# 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) a^{2}x – x = a

D) a^{2}x + 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) a^{2}x – x = a <=>

x(a^{2} -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) a^{2}x + x = a <=>

(a^{2} + 1)x = a

In this case a^{2} + 1 ≠ 0 for every а, because it is a sum of on positive number(1) and one negative number

(a^{2} ≥ 0) therefore x = a/a^{2} + 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) (b^{2} + 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) b^{2} + 1 ≠ 0 for every b(why?), therefore

y = (a + 2)/(b^{2} + 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 = x^{2}

B) (x + m) / 2 = 1 – m and (x – m)/3 = 1 - 2m

C) |3 – x| + x^{2} -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 =
x^{2} <=>

[(-1)(x + 1) ]^{2} - 1 = x^{2} <=>

x^{2} + 2x + 1 - 1 =
x^{2} <=>

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 + x^{2} – 5x + 3 = 0 <=>

x^{2} - 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

#### Parametric equations in the forum