Pa.Linear Eq Ex 3.3 Solutions
Question - 1 : - For which value(s) of λ , do the pair of linear equations
Answer - 1 : -
λx + y =λ2 and x + λy = 1 have
(i) no solution?
(ii) infinitely manysolutions?
(iii) a uniquesolution?
Solution:
The given pair oflinear equations is
λx + y = λ2 andx + λy = 1
a1 =λ, b1= 1, c1 = – λ2
a2 =1,b2=λ, c2=-1
The given equationsare;
λ x + y – λ2 =0
x + λ y – 1 = 0
Comparing the aboveequations with ax + by + c = 0;
We get,
a1 = λ,b1 = 1, c1 = – λ 2;
a2 =1, b2 = λ, c2 = – 1;
a1 /a2 =λ/1
b1 /b2 =1/λ
c1 /c2 =λ2
(i) For no solution,
a1/a2 =b1/b2≠ c1/c2
i.e. λ = 1/λ ≠ λ2
so, λ 2 =1;
and λ 2 ≠ λ
Here, we takeonly λ = – 1,
Since the system oflinear equations has no solution.
(ii) For infinitely manysolutions,
a1/a2 =b1/b2 = c1/c2
i.e. λ = 1/λ = λ2
so λ = 1/λ gives λ = + 1;
λ = λ 2 gives λ= 1,0;
Hence satisfying boththe equations
λ = 1 is theanswer.
(iii) For a uniquesolution,
a1/a2 ≠ b1/b2
so λ ≠1/ λ
hence, λ2 ≠ 1;
λ ≠ + 1;
So, all real values ofλ except +1
Question - 2 : - For which value(s) of k will the pair of equationskx + 3y = k – 3
12x + ky = k
have no solution?
Answer - 2 : -
Solution:
The given pair oflinear equations is
kx + 3y = k – 3 …(i)
12x + ky = k …(ii)
On comparing theequations (i) and (ii) with ax + by = c = 0,
We get,
a1 = k,b1 = 3, c1 = -(k – 3)
a2 =12, b2 = k, c2 = – k
Then,
a1 /a2 = k/12
b1 /b2 =3/k
c1 /c2 =(k-3)/k
For no solution of thepair of linear equations,
a1/a2 =b1/b2≠ c1/c2
k/12 = 3/k ≠(k-3)/k
Taking first twoparts, we get
k/12 = 3/k
k2 =36
k = + 6
Taking last two parts,we get
3/k ≠ (k-3)/k
3k ≠ k(k –3)
k2 –6k ≠ 0
so, k ≠ 0,6
Therefore, value of kfor which the given pair of linear equations has no solution is k = – 6.
Question - 3 : - For which valuesof a and b, will the following pair of linearequations have infinitely many solutions?
x + 2y = 1
(a – b)x +(a + b)y = a + b –2
Answer - 3 : -
Solution:
The given pair oflinear equations are:
x + 2y = 1 …(i)
(a-b)x + (a + b)y = a+ b – 2 …(ii)
On comparing with ax +by = c = 0 we get
a1 = 1,b1 = 2, c1 = – 1
a2 =(a – b), b2 = (a + b), c2 = – (a + b – 2)
a1 /a2 = 1/(a-b)
b1 /b2 = 2/(a+b)
c1 /c2 =1/(a+b-2)
For infinitely manysolutions of the, pair of linear equations,
a1/a2 =b1/b2=c1/c2(coincident lines)
so,1/(a-b) = 2/ (a+b) = 1/(a+b-2)
Taking first twoparts,
1/(a-b) = 2/(a+b)
a + b = 2(a – b)
a = 3b …(iii)
Taking last two parts,
2/ (a+b)= 1/(a+b-2)
2(a + b – 2) = (a + b)
a + b = 4 …(iv)
Now, put the value ofa from Eq. (iii) in Eq. (iv), we get
3b + b = 4
4b = 4
b = 1
Put the value of b inEq. (iii), we get
a = 3
So, the values (a,b) =(3,1) satisfies all the parts. Hence, required values of a and b are 3 and 1respectively for which the given pair of linear equations has infinitely manysolutions.
Question - 4 : - Find the value(s)of p in (i) to (iv) and p and q in(v) for the following pair of equations:
(i) 3x – y –5 = 0 and 6x – 2y – p = 0, if thelines represented by these equations are parallel.
Answer - 4 : -
Solution:
Given pair of linearequations is
3x – y – 5 = 0 …(i)
6x – 2y – p = 0 …(ii)
On comparing with ax +by + c = 0 we get
We get,
a1 =3, b1 = – 1, c1 = – 5;
a2 =6, b2 = – 2, c2 = – p;
a1 /a2 = 3/6= ½
b1 /b2 =½
c1 /c2 =5/p
Since, the linesrepresented by these equations are parallel, then
a1/a2 =b1/b2 ≠ c1/c2
Taking last two parts,we get ½ ≠ 5/p
So, p ≠ 10
Hence, the given pairof linear equations are parallel for all real values of p except 10.
Question - 5 : - Find the value(s)of p in (i) to (iv) and p and q in(v) for the following pair of equations:
(ii) – x + py =1 and px – y = 1, if the pair of equationshas no solution.
Answer - 5 : -
Solution:
Given pair of linear equationsis
– x + py = 1 …(i)
px – y – 1 = 0 …(ii)
On comparing with ax +by + c = 0,
We get,
a1 =-1, b1 = p, c1 =- 1;
a2 =p, b2 = – 1, c2 =- 1;
a1 /a2 =-1/p
b1 /b2 =– p
c1 /c2 =1
Since, the linesequations has no solution i.e., both lines are parallel to each other.
a1/a2 =b1/b2≠ c1/c2
-1/p = –p ≠ 1
Taking last two parts,we get
p ≠ -1
Taking first twoparts, we get
p2 = 1
p = + 1
Hence, the given pairof linear equations has no solution for p = 1.
Question - 6 : - Find the value(s)of p in (i) to (iv) and p and q in(v) for the following pair of equations:
(iii) – 3x +5y = 7 and 2px – 3y = 1, if the linesrepresented by these equations are intersecting at a unique point.
Answer - 6 : -
Solution:
Given, pair of linearequations is
– 3x + 5y = 7
2px – 3y = 1
On comparing with ax +by + c = 0, we get
Here, a1 =-3, b1 = 5, c1 = – 7;
And a2 =2p, b2 = – 3, c2 = – 1;
a1 /a2 =-3/ 2p
b1 /b2 =– 5/3
c1 /c2 =7
Since, the lines areintersecting at a unique point i.e., it has a unique solution
a1/a2 ≠ b1/b2
-3/2p ≠ -5/3
p ≠ 9/10
Hence, the linesrepresented by these equations are intersecting at a unique point for all realvalues of p except 9/10
Question - 7 : - Find the value(s)of p in (i) to (iv) and p and q in(v) for the following pair of equations:
(iv) 2x +3y – 5 = 0 and px – 6y – 8 = 0, if thepair of equations has a unique solution.
Answer - 7 : -
Solution:
Given, pair of linearequations is
2x + 3y – 5 = 0
px – 6y – 8 = 0
On comparing with ax +by + c = 0 we get
Here, a1 = 2,b1 = 3, c1 = – 5;
And a2 =p, b2 = – 6, c2 = – 8;
a1 /a2 = 2/p
b1 /b2 =– 3/6 = – ½
c1 /c2 =5/8
Since, the pair oflinear equations has a unique solution.
a1/a2 ≠ b1/b2
so 2/p ≠ – ½
p ≠ – 4
Hence, the pair oflinear equations has a unique solution for all values of p except – 4.
Question - 8 : - Find the value(s)of p in (i) to (iv) and p and q in(v) for the following pair of equations:
(v) 2x + 3y = 7 and 2px + py= 28 – qy, if the pair of equations have infinitely many solutions.
Answer - 8 : -
Solution:
Given pair of linearequations is
2x + 3y = 7
2px + py = 28 – qy
or 2px + (p + q)y – 28= 0
On comparing with ax +by + c = 0,
We get,
Here, a1 =2, b1 = 3, c1 = – 7;
And a2 =2p, b2 = (p + q), c2 = – 28;
a1/a2 =2/2p
b1/b2 =3/ (p+q)
c1/c2 =¼
Since, the pair ofequations has infinitely many solutions i.e., both lines are coincident.
a1/a2 =b1/b2 = c1/c2
1/p = 3/(p+q) = ¼
Taking first and thirdparts, we get
p = 4
Again, taking last twoparts, we get
3/(p+q) = ¼
p + q = 12
Since p = 4
So, q = 8
Here, we see that thevalues of p = 4 and q = 8 satisfies all three parts.
Hence, the pair ofequations has infinitely many solutions for all values of p = 4 and q = 8.
Question - 9 : - Two straight paths are represented by the equations x –3y = 2 and –2x + 6y = 5.Check whether thepaths cross each other or not.
Answer - 9 : -
Solution:
Given linear equationsare
x – 3y – 2 = 0 …(i)
-2x + 6y – 5 = 0 …(ii)
On comparing with ax +by c=0,
We get
a1 =1,b1 =-3, c1 =- 2;
a2 =-2, b2 =6, c2 =- 5;
a1/a2 =– ½
b1/b2 =– 3/6 = – ½
c1/c2 =2/5
i.e., a1/a2 =b1/b2 ≠ c1/c2 [parallellines]
Hence, two straightpaths represented by the given equations never cross each other, because
they are parallel toeach other.
Question - 10 : - Write a pair of linear equations which has the unique solution x =– 1, y =3. How many such pairs can you write?
Answer - 10 : -
Solution:
Condition for the pairof system to have unique solution
a1/a2 ≠ b1/b2
Let the equations be,
a1x + b1y+ c1 = 0
a2x + b2y+ c2 = 0
Since, x = – 1 and y =3 is the unique solution of these two equations, then
It must satisfy theequations –
a1(-1) + b1(3)+ c1 = 0
– a1 +3b1 + c1 = 0 …(i)
and a2(- 1)+ b2(3) + c2 = 0
– a2 +3b2 + c2 = 0 …(ii)
Since for thedifferent values of a1, b1, c1 and a2,b2, c2 satisfy the Eqs. (i) and (ii).
Hence, infinitely manypairs of linear equations are possible.