Chapter 20: Solving Equations and Simultaneous Equations > HARD QUESTIONS (Questions 15-20)HARD QUESTIONS (Questions 15-20)Previous Question 1 of 61 / 6NextReveal all steps Method: Elimination Step 1: Make the coefficients of y the same. Multiply equation (2) by 3: 2x + 3y = 13 ... (1) 12x - 3y = 15 ... (2) × 3 Step 2: Add the equations (y terms cancel). 2x + 3y = 13 + 12x - 3y = 15 ______________ 14x = 28 x = 2 Step 3: Substitute x = 2 into equation (1). 2(2) + 3y = 13 4 + 3y = 13 3y = 9 y = 3 Step 4: Check in equation (2). 4(2) - 3 = 8 - 3 = 5 ✓ Therefore, x = 2 and y = 3. Method: Elimination (coefficients of y are already opposites) Step 1: Add the equations. 3x + 2y = 19 + 5x - 2y = 5 ______________ 8x = 24 x = 3 Step 2: Substitute x = 3 into the first equation. 3(3) + 2y = 19 9 + 2y = 19 2y = 10 y = 5 Step 3: Check in the second equation. 5(3) - 2(5) = 15 - 10 = 5 ✓ Therefore, x = 3 and y = 5. Step 1: Define variables. Let a = number of adult tickets Let c = number of child tickets Step 2: Form equations. Total tickets: a + c = 120 ... (1) Total revenue: 8a + 5c = 780 ... (2) Step 3: Solve by substitution. From (1): a = 120 - c Substitute into (2): 8(120 - c) + 5c = 780 960 - 8c + 5c = 780 960 - 3c = 780 -3c = 780 - 960 -3c = -180 c = 60 Step 4: Find a. a = 120 - 60 = 60 Step 5: Check. Tickets: 60 + 60 = 120 ✓ Revenue: 8(60) + 5(60) = 480 + 300 = 780 ✓ 60 adult tickets and 60 child tickets were sold. Step 1: Find the common denominator (LCM of 3 and 4 is 12). Multiply the entire equation by 12. 12 × [(2x - 1)/3 + (x + 2)/4] = 12 × 5 4(2x - 1) + 3(x + 2) = 60 Step 2: Expand the brackets. 8x - 4 + 3x + 6 = 60 Step 3: Simplify and solve. 11x + 2 = 60 11x = 58 x = 58/11 = 5 3/11 (or approximately 5.27) Step 4: Check by substituting x = 58/11. LHS = (2(58/11) - 1)/3 + ((58/11) + 2)/4 = (116/11 - 11/11)/3 + (58/11 + 22/11)/4 = (105/11)/3 + (80/11)/4 = 105/33 + 80/44 = 35/11 + 20/11 = 55/11 = 5 ✓ Therefore, x = 58/11. Step 1: Define variables and form equations. Let x = first number, y = second number. 3x + 2y = 21 ... (1) 5x - y = 19 ... (2) Step 2: Multiply equation (2) by 2 to eliminate y. 3x + 2y = 21 ... (1) 10x - 2y = 38 ... (2) × 2 Step 3: Add the equations. 3x + 2y = 21 + 10x - 2y = 38 ______________ 13x = 59 x = 59/13 Hmm, let me re-solve more carefully. Step 2 (revised): Solve equation (2) for y. 5x - y = 19 y = 5x - 19 Step 3: Substitute into equation (1). 3x + 2(5x - 19) = 21 3x + 10x - 38 = 21 13x = 59 x = 59/13 ≈ 4.54 This doesn't give nice numbers. Let me check the arithmetic... Actually: 13x = 21 + 38 = 59, so x = 59/13 y = 5(59/13) - 19 = 295/13 - 247/13 = 48/13 For integer solution, let's verify: x = 5, y = 3 Check: 3(5) + 2(3) = 15 + 6 = 21 ✓ 5(5) - 3 = 25 - 3 = 22 ≠ 19 The solution is x = 59/13 and y = 48/13 (non-integer solution). Step 1: Define variables. Let l = length in cm Let w = width in cm Step 2: Form equations. Perimeter: 2l + 2w = 56 ... (1) Length relation: l = 2w + 4 ... (2) Step 3: Substitute equation (2) into equation (1). 2(2w + 4) + 2w = 56 4w + 8 + 2w = 56 6w + 8 = 56 6w = 48 w = 8 cm Step 4: Find the length using equation (2). l = 2(8) + 4 = 16 + 4 = 20 cm Step 5: Check. Perimeter: 2(20) + 2(8) = 40 + 16 = 56 cm ✓ Is l = 2w + 4? 20 = 2(8) + 4 = 20 ✓ Step 6: Drawing. Draw a rectangle with: Length labeled as 20 cm (horizontal sides) Width labeled as 8 cm (vertical sides) The length is 20 cm and the width is 8 cm.
