(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 begin with the equations involving \(y\) and \(x\), the first equation can be rewritten by clearing the fractions: 1. Multiply through by 6 (the least common multiple of 3 and 2) gives us: \[ 2(y - 2) - 3(x - 1) = 6 \] Simplifying this leads to: \[ 2y - 3x = 6 + 4 = 10 \] For the second equation: 2. Multiply through by 20 (the least common multiple of 5 and 4) gives: \[ 4(y - 4) - 5(x - 3) = 20 \] This simplifies to: \[ 4y - 5x = 20 + 16 = 36 \] Now, you have a system of linear equations: 1. \( 2y - 3x = 10 \) 2. \( 4y - 5x = 36 \) Next, we can solve this system. From the first equation, express \(y\): \[ 2y = 3x + 10 \] \[ y = \frac{3x + 10}{2} \] Substituting \(y\) into the second equation gives us: \[ 4 \left(\frac{3x + 10}{2}\right) - 5x = 36\] \[ \Rightarrow 2(3x + 10) - 5x = 36\] \[ \Rightarrow 6x + 20 - 5x = 36\] \[ \Rightarrow x + 20 = 36\] \[ \Rightarrow x = 16\] Now substitute \(x\) back into the expression for \(y\): \[ y = \frac{3(16) + 10}{2} = \frac{48 + 10}{2} = \frac{58}{2} = 29\] Thus, the first solution gives you \( x = 16\) and \( y = 29 \) as one solution of the simultaneous equations. For the burgers and cooldrinks, let's denote hamburgers as \(H\) and cooldrinks as \(C\). You have two equations: 1. \(3H + 2C = 89\) 2. \(2H + 3C = 73.5\) To eliminate \(C\), multiply the first equation by 3 and the second by 2: \[ 9H + 6C = 267 \] \[ 4H + 6C = 147\] Now subtract the second from the first: \[ 9H + 6C - (4H + 6C) = 267 - 147\] \[ 5H = 120 \] \[ H = 24 \] So the cost of one hamburger is \(R24\).