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USE CODE: SELF10 

Chapter 8: Linear Equations in One Variable Exercise – 8.1



Question: 1



Verify by substitution that:



(i) x = 4 is the root of 3x – 5 = 7



(ii) x = 3 is the root of 5 + 3x = 14



(iii) x = 2 is the root of 3x – 2 = 8x – 12



(iv) x = 4 is the root of 3x/2 = 6



(v) y = 2 is the root of y – 3 = 2y – 5



(vi) x = 8 is the root of (1/2)x + 7 = 11



Solution:



(i) x = 4 is the root of 3x - 5 = 7.



Now, substituting x = 4 in place of ‘x’ in the given equation 3x - 5 = 7,



3(4) – 5 = 7



12 – 5 = 7



7 = 7



Since, LHS = RHS



Hence, x = 4 is the root of 3x - 5 = 7.



(ii). x = 3 is the root of 5 + 3x = 14.



Now, substituting x = 3 in place of ‘x’ in the given equation 5 + 3x = 14,



5 + 3(3) = 14



5 + 9 = 14



14 = 14



Since, LHS = RHS



Hence, x = 3 is the root of 5 + 3x = 14.



(iii). x = 2 is the root of 3x – 2 = 8x – 12.



Now, substituting x = 2 in place of ‘x’ in the given equation 3x –2 = 8x – 12,



3(2) – 2 = 8(2) – 12



6 – 2 = 16 – 12



4 = 4



Since, LHS = RHS



Hence, x = 2 is the root of 3x – 2 = 8x – 12.



(iv) x = 4 is the root of 3x/2 = 6.



Now, substituting x = 4 in place of ‘x’ in the given equation 3x/2 = 6,



(3× 4)/2 = 6



12/2 = 6



6 = 6



Since, LHS = RHS



Hence, x = 4 is the root of 3x/2 = 6.



(v). y = 2 is the root of y – 3 = 2y – 5.



Now, substituting y = 2 in place of ‘y’ in the given equation y – 3 = 2y – 5,



2 – 3 = 2(2) – 5



-1 = 4 – 5



-1 = -1



Since, LHS = RHS



Hence, y = 2 is the root of y – 3 = 2y – 5.



(vi). x = 8 is the root of 12x + 7 = 11.



Now, substituting x = 8 in place of ‘x’ in the given equation 12x + 7 = 11,



12(8) + 7 =11



4 + 7 = 11



11 = 11



Since, LHS = RHS



Hence, x = 8 is the root of 12x + 7 = 11.



Question: 2



Solve each of the following equations by trial and error method:



(i) x + 3 = 12



(ii) x – 7 = 10



(iii) 4x = 28



(iv) x/2 + 7 = 11



(v) 2x + 4 = 3x



(vi) x/4 = 12



(vii) 15/x = 3



(viii) x/18 = 20



Solution:



(i) x + 3 = 12



Here, LHS = x + 3 and RHS = 12




	
		
	
	
		

			x
			LHS
			RHS
			Is LHS = RHS
		

		

			1
			1 + 3 = 4
			12
			No
		

		

			2
			2 + 3 = 5
			12
			No
		

		

			3
			3 + 3 = 6
			12
			No
		

		

			4
			4 + 3 = 7
			12
			No
		

		

			5
			5 + 3 = 8
			12
			No
		

		

			6
			6 + 3 = 9
			12
			No
		

		

			7
			7 + 3 = 10
			12
			No
		

		

			8
			8 + 3 = 11
			12
			No
		

		

			9
			9 + 3 = 12
			12
			Yes
		

	







Therefore, if x = 9, LHS = RHS.



Hence, x = 9 is the solution to this equation.



(ii) x – 7 = 10



Here, LHS = x – 7 and RHS = 10.




	
		
	
	
		

			x
			LHS
			RHS
			Is LHS = RHS
		

		

			9
			9 – 7 = 2
			10
			No
		

		

			10
			10 – 7 = 3
			10
			No
		

		

			11
			11 – 7 = 4
			10
			No
		

		

			12
			12 – 7 = 5
			10
			No
		

		

			13
			13 – 7 = 6
			10
			No
		

		

			14
			14 – 7 = 7
			10
			No
		

		

			15
			15 – 7 = 8
			10
			No
		

		

			16
			16 – 7 = 9
			10
			No
		

		

			17
			17 – 7 = 10
			10
			Yes
		

	







Therefore, if x = 17, LHS = RHS.



Hence, x = 17 is the solution to this equation.



(iii) 4x = 28



Here, LHS = 4x and RHS = 28.







	
		
	
	
		

			x
			LHS
			RHS
			Is LHS = RHS
		

		

			1
			4 x 1 = 4
			28
			No
		

		

			2
			4 x 2 = 8
			28
			No
		

		

			3
			4 x 3 = 12
			28
			No
		

		

			4
			4 x 4 = 16
			28
			No
		

		

			5
			4 x 5 = 20
			28
			No
		

		

			6
			4 x 6 = 24
			28
			No
		

		

			7
			4 x 7 = 28
			28
			Yes
		

	




Therefore, if x = 7, LHS = RHS



Hence, x = 7 is the solution to this equation.



(iv) x/2 + 7 = 11



Here, LHS = x/2 + 7 and RHS = 11.



Since RHS is a natural number, x/2 must also be a natural number, so we must substitute values of x that are multiples of 2.




	
		
	
	
		

			x
			LHS
			RHS
			Is LHS = RHS
		

		

			2
			2/2 + 7 = 8
			11
			No
		

		

			4
			4/2 + 7 = 9
			11
			No
		

		

			6
			6/2 + 7 = 10
			11
			No
		

		

			8
			8/2 + 7 = 11
			11
			Yes
		

	




Therefore, if x = 8, LHS = RHS.



Hence, x = 8 is the solution to this equation.



(v) 2x + 4 = 3x



Here, LHS = 2x + 4 and RHS = 3x.




	
		
	
	
		

			x
			LHS
			RHS
			Is LHS = RHS
		

		

			1
			2(1) + 4 = 6
			3(1) = 3
			No
		

		

			2
			2(2) + 4 = 8
			3(2) = 6
			No
		

		

			3
			2(3) + 4 =10
			3(3) = 9
			No
		

		

			4
			2(4) + 5 = 12
			 3(4) = 12
			Yes
		

	







Therefore, if x = 4, LHS = RHS.



Hence, x = 4 is the solution to this equation.



(vi) x/4 = 12



Here, LHS = x/4 and RHS = 12.



Since RHS is a natural number, x/4 must also be a natural number, so we must substitute values of x that are multiples of 4.




	
		
	
	
		

			X
			LHS
			RHS
			Is LHS = RHS
		

		

			16
			16/4 = 4
			12
			No
		

		

			20
			20/4 = 5
			12
			No
		

		

			24
			24/4 = 6
			12
			No
		

		

			28
			28/4 = 7
			12
			No
		

		

			32
			32/4 = 8
			12
			No
		

		

			36
			36/4 = 9
			12
			No
		

		

			40
			40/4 = 10
			12
			No
		

		

			44
			44/4 = 11
			12
			No
		

		

			48
			48/4 = 12
			12
			Yes
		

	




Therefore, if x = 48, LHS = RHS.



Hence, x = 48 is the solution to this equation.



(vii) 15x = 3



Here, LHS = 15x and RHS = 3.



Since RHS is a natural number, 15x must also be a natural number, so we must substitute values of x that are factors of 15.




	
		
	
	
		

			x
			LHS
			RHS
			Is LHS = RHS
		

		

			1
			15/1= 15
			3
			No
		

		

			3
			15/3 = 5
			3
			No
		

		

			5
			15/5 = 3
			3
			Yes
		

	




Therefore, if x = 5, LHS = RHS.



Hence, x = 5 is the solution to this equation.



(viii) x/18 = 20



Here, LHS = x/18 and RHS = 20.



Since RHS is a natural number, x/18 must also be a natural number, so we must substitute values of x that are multiples of 18.




	
		
	
	
		

			X
			LHS
			RHS
			Is LHS = RHS
		

		

			324
			324/18 = 18
			20
			No
		

		

			342
			342/18 = 19
			20
			No
		

		

			360
			360/18 = 20
			20
			Yes
		

	




Therefore, if x = 360, LHS = RHS.



Hence, x = 360 is the solution to this equation.

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