a) Given: \(\frac{x}{2a+b} = \frac{y}{2b-c} = \frac{z}{2c-a} \)
Prove that \(\frac{2x+y}{4a+4b-c} = \frac{2x+2y+z}{3a+6b}\):
Solution method:
1. Multiply the two terms of the \(1^{\text{st}}\) ratio by 2 and add the antecedents and consequents of the \(1^{\text{st}}\) and the \(2^{\text{nd}}\) ratios.
2. Multiply the terms of the \(1^{\text{st}}\) ratio by 2 and the \(2^{\text{nd}}\) by 2, then add the antecedents and consequents of the three ratios.
1. Multiply the two terms of the \(1^{\text{st}}\) ratio by 2 and add the antecedents and consequents of the \(1^{\text{st}}\) and the \(2^{\text{nd}}\) ratios.
2. Multiply the terms of the \(1^{\text{st}}\) ratio by 2 and the \(2^{\text{nd}}\) by 2, then add the antecedents and consequents of the three ratios.
MR/Nasef.N
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