If I have 4 equations and 3 unknowns, I could solve for the 3 unknowns using the first 3. How does it ensure that the 4th equation is also satisfied? In this case, what should be the usual strategy...
If I have 4 equations and 3 unknowns, I could solve for the 3 unknowns using the first 3. How does it ensure that the 4th equation is also satisfied? In this case, what should be the usual strategy...
Many thanks in advance! Edit: It might be sensible to define 'linearly independent'. A sensible definition (i.e. one that seems to make the statement true) is that two equations and linearly independent if and only if neither equation can be derived from the other using linear operations (addition and multiplication by a non-zero constant).
linear algebra - Number of equations vs number of unknowns ...
Many thanks in advance! Edit: It might be sensible to define 'linearly independent'. A sensible definition (i.e. one that seems to make the statement true) is that two equations and linearly independent if and only if neither equation can be derived from the other using linear operations (addition and multiplication by a non-zero constant).
Actually it is possible to solve two unknown with one equation. But the only way is using the properties of some functions like square roots or use one of the variable multiplying the original linear equation.
algebra precalculus - Can you solve two unknowns with one equation ...
Actually it is possible to solve two unknown with one equation. But the only way is using the properties of some functions like square roots or use one of the variable multiplying the original linear equation.
Unknown is usually employed in equations,so for example you could ask how to solve the equation 2x = 1 2 x = 1,where x x is the unknown. On the other hand the term variable is more used in case of functions.
algebra precalculus - Subtleties of "unknown" vs. "variable ...
Unknown is usually employed in equations,so for example you could ask how to solve the equation 2x = 1 2 x = 1,where x x is the unknown. On the other hand the term variable is more used in case of functions.
In general, how do you solve a system with more equations than unknowns? I know that if I select the equations to match them with the number of unknowns, there may be zero or many solutions dependi...
how to solve a system with more equations than unkowns?
In general, how do you solve a system with more equations than unknowns? I know that if I select the equations to match them with the number of unknowns, there may be zero or many solutions dependi...
"I know that in order to solve three unknowns three equations are needed, so I'm unsure if this can be solved" well, since this doesn't have 3 equations you know it can not. (Although the "3 equations- 3 unknowns" is a slight oversimplification.)
How to solve a system of two equations in three unknowns
"I know that in order to solve three unknowns three equations are needed, so I'm unsure if this can be solved" well, since this doesn't have 3 equations you know it can not. (Although the "3 equations- 3 unknowns" is a slight oversimplification.)
What's the origin of the problem? Cutting down to 6 variables might be a mistake; sometimes the expression in terms of the full set of unknowns is much easier to handle. As things stand there are certainly ways of studying the problem, but knowing the origin and how these equations were derived might give a much easier way of proceeding.
Solving a System of Six unknowns using Six Equations
What's the origin of the problem? Cutting down to 6 variables might be a mistake; sometimes the expression in terms of the full set of unknowns is much easier to handle. As things stand there are certainly ways of studying the problem, but knowing the origin and how these equations were derived might give a much easier way of proceeding.
The quadratic equation 2x2 − px − 4 = 0 2 x 2 − p x − 4 = 0 where p p is a positive constant, has 2 solutions that differ by 6. What is the value of p p? I attempted to factorise to get (2x-2) (x+2)=0 and (2x-4) (x+1)=0 but the solutions don’t differ correctly.
Solve quadratic with unknown constant in question [closed]
The quadratic equation 2x2 − px − 4 = 0 2 x 2 − p x − 4 = 0 where p p is a positive constant, has 2 solutions that differ by 6. What is the value of p p? I attempted to factorise to get (2x-2) (x+2)=0 and (2x-4) (x+1)=0 but the solutions don’t differ correctly.
What is the name of an equation, where the unknown is one of the limits of integration? Is there a theory that studies such equations, standard methods of solution? The simplest example is the equ...
Integral Equation Unknown Limits - Mathematics Stack Exchange
What is the name of an equation, where the unknown is one of the limits of integration? Is there a theory that studies such equations, standard methods of solution? The simplest example is the equ...
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