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This equation is an equation only of y'' and y', meaning it is reducible to the general form described above and is, therefore, separable. Since it is a second-order separable equation, collect all x variables on one side and all y' variables on the other to get: (′) (′) =.
To test whether the third equation is linearly dependent on the first two, postulate two parameters a and b such that a times the first equation plus b times the second equation equals the third equation. Since this always holds for the right sides, all of which are 0, we merely need to require it to hold for the left sides as well:
For a system: adding to both sides of an equation the corresponding side of another equation, multiplied by the same quantity. If some function is applied to both sides of an equation, the resulting equation has the solutions of the initial equation among its solutions, but may have further solutions called extraneous solutions.
In solving mathematical equations, particularly linear simultaneous equations, differential equations and integral equations, the terminology homogeneous is often used for equations with some linear operator L on the LHS and 0 on the RHS. In contrast, an equation with a non-zero RHS is called inhomogeneous or non-homogeneous, as exemplified by ...
To solve this kind of equation, the technique is add, subtract, multiply, or divide both sides of the equation by the same number in order to isolate the variable on one side of the equation. Once the variable is isolated, the other side of the equation is the value of the variable. [37] This problem and its solution are as follows: Solving for x
This counterintuitive result occurs because in the case where =, multiplying both sides by multiplies both sides by zero, and so necessarily produces a true equation just as in the first example. In general, whenever we multiply both sides of an equation by an expression involving variables, we introduce extraneous solutions wherever that ...
The mass of some objects on the scale is unknown and represents variables. Solving an equation corresponds to adding and removing objects on both sides in such a way that the sides stay in balance until the only object remaining on one side is the object of unknown mass. [145]
For example, in the simple equation 3 + 2y = 8y, both sides actually contain 2y (because 8y is the same as 2y + 6y). Therefore, the 2y on both sides can be cancelled out, leaving 3 = 6y, or y = 0.5. This is equivalent to subtracting 2y from both sides. At times, cancelling out can introduce limited changes or extra solutions to an equation.