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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 the simple case of a function of one variable, say, h(x), we can solve an equation of the form h(x) = c for some constant c by considering what is known as the inverse function of h. Given a function h : A → B, the inverse function, denoted h −1 and defined as h −1 : B → A, is a function such that
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
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 ...
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]
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