Discussion
If \(\dfrac{6x+2y}{3x+4y}=\dfrac{3}{4}\) and \(\dfrac{x+z}{2x+3z}=\dfrac{3}{7}\)
Then find \(x:y:z\).
Answer: B
From the first ratio, \(4(6x+2y)=3(3x+4y)\) gives \(24x+8y=9x+12y\), so \(15x=4y\) and x : y = 4 : 15. From the second, \(7(x+z)=3(2x+3z)\) gives \(x=2z\), so x : z = 2 : 1 = 4 : 2.
Hence x : y : z = 4 : 15 : 2, option B. The other options do not satisfy both equations.
The two equations are linear and have only the ratios as unknowns. The first gives 15x = 4y and the second gives x = 2z, so with x = 4 we get y = 15 and z = 2. Checking, (24 + 30)/(12 + 60) = 54/72 = 3/4 and (4 + 2)/(8 + 6) = 6/14 = 3/7, as required.
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