6.5 Two litres of oil and 30 litres of petrol cost R450 petrol cost R380. Calculate the price of petrol and uestion 7 \( .1 \quad \) Calculate the LCM of: 7.1.1 \( a b c ; a^{2} b ; b c^{2} \) 7.1.2 \( 2 a^{2}+2 a-4 ; 3 a^{2}-3 ; 6 a+12 \)

Let's break down the problem step by step. ### Problem 6.5: Cost of Oil and Petrol **Known Conditions:** 1. The total cost of 2 litres of oil and 30 litres of petrol is R450. 2. The cost of petrol alone is R380. **Let:** - \( x \) = cost of 1 litre of oil - \( y \) = cost of 1 litre of petrol **From the information given, we can set up the following equations:** 1. \( 2x + 30y = 450 \) (Equation 1) 2. \( 30y = 380 \) (Equation 2) **Step 1: Solve Equation 2 for \( y \):** \[ y = \frac{380}{30} \] **Step 2: Calculate \( y \):** \[ y = \frac{380}{30} = 12.67 \] **Step 3: Substitute \( y \) back into Equation 1 to find \( x \):** \[ 2x + 30(12.67) = 450 \] \[ 2x + 380.1 = 450 \] \[ 2x = 450 - 380.1 \] \[ 2x = 69.9 \] \[ x = \frac{69.9}{2} = 34.95 \] **Final Results:** - Cost of 1 litre of oil \( x \) = R34.95 - Cost of 1 litre of petrol \( y \) = R12.67 --- ### Problem 7: Calculate the LCM #### 7.1.1 LCM of \( abc, a^2b, bc^2 \) **Step 1: Identify the highest powers of each variable:** - For \( a \): highest power is \( a^2 \) - For \( b \): highest power is \( b^1 \) (from \( abc \)) - For \( c \): highest power is \( c^2 \) **Step 2: Write the LCM:** \[ \text{LCM} = a^2 b c^2 \] #### 7.1.2 LCM of \( 2a^2 + 2a - 4, 3a^2 - 3, 6a + 12 \) **Step 1: Factor each expression:** 1. \( 2a^2 + 2a - 4 = 2(a^2 + a - 2) = 2(a - 1)(a + 2) \) 2. \( 3a^2 - 3 = 3(a^2 - 1) = 3(a - 1)(a + 1) \) 3. \( 6a + 12 = 6(a + 2) \) **Step 2: Identify the unique factors:** - From \( 2(a - 1)(a + 2) \): factors are \( 2, (a - 1), (a + 2) \) - From \( 3(a - 1)(a + 1) \): factors are \( 3, (a - 1), (a + 1) \) - From \( 6(a + 2) \): factors are \( 6, (a + 2) \) **Step 3: Take the highest power of each factor:** - \( 2 \) from \( 2 \) - \( 3 \) from \( 3 \) - \( (a - 1) \) from \( 3(a - 1)(a + 1) \) - \( (a + 2) \) from \( 6(a + 2) \) - \( (a + 1) \) from \( 3(a - 1)(a + 1) \) **Step 4: Write the LCM:** \[ \text{LCM} = 6(a - 1)(a + 2)(a + 1) \] ### Summary of Results: - Cost of oil: R34.95 - Cost of petrol: R12.67 - LCM of \( abc, a^2b, bc^2 \): \( a^2bc^2 \) - LCM of \( 2a^2 + 2a - 4, 3a^2 - 3, 6a + 12 \): \( 6(a - 1)(a + 2)(a + 1) \)