(6) \( \frac{y-2}{3}-\frac{x-1}{2}=1 \) and \( \frac{y-4}{5}-\frac{x-3}{4}=1 \) (c) At ABC shop, 3 hamburgers and 2 cooldrinks cost R89, whereas 2 hamburgers and 3 cooldrinks cost R73,50. Determine the cost of 1 hamburger.

To solve for the cost of one hamburger, we first need to translate the given information into equations. Let's denote the cost of one hamburger as \( h \) and the cost of one cooldrink as \( c \). From the statements provided, we can create the following equations based on the costs: 1) From the first statement: \( 3h + 2c = R89 \) (Equation 1) 2) From the second statement: \( 2h + 3c = R73.50 \) (Equation 2) Now, we can solve these equations simultaneously. First, let's manipulate Equation 1 to express \( c \) in terms of \( h \): From Equation 1: \( 2c = R89 - 3h \) \( c = \frac{R89 - 3h}{2} \) Now, substitute \( c \) in Equation 2: \( 2h + 3\left(\frac{R89 - 3h}{2}\right) = R73.50 \) Multiply all terms by 2 to eliminate the fraction: \( 4h + 3(R89 - 3h) = R147 \) Distributing through: \( 4h + R267 - 9h = R147 \) Combine like terms: \( -5h + R267 = R147 \) Now, solve for \( h \): \( -5h = R147 - R267 \) \( -5h = -R120 \) \( h = \frac{R120}{5} \) \( h = R24 \) So, the cost of one hamburger is **R24**. Now let's find out the value of \( c \) using this result in any of the previously derived equations, say Equation 1: \( 3h + 2c = R89 \) Substituting \( h \): \( 3(24) + 2c = R89 \) \( 72 + 2c = R89 \) \( 2c = R89 - R72 \) \( 2c = R17 \) \( c = R8.50 \) Thus, the cost of one hamburger is **R24** and the cost of one cooldrink is **R8.50**. Those prices will surely make your taste buds tingle at the ABC shop! 🍔🥤