0 → (x + 6)(x – 2) = 0 x + 6 = 0 or x – 2 = 0 x = -6 or x = 2 Y Intercept, x = 0 at x = 0 y = (0 + 6)(0 – 2) = 6 X (-2) = -12 x value of turning pt = -6 + 2 = -2 at x = -2 y = (-2 + 6)(-2 – 2) = 4 X (-4) = -16 2 2 -6 -12 (-2 , -16)

  9. 1.2 1.2 1.2 9 A parabola has equation y = (x + 2)2 – 3. (a) Write down the equation of its axis of symmetry.(b) Write down the coordinates of the turning point on the parabola and state whether it is a maximum or minimum. x = -2 1.2 Out of 9 2 3 Minimum (-2 , -3)

  10. 1.3 10 Solve the equation (x – 5)(x + 2) = 0. x – 5 = 0 or x + 2 = 0 x = 5 or x = -2

  11. 1.3 1.3 1.3 1.3 ax2 + bx + c = 0 a = 1 b = 5 c = -2 + 11 Solve the equation x2 + 5x – 2 = 0 using the quadratic formula. -b √ x = b2 – 4ac – 2a + -5 √ 52 – 4X1X(-2) – 2X1 + -5 √ 33 – 2 + -5 ( ) √ -5 – 33 ( ) √ 33 OR 2 2 or = -5.4 = 0.4

  12. 1.3 1.3 12 Determine the nature of the roots of the equation 2x2 – 5x + 3 = 0 using the discriminant. a = 2, b = -5, c = 3 b2 – 4ac = (-5)2 – 4 X 2 X 3 = 25 – 24 = 1 Two real roots as b2 – 4ac > 0 1.3 Out of 7 # 2.2

  13. 1.4 1.4 0.75m 1.25m 1m 13 In a health and safety inspection this scaffold must have a right angle to be passed as safe. Will it meet the health and safety requirements? If it is a RAT then 0.752 + 12 = 1.252 LHS 0.752 + 12 = 1.5625 RHS 1.252 = 1.5625 LHS = RHS, therefore it is a RAT, so will meet H + S requirements. # 2.2 2.2 Out of 2

  14. 1.4 1.4 1.4 14 The diagram below shows the design stages of a kite. Kite STUV and a circle with centre T are detailed in the diagram. SV is the tangent to the circle at S and UV is the tangent to the circle at U. Given that angle STU is 124°, calculate angle SVU. LSVU = 360 – (90 + 90 + 124) = 560 1240

  15. 1.4 1.4 15 The large watering can holds 10 litres. It is 50cm high. There is a similar small version which is 30cm high. How much will it hold? Linear Scale Factor = 30/50 = 0.6 Volume Scale factor = 0.63 Volume Small Can= 0.63 X 10 = 2.16 litres

  16. 1.4 16 Here is a regular pentagon Calculate the size of the shaded angle Angle at centre: 360 ÷ 5 = 720 Angles at feet: (180 – 72) ÷ 2 = 540 Shaded angle: 2 X 54 = 1080 # 2.1 720 540 540 1.4 Out of 8 2.1 Out of 2

  17. 1.5 1.5 2 17 Sketch the graph of y = 2cosx° for 900 1800 2700 3600 -2 .

  18. 1.5 18 Write down the period of the graph of the equation y = sin 3x°. period = 360 ÷ 3 = 1200

  19. 1.5 1.5 1.5 1.5 Out of 6 4sinx – 2 = 0 A Change to sinx= S 19 Solve the equation 4 sin x° – 2 = 0 i ii 4sinx = 2 a0 180 – a sinx = 2/4 sinx = 0.5 Find Acute Angle iv iii Sin-1(0.5) = 300 360 – a 180 + a C T Quadrants? Sin Positive→ i and ii 180 – 30 x = 300 or Solutions = 1500 = 300

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