Question 11a (WASSCE 2012 November) With GH \( \propto 184 \), I can buy \( x \) packets of biscui If I can buy two more packets of the same biscuit with GH \( ¢ 230 \), find the value of \( x \).

To solve this problem, we can set up a proportion based on the given information. Let's denote the number of packets of biscuits that can be bought with GH¢ 184 as \( x \) and the number of packets that can be bought with GH¢ 230 as \( x + 2 \). Given that GH¢ 184 can buy \( x \) packets of biscuits, we can set up the proportion: \[ \frac{184}{x} = \frac{230}{x + 2} \] Now, we can solve this proportion to find the value of \( x \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{184}{x}=\frac{230}{x+2}\) - step1: Find the domain: \(\frac{184}{x}=\frac{230}{x+2},x \in \left(-\infty,-2\right)\cup \left(-2,0\right)\cup \left(0,+\infty\right)\) - step2: Cross multiply: \(184\left(x+2\right)=x\times 230\) - step3: Simplify the equation: \(184\left(x+2\right)=230x\) - step4: Rewrite the expression: \(46\times 4\left(x+2\right)=46\times 5x\) - step5: Evaluate: \(4\left(x+2\right)=5x\) - step6: Expand the expression: \(4x+8=5x\) - step7: Move the variable to the left side: \(4x+8-5x=0\) - step8: Subtract the terms: \(-x+8=0\) - step9: Move the constant to the right side: \(-x=0-8\) - step10: Remove 0: \(-x=-8\) - step11: Change the signs: \(x=8\) - step12: Check if the solution is in the defined range: \(x=8,x \in \left(-\infty,-2\right)\cup \left(-2,0\right)\cup \left(0,+\infty\right)\) - step13: Find the intersection: \(x=8\) The value of \( x \) is 8.