# Parametric equations

Equation which except the unknown quantity contains another letter which can take different values from some multitude is called parametric equation. This letter taking part in the equation is called parameter. Actually with every parametric equation is written a multitude of equations. We will observe the solutions of simple parametric equations and module parametric equations.

Problem 1 Solve the equation in reference to x
A) x + a = 7
B) 2x + 8a = 4
C) x + a = 2a – x
D) ax = 5
E) a – x = x + b
F) ax = 3a

Solution:

A) x + a = 7 <=> x = 7 – a, a solution to the given equation is found.
In different values of the parameter, the solutions are x = 7 – a

B) 2x + 8a = 4 <=> 2x = 4 - 8a <=> x = 2 – 4a

C) x + a = 2a – x <=> x + x = 2a – a <=> 2x = a <=> x = a/2

D) ax = 5, when a is different from 0 we can divide both sides by a and we get x = 5
If a = 0, we get equation from the kind 0.x = 5, which has no solution;

E) a – x = x + b <=> a – b = x + x <=> 2x = a – b <=> x = (a – b)/2

F) When a = 0 the equation ax = 3a is equal to 0.x = 0
Therefore every x is a solution. If a is different from 0, then
ax = 3a <=> x = 3a/a <=> x = 3

Problem 2 If a is a parameter solve the equation:
A) (a + 1)x = 2a + 3
B) 2a + x = ax + 4
C) a2x – x = a
D) a2x + x = a

Solution:

A) If a + 1 is different from 0, i.e.. a ≠ -1,
then x = (2a + 3)/(a + 1);
if a + 1 = 0, i.e. a = - 1
the equation get this look 0.x = (2).(-1) + 3 <=>
0.x = 1, which has no solution;

B) 2a + x = ax + 4 <=>
x – ax = 4 - 2a <=>
(1 – a).x = 2(2 – a)
If (1 – a) ≠ 0, i.e. a ≠ 1; the solution is
x = 2(2 - a) / (1 - a);
if a = 1 the equation is 0.x = 2(2 - 1) <=>
0.x = 2, which has no solution

C) a2x – x = a <=>
x(a2 -1) = a <=>
(a - 1)(a + 1)x = a
If a - 1 ≠ 0 and a + 1 ≠ 0 i.e. a ≠ 1, -1,
the solution is x = a/(a - 1)(a + 1)
If a = 1 or a = -1, the equation is 0.x = ±1, which has no solution

D) a2x + x = a <=>
(a2 + 1)x = a
In this case a2 + 1 ≠ 0 for every а, because it is a sum of on positive number(1) and one negative number
(a2 ≥ 0) therefore x = a/a2 + 1

Problem 3 If a and b are parameters solve the equation:
A) ax + b = 0
B) ax + 2b = x
C) (b - 1)y = 1 – a
D) (b2 + 1)y = a + 2

Solution:

A) ax + b = 0 <=> ax = -b
If a ≠ 0, the solution is x = -b/a.
If a = 0, b ≠ 0, the equation get the look 0.x = -b and has no solution.
If a = 0 and b = 0, the equation is 0.x = 0 and every x is solution;

B) ax + 2b = x <=> ax – x = -2b <=> (a - 1)x = -2b
If a - 1 ≠ 0, i.e. a ≠ 1, the solution is x = -2b/a - 1
If a - 1 = 0, i.e. a = 1, but b ≠ 0, the equation is 0.x = - 2b and has no solution

C) if b - 1 ≠ 0, i.e. b ≠ 1,
the solution is y = (1 – a)/(b - 1)
If b - 1 = 0, i.e. b = 1, but 1 – a ≠ 0,
i.e. a ≠ 1, the equation is 0.y = 1 – a and has no solution.
If b = 1 and a = 1 the equation is 0.y = 0 and every y is solution

D) b2 + 1 ≠ 0 for every b(why?), therefore
y = (a + 2)/(b2 + 1) is solution to the equation.

Problem 4 For which values of x the following expressions have equal values :
A) 5x + a and 3ax + 4
B) 2x - 2 and 4x + 5a

Solution:

To have equal values we must find the solutions of the equations
5x + a = 3ax + 4 and 2x – 2 = 4x + 5a

A) 5x + a = 3ax + 4 <=>
5x - 3ax = 4 – a <=>
(5 - 3a)x = 4 – a
If 5 - 3a ≠ 0, i.e. a ≠ 5/3, the solution is x = (4 – a)/(5 - 3a)
If 5 - 3a = 0, i.e. a = 5/3, the equation is 0.x = 4 – 5/3 <=>
0.x = 7/3, which has no solution

B) 2x - 2 = 4x + 5a <=>
-2 - 5a = 4x - 2x <=>
2x = - 2 - 5a <=>
x = -(2 + 5a)/2

Problem 5 Solve the parametric equation:
A) |ax + 2| = 4
B) |2x + 1| = 3a
C) |ax + 2a| = 3

Solution:

A) |ax + 2| = 4 <=> ax + 2 = 4 or ax + 2 = -4 <=>
ax = 2 or ax = - 6
If a ≠ 0, the equations are x = 2/a or x = -6/a
If a = 0, the equations have no solutions

B) If a < 0, the equation |2x + 1| = 3a has no solution.
If a > 0 it is equivalent to 2x + 1 = 3a
or 2x + 1 = -3a <=> 2x = 3a - 1 <=> x = (3a - 1)/2 or
2x = -3a - 1 <=> x = (3a - 1)/2 = -(3a + 1)/2

C) |ax + 2a| = 3 <=> ax + 2a = 3 or ax + 2a = - 3,
and we find ax = 3 - 2a or ax = -3 - 2a
If a = 0 there are no solutions, if a ≠ 0
they are x = (3 - 2a)/a and x = -(3 + 2a)/a

Problem 6 Solve the equation 2 – x = 2b – 2ax, where a and b are real parameters. Find for which values of a the equation has for solution a natural number, if b = 7

Solution:

We write the given equation like this (2a - 1)x = 2(b - 1)
The following cases are possible:
If 2a - 1 ≠ 0, i.e. a ≠ 1/2, the equation has single solution
x = 2(b - 1)/(2a - 1)
If a = 1/2 and b = 1, the equation gets the kind 0.x = 0 and every x is solution
If a = 1/2 and b ≠ 1, we get 0.x = 2(b - 1), where 2(b - 1) ≠ 0
In this case the equation has no solution.
If b = 7 and a ≠ 1/2 the single solution is
x = 2(7 - 1)/(2a - 1) = 12/(2a - 1)
If a is whole number, then 2a - 1 is also whole number and the solution
x = 12/(2a - 1) is natural number when
2a - 1 is positive divisor of 12.
To have a as whole number, the divisor of 12 mu be odd.But the only whole positive odd numbers divisible by 12 are 1 and 3.
Therefore 2a - 1 = 3 <=> a = 2 or 2a - 1 = 1 <=>
a = 1 a = 2 or 2a - 1 = 1 <=> a = 1

Problem 7 Solve the equation |ax - 2 – a| = 4, where a is parameter. Find for which values of а the roots of the equation are whole negative numbers.

Solution:

From the definition for module we get
|ax - 2 – x| = 4 <=> ax - 2 – x = 4 or ax - 2 – x = - 4
From the first equality we get x(a - 1) - 2 = 4 <=>
(a - 1)x = 4 + 2 <=> (a - 1)x = 6
From the second we have (a - 1)x = -2
If a - 1 = 0, i.e. a = 1, the last equations have no solutions.
If a ≠ 1 we find x = 6/(a - 1) or x = -2/(a - 1)
To have these roots as whole negative numbers must:
For the first equality a - 1 to be negative divisors of 6, and for the second positive divisors of 2
So a - 1 = -1; -2; -3; - 6 or a - 1 = 1; 2
We get a - 1 = -1 <=> a = 0; a - 1 = -2 <=>
a = -1; a - 1 = -3 <=> a = -2; a - 1 = -6 <=> a = -5
or a - 1 = 1 <=> a = 2; a - 1 = 2 <=> a = 3
So a = -5; -2; -1; 0; 2; 3 are solutions of the problem.

Problem 8 Solve the equation:
A) 3ax – a = 1 – x, where a is parameter;
B) 2ax + b = 2 + x, where a and b are parameters

Solution:

A) 3ax + x = 1 + a <=> (3a + 1)x = 1 + a.
If 3a + 1 ≠ 0, i.e. a ≠ -11 /3 /3 , the solution is
x = (1 + a)/(3a + 1)
If a = - 1/3 the equations gets the look 0.x = 1.1/3, which has no equation.

B) 2ax – x = 2 – b <=> (2a - 1)x = 2 – b
If 2a - 1 ≠ 0, i.e. a ≠ 1/2, x = (2 – b)/(2a - 1) is the solution.
If a = 1/2 the equation gets the look 0.x = 2 – b
Then, if b = 2, every x is solution, if b ≠ 2, the equation has no solution.

Problem 9 The equation 6(kx - 6) + 24 = 5kx is given, where к is whole number. Find for which values of k the equation :
A) has for root -4/3
B) no solutions;
C) has for root natural number.

Solution:

We remake the equation to 6kx - 36 + 24 = 5kx <=> kx = 12

A) If x = - 4/3, for k we get the equation - 4/3k = 12 <=> k = - 9

B) The equation kx = 12 has no solutions when k = 0

C) When k ≠ 0 the root is x = 12/k and he is a natural number, if k is whole positive number, dividing 12, i.e. k = 1, 2, 3, 4, 6, 12

Problem 10 Solve the equation:
A) 2ax + 1 = x + a, where a is parameter;
B) 2ax + 1 = x + b, where a и b are parameters.

Solution:

A) 2ax + 1 = x + a <=> 2ax – x = a - 1 <=>
(2a - 1)x = a - 1
If 2a - 1 ≠ 0, i.e. a ≠ 1/2, the only solution of the equation is
x = (a - 1)/(2a - 1)
If 2a - 1 = 0, i.e. a = 1/2, the equation take the look
0.x = 1/2 - 1 <=> 0.x = - 1/2, which has no solution

B) 2ax + 1 = x + b <=>
2ax – x = b - 1 <=>
(2a - 1)x = b - 1
If 2a - 1 ≠ 0, i.e. a ≠ 1/2, the solution is
x = (b - 1)/(2a - 1)
If a = 1/2, the equation is equivalent to 0.x = b - 1
If b = 1 every x is solution, if b ≠ 1 there is no solution.

Problem 11 The equation 3(ax - 4) + 4 = 2ax is given, where the parameter a is whole number. Find for which values of a the equation has for root:
А) (-2/3)
B) whole number
C) natural number

Solution:

A) If x = -2/3 is solution to the equation it is valid that
3[a(-2/3) - 4] + 4 = 2a(-2/3) <=>
-2a - 12 + 4 = -4a/3 <=>
4a/3 - 2a = 8 <=> (4a - 6a)/3 = 8 <=>
-2a/3 = 8 <=> a = -12

B) 3(ax - 4) + 4 = 2ax <=> 3ax - 2ax = 12 - 4 <=> ax = 8
If a ≠ 0 the solution of the equation is x = 8/a, it is whole number if it is divisor of 8.
Therefore; ±2; ±4; ±8
If a=0 the equation has no solution

C) To have natural(whole positive) number for the solution x=8/a the number must a=1, 2, 4, 8

Problem 12 The equation 2 – x = 2b – 2ax is given, where a and b are parameters. Find for which values of a the equation has for solution a natural number, if b = 7

Solution:

In the equation we substitute b = 7 and we get 2 – x = 2.7 - 2ax <=>
2ax – x = 14 – 2 <=> (2a - 1)x = 12
If 2a -1 ≠ 0, i.e. a ≠ 1/2, the equation is
x = 12/(2a - 1) and it will be natural number, if the denominator 2a - 1 is positive divisor of 12 and except that, to have а as whole number it is necessary 2a - 1 to be odd number.
Therefore 2a - 1 can be 1 or 3
From 2a - 1 = 1 <=> 2a = 2 <=> a = 1 and 2a - 1 = 3
<=> 2a = 4 <=> a = 2

Problem 13 The function f(x) = (3a - 1)x - 2a + 1 is given, where a is parameter. Find for which values of a the graphic of the function:
А) crosses the abscissa axis;
B) don’t cross the abscissa axis

Solution:

To be able to cross the graphic of a function the abscissa axis, needs the equation
(3a - 1).x -2a + 1 = 0 to have solution and will not cross it if there is no solution.
From the given equation we get (3a - 1)x = 2a - 1
If 3a - 1 ≠ 0, i.e. a ≠ 1/3, the equation have solution
x = (2a – 1)/(3a - 1), therefore the graphic of the function crosses the abscissa axis.
If a = 1/3 we get 0.x = 2/3 - 1 <=> 0.x = -1/3, which has no solution .
Therefore if a = 1/3 the graphic do not cross the abscissa axis.

Problem 14 Solve the parametric equation:
A) |x -2| = a
B) |ax -1| = 3
C) |ax - 1| = a - 2

Solution:

A) if a < 0 the equation has no solution, if a > 0 we get:
|x - 2| = a <=> x - 2 = a or x - 2 = -a
From x - 2 = a => x = a + 2, and from
x - 2 = -a => x = 2 – a
If a = 0, then x - 2 = 0 or x = 2

B) |ax - 1| = 3 <=> ax - 1 = 3 or ax - 1 = -3
from where ax = 4 or ax = - 2
If a ≠ 0 the solutions are x = 4/a or x = - 2/a
If a = 0 there is no solution

C) if a - 2 < 0, i.e. a < 2, the equation has no solution
If a - 2 > 0, i.e. a > 2 we get
|ax - 1| = a - 2 <=> ax - 1 = a - 2 or ax - 1 = 2 – а
So we get ax = a - 1 or ax = 3 – a
Because a > 2, a ≠ 0, therefore
x = (a - 1)/a or x = (3 – a)/a.
If a = 2 the equation is equivalent with
2x - 1 = 0 <=> 2x = 1 <=> x = 1/2

Problem 15 Find for which values of the parameter m (a) the two equations are equivalent:
A) (x + m) / 2 = 1 – m and (-x - 1) 2 - 1 = x2
B) (x + m) / 2 = 1 – m and (x – m)/3 = 1 - 2m
C) |3 – x| + x2 -5x + 3 = 0 и ax + 2a = 1 + x, if x > 3

Solution:

A) We will solve the second equation. We remake it in this way
(-x - 1)2 - 1 = x2 <=>
[(-1)(x + 1) ]2 - 1 = x2 <=>
x2 + 2x + 1 - 1 = x2 <=>
2x = 0 <=> x = 0
For the first one we get
(x + m)/2 = 1 – m <=> x + m = 2 - 2m <=> x = 2 - 3m
The two equations are equivalent if they have the same roots, i.e.
2 - 3m = 0 <=> m = 2/3

B) For the first equation the solution is х = 2 - 3m and for the second we get
x – m = 3 - 6m <=> x = 3 – 5m
They have the same roots when
2 - 3m = 3 - 5m <=> 5m - 3m = 3 - 2 <=> 2m = 1 <=> m = 1/2

C) Because x > 3, 3 – x < 0, therefore
|3 – x| = -(3 – x) = x - 3
The first equation gets the look x - 3 + x2 – 5x + 3 = 0 <=>
x2 - 4x – 0 <=> x(x - 4) = 0 <=>
x = 0 or x = 4
By condition x > 3, therefore only x = 4 is solution. For the second equation we get
ax – x = 1 - 2a <=> (a - 1)x = 1 - 2a
If a - 1 = 0 there is no solution(Why?), if a - 1 ≠ 0, i.e. a ≠ 1, the solution is
x = (1 - 2a)/(a - 1) The two equations will be equivalent if 4 = (1 - 2a)/(a - 1) <=> 4(a - 1) = 1 - 2a <=> 4a + 2a = 1 + 4 <=> 6a = 5 <=> a = 5/6

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