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Conditions for a system of linear equations to have infinite solutions

The number of...

Question

The number of integral solution to the system of equation $a_{1} + a_{2} + a_{3} + a_{4} + a_{5} = 25$ and $a_{1} + a_{2} + a_{3} = 10$ are

2^{8} + 2^{5}

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2^{10} + 2^{5}

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2^{25} + 2^{10}

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Infinite

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Solution

The correct option is D Infinite
We have, $a_{1} + a_{2} + a_{3} + a_{4} + a_{5} = 25$ ..........(1)
and $a_{1} + a_{2} + a_{3} = 10$ .........(2)
By, $(2) - (1)$
$a_{4} + a_{5} = 15$ ....(3)
Let $a_{4} = x, a_{5} = y$
$\Rightarrow x + y = 15$ .......(4)
Here $x$ and $y$ are integers.
So number of integral solution will be infinite, since we have one equation and two unknowns

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Similar questions

a_{1} + a_{2} + a_{3} + a_{4} + a_{5} = 25

a_{1} + a_{2} + a_{3} = 10

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

a_{1}, a_{2}, a_{3}, a_{4}, a_{5}

6 x^{5} - 41 x^{4} + 97 x^{3} - 97 x^{2} + 41 x - 6 = 0,

| a_{1} | \leq | a_{2} | \leq | a_{3} | \leq | a_{4} | \leq | a_{5} |,

a_{1}

a_{2}

a_{3}

a_{2}

a_{3}

a_{4}

a_{3}

a_{4}

a_{5}

a_{1}

a_{3}

a_{5}

a_{1} a_{2} a_{3} a_{4} a_{5}

9

a_{1} + a_{2} + a_{3} + a_{4} + a_{5}

9

a_{1} - a_{2} + a_{3} - a_{4} + a_{5}

9

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