In the following, find the coordinates of the point whose abscissa is the solution of the first equation and ordinate is the solution of the second equation:
2a3−1=a2;15−4b7=2b−13
Claim : Solution is (6,2)
If true then enter 1 and if false then enter 0
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Solution
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First equation:
2a3−1=a2
⇒2a3−a2=1
On solving we get,
⇒a=6 Second equation:
15−4b7=2b−13
On cross - multiplying, we get,
45−12b=14b−7
⇒b=2 Hence, required point =(a,b)=(6,2)
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Claim : Solution is (6,2)
If true then enter 1 and if false then enter 0
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