What is the answer to solvefor a,4+a_{n-1}a_{1}=2 ?

The answer to solvefor a,4+a_{n-1}a_{1}=2 is True for all a;\quad 4+a_{n-1}a_{1}=2

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solvefor a,4+a_{n-1}a_{1}=2

solve for

a, 4 + a_{n - 1} a_{1} = 2

True for all a; 4 + a_{n - 1} a_{1} = 2

Solution steps

4 + a_{n - 1} a_{1} = 2

Both sides are equal

True for all a; 4 + a_{n - 1} a_{1} = 2

3 a + 4 + b + 2 a + 5 + b

9 a^{2} + 25 a + 16 = a

2 (1 + x^{2}) (2 x^{2} - 2 x x^{2} + 1 + 2)

- 15 \geq 8 k + 9

2 \leq x \leq 7

What is the answer to solvefor a,4+a_{n-1}a_{1}=2 ?
The answer to solvefor a,4+a_{n-1}a_{1}=2 is True for all a;\quad 4+a_{n-1}a_{1}=2