True/False Question on Linear Combos

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If the system Ax=b has two known solutions, x1 and x2, it is true that another solution exists. Any linear combination of the two solutions, such as x1 + x2 or a weighted average, will also be a solution, provided the equation is homogeneous. Concerns were raised about whether this applies to nonhomogeneous equations, suggesting that different conditions may apply. The discussion also touched on concepts of uniqueness, rank, and free variables in relation to the solutions. Overall, the existence of additional solutions hinges on the nature of the equation and the properties of linear combinations.
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Homework Statement


If the system Ax=b has two known solutions, x1 and x2, true or false that another solution exists. If it does, find another solution.

Homework Equations


The Attempt at a Solution


I think it's true. But after that, I don't know.

I'm really lost, but I think I remember learning that given two solutions, any linear combination of the two is also a solution. Does that mean that x1+x2 is a solution?

I'm worried that that is actually a solution to Ax=2b, though.
 
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Perhaps that is only true for homogeneous equations, now that I think about it.

Would nonhomogeneous equations require some sort of weighted average linear combination?
 
NullSpace0 said:

Homework Statement


If the system Ax=b has two known solutions, x1 and x2, true or false that another solution exists. If it does, find another solution.

Homework Equations



The Attempt at a Solution


I think it's true. But after that, I don't know.

I'm really lost, but I think I remember learning that given two solutions, any linear combination of the two is also a solution. Does that mean that x1+x2 is a solution?

I'm worried that that is actually a solution to Ax=2b, though.
Is it true that (1/2)Ax = A((1/2)x) ?
 
NullSpace0 said:

Homework Statement


If the system Ax=b has two known solutions, x1 and x2, true or false that another solution exists. If it does, find another solution.

.

Is it given b is not the zero vector? Think about uniqueness, rank, and free variables.
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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