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Q1. Any solution of the linear equation 2x + ny = 9 in two variables, is of the form:

• 1)
• 2) _; where n is a real number
• 3) ; where n is a real number
• 4)

Solution

_{2x + ny = 9 2x = 9 x = Therefore, solution is of the form :;} where n is a real number

Q2. If the line represented by the equation 3x + y = 8 passes through the points (2,2), then the value of is

• 1) 3
• 2) 1
• 3) 0
• 4) 4

Solution

3x + y = 8 3(2) + (2) = 8 6 + 2 = 8 2 = 2 = 1

Q3. The point of the form (a, a) always lies on

• 1) line y = x
• 2) y - axis
• 3) x - axis
• 4) line x + y = 0

Solution

The point of the form (a, a) always lies on the line x - y = 0 or y = x As for x = a, we have: x = y a = y y = a

Q4. The food charges in a hostel are as follows: For the first day, the charges are Rs. 100 and for the subsequent days it is Rs. 50 per day. Taking the number of days as x and total charges as Rs. y, write a linear equation for this information and draw its graph.

Solution

Number of days = x Total charges = y According to question, y = 100 + (x - 1) 50 = 100 + 50x - 50 y = 50x + 50 x 1 2 3 y 100 150 200

Q5. Equation of x-axis is:

• 1) x = 1
• 2) y = x
• 3) y = 0
• 4) x = 0

Solution

Equation of x-axis is y = 0.

Q6. The linear equation 5x - y = 2x - 1 has

• 1) 2 solutions
• 2) Infinitely many solutions
• 3) a unique solution
• 4) no solution

Solution

5x - y = 2x - 1 5x - 2x - y + 1 = 0 3x - y + 1 = 0 A linear equation in two variables has infinitely many solutions.

Q7. If the point (2, -1) lies on the graph of the equation 3x + ky = 4, then the value of k is

• 1) 1
• 2) -1
• 3) 2
• 4) -2

Solution

3x + ky = 4 Point (2, -1) lies on the graph of this equation. 3(2) + k(-1) = 4 6 - k = 4 k = 2

Q8. The sum of twice the odd number and  the even number is 10.Write  a linear equation for this information. Also find the odd number if the even number is 4

Solution

Let  the odd number be x  and an even number be y                                     Twice the odd number and even number is 10.   So,             Now, even number =4   So the odd number is 3

Q9. Which of the following is not the solution of linear equation 

• 1)
• 2)
• 3)
• 4)

Solution

Let us check  whether  is   a solution  of   linear equation    Substituting x = and y = in the LHS of equation we get  

Q10. Which  of the following is the solution of linear equation   

• 1) (1,5)
• 2) (-1, 5)
• 3) (-1, -5)
• 4) (1, -5)

Solution

Substitute (x,y) as(-1,5)  in the linear equation Since L.H.S=R.H.S, so (-1,5) is the solution of 

Q11. I am three times as old as my son. Five years later, I shall be two and a half times as old as my son. How old I am and how old is my son?

Solution

Let age of my son = x years My age = 3x After five yrs: 3x + 5 = (x + 5) 6x + 10 = 5x + 25 x = 15Age of son = 15 years Age of father (i.e. my age) = 3 15 yrs = 45 years

Q12. For what value of k, x = 2 and y = -1 is a solution of x + 3y - k = 0?

• 1) -1
• 2) 3
• 3) 2
• 4) -2

Solution

x + 3y - k = 0 Substituting the given values of x and y, 2 + 3(-1) - k = 0 2 - 3 - k =0 -1 - k = 0 k = -1

Q13.   Laxmi purchases some bananas and some oranges .Each banana costs Rs.2 while each orange costs Rs.3. If the total amount paid by Laxmi was Rs.30 and the number of oranges purchased by her was 6, then how many bananas did she purchase?

Solution

 Let us assume that Laxmi purchased x bananas and y oranges.   Since each banana costs Rs.2, x bananas cost Rs.2 × x = Rs.2x   Similarly, each orange costs Rs.3. Thus, y oranges cost Rs.3 × y = Rs.3y   Thus, the total amount paid by Laxmi is Rs. (2x + 3y), which equals Rs.30 Thus, we can express the given information in the form of a linear equation as 2x + 3y =3 0 Now, we know that Laxmi purchased 6 oranges, i.e., the value of y is 6. Substitute  this value of y in the equation 2x + 3y = 30, thereby reducing it to a linear equation in one variable. We can then solve the equation to obtain the value of x.     Thus, we see that the value of x is 6, i.e., Laxmi purchased 6 bananas.

Q14. Solve for x:

Solution

Q15. Draw the graph of the linear equation 2x + 3y = 12 (i) Write the co-ordinates of a point where graph intersects x-axis. (ii) From the graph show whether points (3, 2) and (-3, 6) are the solutions of the equation.

Solution

2x + 3y = 12 x 0 3 -3 y 4 2 6 (i) Co-ordinates of point where graph intersects x-axis is : As the point is on x axis, so,y=0. so 2x=12 x=6 the point on x- axis will be:(6,0). (ii) Point (3,2) and (-3,6) lie on graph (3, 2) and (-3, 6) are solutions of the equation.

Q16. The linear equation 2x + 5y = 8 has:

• 1) Infinitely many solution
• 2) Two solution
• 3) A unique solution
• 4) No solution

Solution

A linear equation has infinitely many solutions.

Q17. Express 3x = 5y in the form of ax + by + c = 0 and hence indicate the values of a, b and c.

Solution

3x - 5y + 0 = 0 a = 3, b = -5, c = 0

Q18. If the point (-1, -5) lie on the graph of 3x = ay + 7, then find the value of 'a'.

Solution

(-1, -5) lies on the graph of 3x = ay + 7 3(-1) = a(-5) + 7 -3 = -5a + 7 5a = 7 + 3

Q19. If a linear equation has solutions (-2,2), (0,0) (2,-2) then it is of the form:

• 1) y - x = 0
• 2) -2x + y = 0
• 3) x + y = 0
• 4) -x + 2y = 0

Solution

For (-2, 2): -2 + 2 = 0 For (0, 0): 0 + 0 = 0 For (2, -2): 2 + (-2) = 0 Hence, x + y = 0 is the equation for which given points are the solution.

Q20. Which of the following is a solution of the equation 4x + 3y = 16?

• 1) (1, 3)
• 2) (2, 3)
• 3) (1, 4)
• 4) (2, 4)

Solution

4x + 3y = 16 For (2, 3): LHS = 4(2) + 3(3) = 8 + 9 =17 RHS For (1, 4): LHS = 4(1) + 3(4) = 4 + 12 = 16 = RHS For (2, 4): LHS = 4(2) + 3(4) = 8 + 12 = 20 RHS For (1, 3): LHS = 4(1) + 3(3) = 4 + 9 = 13 RHS Therefore, (1, 4) is the solution.

Q21. If the point (2k - 3, k + 2) lies on the graph of the equation 2x + 3y + 15 = 0, find value of k.

Solution

Since (2k - 3, k + 2) lies on 2x + 3y + 15 = 0 2(2k - 3) + 3(k + 2) + 15 = 0 4k - 6 + 3k + 6 + 15 = 0 7k + 15 = 0

Q22. Draw the graph of  following  linear equation:               ?                       (Take  )

Solution

Let us simplify the given linear equation                 Let us find the solutions in ordered pair in the form of table  as follows:   X 2 6 -2 -4 Y -1.4142 -5.4142 2.5858 4.5858        

Q23. Any point on the x-axis is of the form:

• 1) (x, x)
• 2) (x, y)
• 3) (x, 0)
• 4) (0, y)

Solution

For any point to lie on x-axis, value of y will always be 0. Therefore, (x, 0) will the point.

Q24. Point P(2,-3) lies on the line represented by the equation:

• 1) x + 2y = 0
• 2) 2x + 2y = 0
• 3) 2x + y = 1
• 4) x + y = 1

Solution

Consider the equation 2x + y = 1. For x = 2 and y = -3, LHS = 2(2) + (-3) = 4 - 3 =1 = RHS Therefore, P(2, -3) lies on the linerepresented by the equation 2x + y = 1.

Q25. Which of the following is not a linear equation?

• 1) ax + 0y + c = 0
• 2) 0x + 0y + c=0
• 3) ax + by + c = 0
• 4) 0x + by + c = 0

Solution

A linear equation is one which can be expressed as y = mx + c. Since in equation 0x + 0y + c = 0, x and y do not exist hence it is not a linear equation.

Q26. Which of the following is a linear equation in one variable?

• 1) 2x +3y = 0
• 2) x + 5 = 6
• 3) 5x = y² + 3
• 4) x² = 5x + 3

Solution

A linear equation in one variable involve only one variable is of the form ax + b = 0, where a 0. Clearly, from amongst the given options. x + 5 = 6 is a linear equation in one variable.

Q27. Express y in terms of x, given that 2x - 5y = 7. Check whether the point (-3, -2) is on the given line.

Solution

2x - 5y = 7 5y = 2x - 7 _{For (-3, -2): 2x - 5y = 7 LHS = 2(-3) -5(-2) = -6 + 10 = 4 RHSThe point (-3, -2) is not on the given line.}

Q28. The taxi fare for the first km is Rs. 10 and fare for subsequent distance is Rs. 6 per km. If the distance covered is x km and total fare is Rs. y, write a linear equation for this statement and draw its graph.

Solution

Given, Taxi fare for first kilometer = Rs. 10 Taxi fare for subsequent distance = Rs. 6 Total distance covered = x Total fare = y Since the fare for first kilometer = Rs. 10 According to problem, Fare for (x - 1) kilometer = 6(x-1) So, the total fare = 6(x-1) + 10 y = 6(x-1) + 10 y = 6x - 6 + 10 y = 6x + 4 Hence, y = 6x + 4 is the required linear equation. x -1 0 1 y -2 4 10

Q29. The linear equation x = 5 can be written in two variables as:

• 1) 1.x + 5 = 0
• 2) 1.x + 0.y + (-5) = 0
• 3) 0.x + 1.y + (-5) = 0
• 4) 1.x + 1.y + (-5) = 0

Solution

x = 5 (1)x + (0)y = 5 1.x + 0.y + (-5) = 0

Q30. The point lying on the equation 2x - y = 5 is:

• 1) (2, -1)
• 2) (3, 4)
• 3) (6, 1)
• 4) (-3, 1)

Solution

Consider the point (2, -1). LHS = 2(2) - (-1) = 4 + 1 = 5 = RHS Since, LHS = RHS, point (2, -1) lies on the equation 2x - y = 5.

Q31. The number of line(s) passing through a point (3,4) is (are):

• 1) Infinite
• 2) Two
• 3) Only one
• 4) Three

Solution

Infinite numbers of lines pass through a single point i.e. (3, 4) here.

Q32. Find four solutions of the equation 2x - y = 4.

Solution

2x - y = 4 Or, y = 2x - 4 …………equation (1) Putting x = 0 in equation (1), y = (2 x 0) - 4 = 0 - 4 = -4 Putting x = 1 in equation (1), y = (2 x 1) - 4 = 2 - 4 = -2 Putting x = 2 in equation (1), y = (2 x 2) - 4 = 4 - 4 = 0 Putting x = 3 in equation (1), y = (2 x 3) - 4 = 6 - 4 = 2 Hence, four solutions of equation 2x - y = 4 are (0, -4), (1, -2), (2, 0), (3, 2).

Q33. Sum of the digits of a two digit number is 12. If 18 is added to the original number the digits interchange their places. Write two linear equations representing these situations.

Solution

Let the unit's digit be x and the ten's digit be y. Sum of the digits of a two digit number is given to be 12. x + y = 12 …….(i) Original number = 10y + x Number obtained on reversing the digits = 10x + y According to the question: 10y + x + 18 = 10x + y 9x - 9y = 18 x - y = 2 …….(ii) (i) and (ii) are the required equations.

Q34. Give two solutions of the equation x + 3y = 8.

Solution

Solution 1: For x = 1, x + 3y = 8 1 + 3y = 8 3y = 7 y = (x, y) = (1, ) Solution 2: For x = 2, x + 3y = 8 2 + 3y = 8 3y = 6 y = 2 (x, y) = (2, 2)

Q35. Sum of the digits of a two digit number is 14. If we add 18 to the original number, the digits interchange their places. Write two equations for these two statements.

Solution

Let the unit's digit of the number be x and ten's digit be y. x + y = 14 ……….(i) Original number = 10y + x Number obtained on reversing the digits = 10x + y According to question, 10y + x + 18 = 10x+ y 9x - 9y = 18 x - y = 2 ………..(ii)

Q36. Draw the graph of the equation 3x + y = 5 and write the co-ordinates of the points where the line intersects x-axis and y-axis.

Solution

3x + y = 5x 0 1 2 y 5 2 -1 Graph meets y-axis at (0,5) and x-axis at

Q37. By graphical method , find the solutions of the linear equation .

Solution

Let us simplify the above linear equation       Let us find the solutions in ordered pair in the form of table  as follows:   x -5 -1 3 7 Y -1 -0.5 0 0.5              

Q38. Draw the graphs of the equations 2x - y = 3 and 3x +2y = 1 on the same coordinate axes. Also, find the point of intersection of the two lines from the graphs.

Solution

2x - y = 3 x -1 1 3 y -5 -1 3 3x + 2y = 1 x -1 1 3 y 2 -1 -4 Point of intersection of the two lines is (1,-1).

Q39. Ram invested some amount at 6% simple interest and some amount at 9% simple interest. At the end of the year, he earned Rs.1200 as interest on the two amounts. Write  a linear equation in two variables to represent the above statement.

Solution

Let the amount invested at 6% simple interest be Rs.x and that invested at 9% simple interest be Rs.y. Thus, interest earned at the end of first year on Rs.x = 6% of Rs.x = Rs. Similarly, interest earned at the end of first year on Rs.y = 9% of Rs.y = Rs. Thus, total amount earned as interest , which equals Rs.1200 Thus, the given information can be represented as a linear equation as:

Q40. The parking charges of a car in a parking lot are Rs. 30 for the first two hours and Rs. 10 for subsequent hours. Take total parking time to be x hours and total charges as Rs. y, write a linear equation in two variables to express the above statement. Draw a graph for the linear equation and find the charges for five hours using the graph.

Solution

Parking charges for first 2 hrs = Rs 30 Parking charges for subsequent hrs = Rs 10 Total park time = x hours Total charges = Rs y y = 30 + 10(x- 2) y = 30 + 10x - 20 10x - y = -10 x 1 2 3 y 20 30 40   From the graph, it is clear that for five hours, parking charges will be Rs 60.

Q41. After 5 years, the age of father will be two times the age of the son. Write a linear equation in two variables to represent this statement.

Solution

Let father's present age be = x years Son's present age = y years After 5 years father's age will be = (x + 5) years After 5 years son's age will be = (y + 5) years According to the question: x + 5 = 2(y + 5) x + 5 = 2y + 10 x - 2y = 10 - 5 x - 2y = 5

Q42. Find the value of a, for which the equation 2x + ay = 5 has (1,-1) as a solution. Find two more solutions for the equation obtained.

Solution

2x + ay = 5 As (1,-1) is a solution of the equation, so it will satisfy this equation. 2(1) + a(-1) = 5 2 - a = 5 a = 2 - 5 = -3 Equation is 2x - 3y = 5 For x = 2, we have: 2(2) - 3y = 5 4 - 3y = 5 3y = -1 y = - For x = 4, we have: 2(4) - 3y = 5 8 - 3y = 5 3y = 3 y = 3 Solutions are (2,-) and (4,3).

Q43. A fraction is such that if the numerator is multiplied by 3 and the denominator is reduced by 3, we get 15. Write linear equation to represent this information and also draw graph of the linear equation.

Solution

  Let the numerator be x and the denominator be y                         Numerator is multiplied by  3=3x                         Denominator is reduced by 3=y-3                           New fraction formed=     Let us find the solutions in ordered pair in the form of table  as follows:     X 0 -5 2 -4 Y 3 2 3.6 2.2            

Q44. Which of the following pair is a solution of the equation 2x - 3y = 7

• 1) (1,5)
• 2) (2,-3)
• 3) (0,2)
• 4) (5,1)

Solution

LHS: 2(5) - 3(1) = 10 - 3 = 7 = RHS (5,1) is the solution of the equation 2x - 3y = 7.

Q45. Ashish and Deepak contribute for a charity. The contribution of Ashish is the contribution of Deepak. Write a linear equation to represent the above and drawn the graph. From the graph, find the contribution of Ashish if Deepak contributes Rs. 50.

Solution

y= x 10 20 50 y 4 8 20 From the graph, contribution of Ashish when Deepak contributes Rs 50 = Rs 20

Q46. Find the coordinates of the points where the line 2x - y = 3 meets both the axis.

Solution

On x axis, y = 0 2x - 0 = 3 On y axis, x = 0 0 - y = 3 y = -3 Point = (0,-3) Therefore, required points are and (0, -3).

Q47. Solve the equation 2x + 1 = x - 3 and represent the solution on: (i) The number line (ii) The Cartesian plane.

Solution

2x + 1 = x- 3 2x - x = -3 - 1 x = -4 (i) Representation on number line (ii) Representation on Cartesian plane

Q48. Draw the graph of the equation 3x + 4y = 12 and find the coordinates of the points of intersection of the graph with axes.

Solution

3x + 4y = 12 x 0 2 4 6 y 3 1.5 0 -1.5 Co-ordinates of point of intersection with x-axis = (4,0) Co-ordinates of point of intersection with y-axis = (0,3)

Q49. Solve for x:

Solution

Q50. A fraction is such that if the numerator is multiplied by 3 and the denominator is reduced by 3, we get 18/11.  Write linear equation to represent this information.

Solution

Let the numerator be x and the denominator be y                         Numerator is multiplied by  3=3x                         Denominator is reduced by 3=y-3                           New fraction formed=

Q51. From the choices given below, choose the equation whose graph is given in figure (a) 2x + y = 6 (B) y = 2x+ 4 (c) 2 (x- 1) + 3y = 4 (D) 2x - 3y = 6

Solution

For (-3, 4) (a) 2x + y = 6 LHS = 2(- 3) + 4 = -6 + 4 = -2 -2 6 (b) y = 2x + 4 4 = 2(-3) + 4 4 -6 + 4 (c) 2 (x - 1) + 3y = 4 2(-3-1) + 3(4) = 4 2(-4) + 12 = 4 -8 + 12 = 4 4 = 4 (d) 2x - 3y = 6 2(-3) -3(4) 6 Since, (-3, 4) is solution of 2(x-1) + 3y = 4 Given graph is of equation 2 (x - 1) + 3y = 4

Q52. The graph of the equation of the form y=mx is a line which always passes through:

• 1) (0, y)
• 2) (0, m)
• 3) (x, 0)
• 4) (0, 0)

Solution

(0, 0). The equation of a straight line with gradient m passing through the origin is given by y = mx.

Q53. Express the equation 5x = -y in the general form and indicate the values of a, b and c.

Solution

5x = -y 5x + y = 0 Comparing this equation with general form ax + by + c = 0, we have: a = 5, b = 1, c = 0

Q54. Solve for x :

Solution

Solve 6(13x- 47) = 30(x - 5) 13x - 47 = 5(x - 5) 13x - 47 = 5x - 25 13x - 5x = -25 + 47 8x = 22

Q55. Express y in terms of x from the equation 3x + 2y = 8 and check whether the point (4, -2) lies on the line.

Solution

2y = 8 - 3x For pt. (4, -2) 3 4 + 2 (-2) = 8 12 - 4 = 8 8 = 8 L.H.S. = R.H.S. (4, -2) lies on the line

Q56. A and B are two friends. A is elder to B by 5 years. B's sister C is half the age of B while A's father D is 8 years older than twice the age of B. If the present age of D is 48 years, find the present ages of A, B and C.

Solution

Let the age of the sister C be x years Age of B = 2x years Age of A = (2x + 5) years Age of D, A's father = [2 (2x) + 8] = 4x + 8 According to the question: 4x + 8 = 48 x = 10 Therefore, Age of C = x = 10 yrs Age of B = 2x = 20 yrs Age of A = 2x + 5 = 25 yrs.

Q57. The sum of the 10% of one number and 20% of the other number is equal to the 5 more then the 15% of the  sum of two numbers. Write  a linear equation  in two variables to represent above  statement .

Solution

Let the first  number be x                                Let the second  number be y 10% of first number= 20% of second number= According to the given information    

Q58. Given geometric representation of x + 3 = 0 as an equation (i) in one variable (ii) in two variables.

Solution

(i) x + 3 = 0 x= -3 (ii) x + 3 = 0 can be written as 1x + 0y + 3 = 0 x -3 -3 -3 y -2 0 2

Q59. Sita bought 18 apples for Rs 45 and 12 oranges for Rs 25. What will be the linear equation  of above statement ?

Solution

Let the apple be denoted by  A  Oranges be denoted by  O.   Sita bought 18 apples which is equal to 18A and  12 oranges  which is equal to 12O Total cost of 18 apples and  12 oranges=Rs45+Rs25=Rs70 Thus, the given information can be represented as a linear equation as: 18A+12O=70

Q60.  Find the value of  ‘ k’ if x=1 and y= is a solution of the equation ?

Solution

Substituting  x=1 and      y=   in  above equation we get

Q61. Find four solutions of the following linear equation in two variables. 2 (x + 3) -3 (y - 1) = 0

Solution

2 (x + 3) -3 (y - 1) = 0 2x + 6 - 3y + 3 = 0 2x - 3y = -9 Solutions can be calculated as below: (i) When x = 0, we have: -3y = -9 y = 3 (x, y) = (0, 3) (ii) When y = 0, we have: 2x = -9 x = -4.5 (x, y) = (-4.5, 0) (iii) When y = 1, we have: 2x - 3 = -9 2x = -6 x = -3 (x, y) = (-3, 1) (iv) When y = -1, we have: 2x - (-3) = -9 2x + 3 = -9 2x = -12 x = -6 (x, y) = (-6, -1) The four solutions of the given equation are (0, 3), (-4.5, 0), (-3, 1), (-6, -1).

Q62.  The sum of the twice the first number and thrice of the second number is equal to the 12 more than the fifty percent of first number. Represent the above information in the form of linear equation and also draw the graph

Solution

 Let the first  number be x             Second number be y                         According to the given information we have                   Let us plot the graph of the equation    Let us find the solutions in ordered pair in the form of table  as follows:     X 4 -2 2 4 Y 6 5 3 2        

Q63. Two pens and three pencils together cost Rs. 20. Represent this statement as a linear equation in two variables and given two solutions for it. Also verify if (4,2) is a solution of the equation formed.

Solution

Let cost of pen be x. And cost of pencil by y. 2x + 3y = 20 Put x = 2 2x + 3y = 20 2(2) + 3y = 20 4 + 3y = 20 3y = 16 y = (x, y) = (2, ) Put x = 3 2x + 3y = 20 2(3) + 3y = 20 6 + 3y = 20 3y = 14 y = (x, y) = (3, ) Put x = 4 and y = 2 LHS = 24 + 32 = 8 + 6 = 14 20 (4, 2) is not a solution of the given equation.

Q64. Solve for x:

Solution

Q65.  If (2,- 5) point is given, find the equation of a line on which it lies. How many such equations are there? Draw  the graph of one such equation.

Solution

There are  infinitely many linear equations which are satisfied by the coordinates of the point  (2,- 5).   For example, line  passing through the point (2,- 5)  are satisfied by  some of the Linear equations are as follows:     Let us draw the graph of the  equation .   Simplify the equation  we get     Let us find the solutions in ordered pair in the form of table  as follows:     X -4 -2 2 0 4 Y 10 5 -5 0 -10  

Q66. Express y in terms x in equation 2x - 3y = 12. Find the points where the line represented by this equation cuts x-axis and y-axis.

Solution

2x - 3y = 12 3y = 2x - 12 On x-axis (6,0) is the point where the line will cut the x - axis. On y-axis x = 0 (0, -4) is the point where the line will cut the y- axis.

Q67. John age is 10 years less than three times his son’s age. If John son is 20 years old, then how old is John?

Solution

Let us assume that John age is x years and his son’s age is y years. A linear equation in two variables can be formed to represent the given statement as  x = 3y – 10   We know that the age of John’s son is 20 years. This implies that the value of y is 20.             Substituting this value of y in the equation x = 3y – 10, we get:             x = 3 × 20 – 10             Þ x = 60 – 10             Þ x = 50             Thus, John’s age is 50 years.

Q68. 4 years ago, age of a mother was 3 times the age of her daughter. Write a linear equation to represent this situation and draw its graph.

Solution

Let present age of daughter be x years. 4 years ago age of daughter was (x - 4) years _{Let present age of mother be y years.} Mother's age, 4 years ago = (y - 4) years _{According to the question: y - 4 = 3(x - 4)} y - 4 = 3x - 12 3x - y = 8 x 1 2 3 y -5 -2 1

Q69. Draw the graphs of the equations x + y - 10 = 0 and x - y + 4 = 0 on the same graph paper.

Solution

For x + y - 10 = 0 x 1 2 3 y 9 8 7 For x - y + 4 = 0 x 6 2 1 y 10 6 5

Q70. One tenth of the number  is equivalent to 20 more than the 10% of the other number. Represent this information in the equivalent linear equation of two variables

Solution

Let the one number be x             Let the other number be y According to the given information,  

Q71. Solve the following equation:

Solution

4x -4 - 3x + 6 = 12 x + 2 = 12 x = 12 - 2 = 10

Q72. If one of the angle of the triangle is  equal to the sum of the other two angles .Express this information into equivalent linear equation of two variables  and  also find the angles of triangle id  measure of one of the angle is 60 degree.

Solution

  Let  the first angle = x             Let the second angle=y             According to the given information, Third angle=x+y                         We know that sum of angles in  a triangle is equal to 180 degree.             So,             X+y+(x+y)=180             2x+2y=180             X+y=90                                    (i)              Now measure of one angle=60 degree.             Substituting in (i) we get             60+y=90             y=90-60=30             Third angle=x+y=60+30=90             Thus angles of triangle are 60, 30 and 90 degrees

Q73. Draw the graphs of each of the equation x - 2y - 3 = 0 and 4x + 3y - 1 = 0 on the same graph.

Solution

x - 2y - 3 = 0 x 5 1 3 y 1 -1 0 4x + 3y - 1 = 0 x -2 1 -5 y 3 -1 7