PROBLEMS ON SOLVING SYSTEM OF EQUATIONS WITH SPECIAL CASES

Problem 1 :

For what value of k the pair of equations

x + (k + 1) y = 5

(k + 1)x + 9y = 8k - 1

has infinitely many solutions.

Solution :

x + (k + 1) y = 5

(k + 1)x + 9y = 8k - 1

Since the system of equation has infinitely many solution, then 

m₁ = m₂

b₁ = b₂

From x + (k + 1) y = 5

y = -x/(k + 1) + 5/(k + 1)

m₁ = -1/(k + 1) and b₁ = 5/(k + 1)

From (k + 1)x + 9y = 8k - 1

y = -(k + 1)/9 + (8k - 1)/(k + 1)

m₂ = -(k + 1)/9 and b₂ = (8k - 1)/(k + 1)

Equating slopes, we get

-1/(k + 1) = -(k + 1)/9

9 = (k + 1)²

So, the value of k are -4 and 2.

Problem 2 :

Find the value of k for which the pair of equations

2x + 3y = 7

(k - 1)x + (k + 2)y = 3k

has infinitely many solutions.

Solution :

2x + 3y = 7

(k - 1)x + (k + 2)y = 3k

Since the system of equation has infinitely many solution, then 

m₁ = m₂

b₁ = b₂

From 2x + 3y = 7

y = -2x/3 + 7/3

m₁ = -2/3 and b₁ = 7/3

From (k - 1)x + (k + 2)y = 3k

y = -(k - 1)x/(k + 2) + 3k/(k + 2)

m₂ = -(k - 1)/(k + 2) and b₂ = 3k/(k + 2)

Equating slopes, we get

-2/3 = -(k - 1)/(k + 2)

2(k + 2) = 3(k - 1)

2k + 4 = 3k - 3

2k - 3k = -3 - 4

-k = -1

k = 1

So, the value of k is 1.

Problem 3 :

For what value of k the pair of equations

kx + 2y = 5

3x - 4y = 10

has no solution.

Solution :

kx + 2y = 5

3x - 4y = 10

Since the system of equations has no solutions, slopes will be equal and y-intercepts will be different.

m₁ = m₂

From kx + 2y = 5

2y = -kx + 5

y = (-k/2)x + (5/k)

m₁ = -k/2

From 3x - 4y = 10

4y = 3x - 10

y = (3/4)x - (10/4)

m₂ = 3/4

-k/2 = 3/4

-4k = 6

k = -6/4

k = -3/2

So, the value of k is -3/2.

Problem 4 :

For what value of k the pair of equations

3x + y = 1

(2k - 1)x + (k - 1)y = (2k + 1)

has no solution.

Solution :

3x + y = 1

(2k - 1)x + (k - 1)y = (2k + 1)

Since the system of linear equations has so solution, their slopes and y-intercepts will be equal.

m₁ = m₂

From 3x + y = 1

y = -3x + 1

m₁ = -3

From (2k - 1)x + (k - 1)y = (2k + 1)

(k - 1)y = - (2k - 1) x + (2k + 1)

y = [- (2k - 1)/(k - 1)]x + (2k + 1)/(k - 1)

m₁ = - (2k - 1)/(k - 1)

Equating the slopes, we get

-3 = - (2k - 1)/(k - 1)

3(k - 1) = 2k - 1

3k - 3 = 2k - 1

3k - 2k = -1 + 3

k = 2

So, the value of k is 2.

Problem 5 :

Show that the system of equations

3x + 4y = 8

6x + 8y = 10

is inconsistent.

Solution :

3x + 4y = 8

6x + 8y = 10

Since the system is inconsistent, it has no solution.

m₁ = m₂

From 3x + 4y = 8

4y = -3x + 8

y = (-3/4)x + (8/4)

m₁ = -3/4

From 6x + 8y = 10

8y = -6x + 10

y = (-6/8)x + (10/8)

y = (-3/4)x + (5/4)

m₂ = -3/4

Since the slopes are equal, then it is inconsistent.

Problem 6 :

For what value of k for which the pair of equations.

2x + 5y = 0

kx + 10y = 0

has a non zero solution.

Solution :

2x + 5y = 0 ------(1)

kx + 10y = 0 ------(2)

From (1), 5y = -2x

y = (-2/5)x

m₁ = -2/5

From (2), 10y = -kx

y = (-k/10)x

m₂ = -k/10

m₁ = m₂

-2/5 = -k/10

2/5 = k/10

20 = 5k

k = 20/5

k = 4

So, the value of k is 4.