Linear Equations
NCERT Exercise 3.5
Part 1
Question 1: Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
(a) `x – 3y – 3 = 0` and `3x – 9y – 2 = 0`
Solution:
`(a_1)/(a_2)=1/3`
`(b_1)/(b_2)=(-3)/(-9)=1/3`
`(c_1)/(c_2)=3/2`
It is clear that;
`(a_1)/(a_2)=(b_1)/(b_2)≠(c_1)/(c_2)`
Hence, there will be no solution for the given pair of linear equations.
(b) `2x + y = 5` and `3x + 2y = 8`
Solution:
`(a_1)/(a_2)=2/3` and `(b_1)/(b_2)1/2`
It is clear that;
`(a_1)/(a_2)≠(b_1)/(b_2)`
Hence, there will be unique solution for the given pair of linear equations.
From cross-multiplication method, we know;
`(x)/(b_1c_2-b_2c_1)=(y)/(c_1a_2-c_2a_1)=(1)/(a_1b_2-a_2b_1)`
Or, `(x)/(-8+10)=(y)/(-15+15)=(1)/(4-3)`
Or, `x/2=y/1=1`
Or, `x=1` and `y=2`
(c) 3x – 5y = 20 and 6x – 10y = 40
Solution: a1 = 3, b1 = - 5, c1 = - 20
a2 = 6, b2 = - 10, c2 = - 40
`(a_1)/(b_1)=3/6=1/2`
`(b_1)/(b_2)=(-5)/(-10)=1/2`
`(c_1)/(c_2)=(-20)/(-40)=1/2`
It is clear that;
`(a_1)/(a_2)=(b_1)/(b_2)=(c_1)/(c_2)`
Hence, there will be infinitely many solutions for the given pair of linear equations.
(d) x – 3y – 7 = 0 and 3x – 3y – 15 = 0
Solution: a1 = 1, b1 = - 3, c1 = - 7
a2 = 3, b2 = - 3, c2 = - 15
`(a_1)/(a_2)=1/3` and `(b_1)/(b_2)=(-3)/(-3)=1`
It is clear that;
`(a_1)/(a_2)≠(b_1)/(b_2)`
Hence, there will be unique solution for the given pair of linear equations.
From cross-multiplication method, we know;
`(x)/(b_1c_2-b_2c_1)=(y)/(c_1a_2-c_2a_1)=(1)/(a_1b_2-a_2b_1)`
Or, `(x)/((-3xx-15)-(-3xx-7))`
`=(y)/((-7xx3)-(-15xx1))`
`=(1)/((1xx-3)-(3xx-3)`
Or, `(x)/(45-21)=(y)/(-21+15)=(1)/(-3+9)`
Or, `(x)/(24)=(y)/(-6)=(1)/(6)`
Or, `x=(24)/(6)=4`
And `y=(-6)/(6)=-1`
Question 2: For which values of a and b does the following pair of linear equations have an infinite number of solutions?
2x + 3y = 7 and (a – b)x + (a + b)y = 3a + b – 2
Solution: a1 = 2, b1 = 3, c1 = - 7
a2 = (a – b), b2 = (a + b), c2 = - (3a + b – 2)
For infinite number of solutions, the equations should fulfill following criterion:
`(a_1)/(a_2)=(b_1)/(b_2)=(c_1)/(c_2)`
Or, `(2)/(a-b)=(3)/(a+b)`
`=(-7)/(-(3a+b-2))`
Or, `2(a + b) = 3(a – b)`
Or, `2a + 2b = 3a – 3b`
Or, `2a + 5b = 3a`
Or, `a = 5b` --------(1)
Similarly, `6a + 2b – 4 = 7a – 7b`
Or, `6a + 2b – 7a + 7b = 4`
Or, `- a + 9b = 4`
Or, `a = 9b + 4` --------(2)
From equations (1) and (2), it is clear;
`5b = 9b – 4`
Or, `4b = 4`
Or, `b = 1`
Substituting the value of b in equation (1), we get;
`a = 5b = 5`
Hence, `a = 5` and `b = 1`
Question 3: For which value of k will the following pair of linear equations have no solution?
3x + y = 1 and (2k – 1)x + (k – 1)y = 2k + 1
Solution: a1 = 3, b1 = 1, c1 = 1
a2 = (2k – 1), b2 = (k – 1), c2 = - (2k + 1)
For no solution, the equations should fulfill following criterion:
`(a_1)/(a_2)=(b_1)/(b_2)≠(c_1)/(c_2)`
Or, `(3)/(2k-1)=(1)/(k-1)≠(1)/(2k+1)`
Or, `3(k – 1) = 2k – 1`
Or, `3k – 3 = 2k – 1`
Or, `3k = 2k – 1 + 3`
Or, `3k = 2k + 2`
Or, `k = 2`
Question 4: Solve the following pair of linear equations by the substitution and cross-multiplication methods: 8x + 5y = 9 and 3x + 2y = 4.
Solution: Substitution method:
Let us use the first equation to express one variable in terms of another variable;
8x + 5y = 9
Or, 8x = 9 – 5y
Or, `x=(9-5y)/(8)`
Substituting the value of x in second equation, we get;
`3x+2y=4`
Or, `3((9-5y)/(8))+2y=4`
Or, `(27-15y)/(8)+2y=4`
Or, `(27-15y+16y)/(8)=4`
Or, `27+y=32`
Or, `y=32-27=5`
Substituting the value of y in first equation, we get;
`x=(9-5y)/(8)=(9-5xx5)/(8)`
`=(9-25)/(8)=(-16)/(8)=-2`
Hence, x = - 2 and y = 5
Cross-multiplication Method:
8x + 5y = 9 and 3x + 2y = 4
a1 = 8, b1 = 5, c1 = - 9
a2 = 3, b2 = 2, c2 = - 4
From cross-multiplication method, we know;
`(x)/(b_1c_2-b_2c_1)=(y)/(c_1a_2-c_2a_1)=(1)/(a_1b_2-a_2b_1)`
Or, `(x)/((5xx-4)-(2xx-9))`
`=(y)((-9xx3)-(-4xx8))`
`=(1)/((8xx2)-(3xx5))`
Or, `(x)/(-20+18)=(y)/(-27+32)=(1)/(16-15)`
Or, `(x)/(-2)=y/5=1`
Or, x = - 2 and y = 5