cadaVR anatomy by Mozaik Education

Parametric equations

(higher level courseware)

The equations, in which a variable (letter) appears which should be considered as a given number during the solution and the root of the equation should be given with the help of this, are called parametric equations.

Example 1

Let us solve equation \latex{ax=1} where \latex{ a } denotes a real parameter.

Solution

From the equation we can get the unknown by dividing both sides by the value of parameter \latex{ a. } This operation can only be done if the value of the parameter is not \latex{ 0. } Therefore in this case we have to examine the following cases:

If \latex{a = 0}, then we cannot divide, the equation in this case takes the form \latex{0 \times x = 1}, i.e. \latex{0 = 1}, which is a contradiction. In this case we do not get a solution.
If \latex{a\neq0}, then we can do the division, and \latex{x=\frac{1}{a}} is resulting.

Thus if \latex{a\neq0}, then \latex{x=\frac{1}{a}}, if \latex{a=0}, then there is no solution.

Example 2

Let us solve equation \latex{(b-1)x=b^2-1}, where b denotes a real parameter.

Solution

If \latex{b = 1}, then the equation takes the form of \latex{0 \times x = 0}; every real number is a root of this equation (we get an identity).
If \latex{b \neq 1}, then doing the division by \latex{b – 1}:

\latex{x=\frac{b^2-1}{b-1}=\frac{(b+1)(b-1)}{b-1}=b+1}

In this case the equation will have one solution: \latex{x=b+1}.
Thus if \latex{b = 1}, then every real number will be a solution, if \latex{b \neq 1}, then there will be only one solution: \latex{x = b + 1}.

Instead of the coefficients he used vowels, and he denoted the unknown quantities by consonants.

But the system of notation used today is attached to the name of RENÉ DESCARTES \latex{ (1596–1650) }, a French philosopher and mathematician who had a “novelistic” life.

Example 3

Let us solve equation \latex{ax = x + b}, where a and b are real parameters.

Solution

Let us first rearrange the equation:

\latex{ax-x=b},

\latex{(a-1)x=b}.

In the next step we would like to divide by the parametric expression. However in this case we have to pay attention, since this operation can be done only in the determined cases.

If \latex{a = 1}, then the equation is as follows: \latex{0 \times x = b}.
If \latex{b = 0}, then we get an identity for which every real number will be a solution.
If \latex{b\neq0}, then a contradiction is resulting, and the equation will have no solution.
If \latex{a\neq1}, then from the equation: \latex{x=\frac{b}{a-1}}.

Summarised:
If \latex{a=1} and \latex{b=0}, then every real number is a solution of the equation.
If \latex{a=1} and \latex{b\neq0}, then there is no solution.
If \latex{a\neq1}, then we get solution \latex{x=\frac{b}{a-1}}.

Example 4

Let us solve the following equations, where letter \latex{ a } denotes a real parameter.

\latex{\frac{2x-4a}{3}=\frac{4x-3a}{4}}

\latex{\frac{3}{2x-4a}=\frac{4}{4x-3a}}

Solution (a)

In the first step let us multiply both sides by 12:

\latex{4(2x-4a)=3(4x-3a)},
\latex{8x-16a=12x-9a},
\latex{-4x=7a},
\latex{x=-\frac{7a}{4}}.

This is indeed a solution of the equation for any real parameter \latex{ a }; we can make sure of it during checking.

Solution (b)

This equation has a meaning if \latex{2x-4a\neq0} and \latex{4x-3a\neq 0}. Equation b) is the reciprocal of equation a). If two non-zero quantities are equal, then their reciprocals are equal too. In this case we get the previous equation with solution \latex{x=-\frac{7a}{4}}

But here we have to pay attention that only those can be accepted as solutions for which the denominators of fractions in the equation do not take the value of 0.
The denominator on the left side would be equal to 0 if

\latex{2x-4a=0},
\latex{2\times\left(-\frac{7a}{4}\right)-4a=0},
\latex{-\frac{15}{2}a=0}.

It is only possible if \latex{a = 0}. We similarly get that the denominator on the right side can only be equal to 0 if \latex{a = 0}.
So the result is the following: the root of the equation is \latex{x=-\frac{7a}{4}}, if \latex{a\neq0}. If \latex{a=0}, then the equation does not have a solution.

⯁ ⯁ ⯁

So we can state that we can look for the solution of the parametric equations with the help of the previously learnt methods. But when doing the transformations we always have to pay attention to the scope of the operations to be done. (For example we should not divide by zero.) Therefore while doing this type of operation we always have to examine the possible cases, and we have to give the solutions for every allowable value of the parameter.

Exercises

{{exercise_number}}. Solve the following parametric equations.

\latex{ax=a}

\latex{x+b=bx}

\latex{ax=b+1}

\latex{ax=a^2}

{{exercise_number}}. Solve the following parametric equations.

\latex{(a+1)x=a^2-1}

\latex{(a^2-2a+1)x=a-1}

\latex{b^2x+1=b(x+1)}

\latex{a(ax-1)=ax+2}

{{exercise_number}}. From the relations learnt in Physics express all the possible variables as a function of the others. Interpret the physical content of the conditions formulated in each case.

\latex{s=v\times t}

\latex{F=p\times A}

\latex{U=I\times R}

\latex{W=P\times t}

{{exercise_number}}. Solve the equation on the set of real numbers where \latex{ a } and \latex{ b } denote real parameters.

\latex{\frac{x-a}{a}-a=\frac{2x-b}{b}}

{{exercise_number}}. For which values of real parameters \latex{ a } and \latex{ b } will the root of the equation in exercise \latex{ 4 } be

zero;

positive;

negative?

Mathematics 9.

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