# 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:** a_{1} = 3, b_{1} = - 5, c_{1} = - 20

a_{2} = 6, b_{2} = - 10, c_{2} = - 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:** a_{1} = 1, b_{1} = - 3, c_{1} = - 7

a_{2} = 3, b_{2} = - 3, c_{2} = - 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:** a_{1} = 2, b_{1} = 3, c_{1} = - 7

a_{2} = (a – b), b_{2} = (a + b), c_{2} = - (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:** a_{1} = 3, b_{1} = 1, c_{1} = 1

a_{2} = (2k – 1), b_{2} = (k – 1), c_{2} = - (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

a_{1} = 8, b_{1} = 5, c_{1} = - 9

a_{2} = 3, b_{2} = 2, c_{2} = - 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