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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:
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 complete the square, form a squared binomial on the left-hand side of a quadratic equation, from which the solution can be found by taking the square root of both sides. The standard way to derive the quadratic formula is to apply the method of completing the square to the generic quadratic equation a x 2 + b x + c = 0 {\displaystyle ...
Taking the square root of both sides, and isolating x, gives: =. Some sources, particularly older ones, use alternative parameterizations of the quadratic equation such as ax 2 + 2 bx + c = 0 or ax 2 − 2 bx + c = 0 , [ 11 ] where b has a magnitude one half of the more common one, possibly with opposite sign.
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.
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 ...
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: (′) (′) =.
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]