Can Simultaneous Equations Have Multiple Solutions?

AI Thread Summary
Simultaneous equations can have multiple solutions, but the specific equations presented, 4x + 3x = 28 and 2x + 5x = 42, lead to a single variable scenario. The discussion highlights the importance of correctly interpreting the equations, as the original equations lack a second variable. When reformulated as 4x + 3y = 28 and 2x + 5y = 42, the equations can be solved using the Gaussian method. However, if taken literally, the first equations suggest no solution exists unless the constants are equal. Proper formulation is crucial for determining the nature of solutions in simultaneous equations.
problems
Messages
2
Reaction score
0
please help me (Simultaneous Equations)

4x+3x=28
2x+5x=42

how can i sole it from gaussian method :(
 
Physics news on Phys.org


How about you first write down the corresponding (augmented) matrix?
 


Or you sure its not:(you just ave one variable)

4x+3y=28
2x+5y=42

or something else?
 


Right, I missed that. I assumed that was what you meant. When taken literally as you posted, there is no solution (unless 28 = 52).
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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
Back
Top