JEE 2023 — Mathematics PYQJEE | Mathematics | 2023Let the system of linear equations: x+y+kz=2 2x+3y−z=1 3x+4y+2z=k have infinitely many solutions. Then the system: (k+1)x+(2k−1)y=7 (2k+1)x+(k+5)y=10 has:Choose the correct answer:A. infinitely many solutionsB. unique solution satisfying x−y=1C. unique solution satisfying x+y=1 (Correct Answer)D. no solutionCorrect Answer: unique solution satisfying x+y=1ExplanationSolution: Find k for infinite solutions (Δ=0): 123amp;1amp;3amp;4amp;kamp;−1amp;2=0⟹1(6+4)−1(4+3)+k(8−9)=0⟹k=3 Substitute k=3 in the second system: 4x+5y=7 7x+8y=10 Solve for x,y: By solving, we get x=−2 and y=3. Check condition: x+y=−2+3=1. Correct Option: (3) unique solution satisfying x+y=1Ainfinitely many solutionsBunique solution satisfying x−y=1Cunique solution satisfying x+y=1Dno solutionShow Answer SExplanationSolution: Find k for infinite solutions (Δ=0): 123amp;1amp;3amp;4amp;kamp;−1amp;2=0⟹1(6+4)−1(4+3)+k(8−9)=0⟹k=3 Substitute k=3 in the second system: 4x+5y=7 7x+8y=10 Solve for x,y: By solving, we get x=−2 and y=3. Check condition: x+y=−2+3=1. Correct Option: (3) unique solution satisfying x+y=1