Question

15. Find the values of λ for which the equations (2 − λ)x + 2y + 3 = 0, 2x + (4 − λ)y + 7 =

0, 2x + 5y + (6 − λ) = 0 are consistent and find the values of x and y corresponding to each

of these values of λ.​

Answer 1

The value of y is 1 and the value of x is -5 if the value of λ is 1 and the value of x is 0.5 if the value of λ is 12.

Given:

(2 − λ)x + 2y + 3 = 0

2x + (4 − λ)y + 7 =0

2x + 5y + (6 − λ) = 0

To find:

The values of x and y of each values of λ.

Solution:

(2 − λ)x + 2y + 3 = 0              ............................(1)

2x + (4 − λ)y + 7 =0               .............................(2)

2x + 5y + (6 − λ) = 0             ..............................(3)

• We can solve the 3 equations with the help of substitution method, elimination method and cross- multiplication method.
• In substitution method we have to we solve the equation further or we take all the variables on one side except one and put the obtained value of the variable in terms of others variables in other equations.
• In elimination method we have to eliminate one of the variables by performing some operations on the equations.

On subtracting equation 2 from equation 3 we get,

(4-λ)y-5y+7-(6-λ)=0

4y-λy-5y+7-6+λ=0

-λy-y+1+λ=0

-(λ+1)y+(λ+1)=0

-(λ+1)y= -(λ+1)

y= -(λ+1)/-(λ+1)

y=1

So, the value of y is 1.

On, putting the value of y in equation 1 and 2 we get,

(2 − λ)x+5 = 0            ....................................(4)

2x-λ+11=0             ....................................(5)

On solving equation 4 further we get,

(2 − λ)x= -5

x= -5/(2 − λ)

Now, put the value of x in equation 5.

2{-5/(2 − λ)}- λ+11=0

-10/(2 − λ) -λ+11=0

-10-(2 − λ)λ+11(2 − λ)/ (2 − λ)=0

-10-2λ+λ²+22-11λ=0

λ²-13λ+12=0

λ²-12λ-λ+12=0                        (Middle term splitting method)

λ(λ-12)-1(λ-12)=0

(λ-12)(λ-1)=0

λ= 1, 12

Thus, the values of λ are 1 and 12.

Put the values of λ and y in equation 3.

For λ=1 and y=1

2x + 5(1) + (6 − 1) = 0

2x+5+5=0

2x+10=0

2x=-10

x=-10/2

x= -5

So, for λ=1 and y=1, x= -5.

Similarly, for λ=12 and y=1

2x + 5(1) + (6 − 12) = 0

2x+5-6=0

2x-1=0

2x=1

x=1/2

x=0.5

So, for λ=12 and y=1, x= 0.5.

Therefore, the value of y is 1 and the value of x is -5 if the value of λ is 1 and the value of x is 0.5 if the value of λ is 12.

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