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In a park, two paths are represented by \(3x+y=21\) and \(x+y=11\). Where will they meet?

Answer and explanation

Correct answer: Point \(\left(5,6\right)\)

The direct answer is A: \(\left(5,6\right)\). The meeting point must satisfy both equations. Subtract \(x+y=11\) from \(3x+y=21\): the \(y\)-terms cancel, giving \(2x=10\), so \(x=5\). Substitute this into \(x+y=11\): \(5+y=11\), hence \(y=6\). Therefore the paths meet at \(\left(5,6\right)\). Check: in the first equation, \(3(5)+6=21\); in the second, \(5+6=11\), so the point works in both. Option A is correct. Option B reverses the coordinates; \((6,5)\) gives 23 in the first equation, not 21. Option C, \((4,7)\), satisfies the second equation but gives 19 in the first. Option D, \((7,4)\), satisfies the second equation but gives 25 in the first. Always substitute a proposed ordered pair into both equations, because the first number is \(x\) and the second is \(y\).

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