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Page 1 Exercise 3D 1. Show that the following system of equations has a unique solution: 3x + 5y = 12, 5x + 3y = 4. Also, find the solution of the given system of equations. Sol: The given system of equations is: Page 2 Exercise 3D 1. Show that the following system of equations has a unique solution: 3x + 5y = 12, 5x + 3y = 4. Also, find the solution of the given system of equations. Sol: The given system of equations is: 3x + 5y = 12 5x + 3y = 4 These equations are of the forms: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 3, b1= 5, c1 = -12 and a2 = 5, b2 = 3, c2 = -4 For a unique solution, we must have: , i.e., Hence, the given system of equations has a unique solution. Again, the given equations are: On multiplying (i) by 3 and (ii) by 5, we get: 9x + 15y = 36 On subtracting (iii) from (iv), we get: 16x = -16 x = -1 On substituting x = -1 in (i), we get: 3(-1) + 5y = 12 5y = (12 + 3) = 15 y = 3 Hence, x = -1 and y = 3 is the required solution. 2. Show that the following system of equations has a unique solution: 2x - 3y = 17, 4x + y = 13. Also, find the solution of the given system of equations. Sol: The given system of equations is: 2x - 3y - 4x + y - The given equations are of the form a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 2, b1= -3, c1 = -17 and a2 = 4, b2 = 1, c2 = -13 Now, = = and = = -3 Since, , therefore the system of equations has unique solution. Using cross multiplication method, we have Page 3 Exercise 3D 1. Show that the following system of equations has a unique solution: 3x + 5y = 12, 5x + 3y = 4. Also, find the solution of the given system of equations. Sol: The given system of equations is: 3x + 5y = 12 5x + 3y = 4 These equations are of the forms: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 3, b1= 5, c1 = -12 and a2 = 5, b2 = 3, c2 = -4 For a unique solution, we must have: , i.e., Hence, the given system of equations has a unique solution. Again, the given equations are: On multiplying (i) by 3 and (ii) by 5, we get: 9x + 15y = 36 On subtracting (iii) from (iv), we get: 16x = -16 x = -1 On substituting x = -1 in (i), we get: 3(-1) + 5y = 12 5y = (12 + 3) = 15 y = 3 Hence, x = -1 and y = 3 is the required solution. 2. Show that the following system of equations has a unique solution: 2x - 3y = 17, 4x + y = 13. Also, find the solution of the given system of equations. Sol: The given system of equations is: 2x - 3y - 4x + y - The given equations are of the form a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 2, b1= -3, c1 = -17 and a2 = 4, b2 = 1, c2 = -13 Now, = = and = = -3 Since, , therefore the system of equations has unique solution. Using cross multiplication method, we have = = = = = = = = x = , y = x = 4, y = -3 Hence, x = 4 and y = -3. 3. Show that the following system of equations has a unique solution: + = 3, x 2y = 2. Also, find the solution of the given system of equations. Sol: The given system of equations is: + = 3 = 3 2x + 3y = 18 2x + 3y 1 and x 2y = 2 x 2y These equations are of the forms: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 2, b1= 3, c1 = -18 and a2 = 1, b2 = -2, c2 = -2 For a unique solution, we must have: , i.e., Hence, the given system of equations has a unique solution. Again, the given equations are: 2x + 3y x 2y On multiplying (i) by 2 and (ii) by 3, we get: 4x + 6y 3x - 6y On adding (v) from (vi), we get: 7x = 42 x = 6 On substituting x = 6 in (iii), we get: Page 4 Exercise 3D 1. Show that the following system of equations has a unique solution: 3x + 5y = 12, 5x + 3y = 4. Also, find the solution of the given system of equations. Sol: The given system of equations is: 3x + 5y = 12 5x + 3y = 4 These equations are of the forms: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 3, b1= 5, c1 = -12 and a2 = 5, b2 = 3, c2 = -4 For a unique solution, we must have: , i.e., Hence, the given system of equations has a unique solution. Again, the given equations are: On multiplying (i) by 3 and (ii) by 5, we get: 9x + 15y = 36 On subtracting (iii) from (iv), we get: 16x = -16 x = -1 On substituting x = -1 in (i), we get: 3(-1) + 5y = 12 5y = (12 + 3) = 15 y = 3 Hence, x = -1 and y = 3 is the required solution. 2. Show that the following system of equations has a unique solution: 2x - 3y = 17, 4x + y = 13. Also, find the solution of the given system of equations. Sol: The given system of equations is: 2x - 3y - 4x + y - The given equations are of the form a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 2, b1= -3, c1 = -17 and a2 = 4, b2 = 1, c2 = -13 Now, = = and = = -3 Since, , therefore the system of equations has unique solution. Using cross multiplication method, we have = = = = = = = = x = , y = x = 4, y = -3 Hence, x = 4 and y = -3. 3. Show that the following system of equations has a unique solution: + = 3, x 2y = 2. Also, find the solution of the given system of equations. Sol: The given system of equations is: + = 3 = 3 2x + 3y = 18 2x + 3y 1 and x 2y = 2 x 2y These equations are of the forms: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 2, b1= 3, c1 = -18 and a2 = 1, b2 = -2, c2 = -2 For a unique solution, we must have: , i.e., Hence, the given system of equations has a unique solution. Again, the given equations are: 2x + 3y x 2y On multiplying (i) by 2 and (ii) by 3, we get: 4x + 6y 3x - 6y On adding (v) from (vi), we get: 7x = 42 x = 6 On substituting x = 6 in (iii), we get: 2(6) + 3y = 18 3y = (18 - 12) = 6 y = 2 Hence, x = 6 and y = 2 is the required solution. 4. Find the value of k for which the system of equations has a unique solution: 2x + 3y = 5, kx - 6y = 8. Sol: The given system of equations are 2x + 3y 5 = 0 kx - 6y - 8 = 0 This system is of the form: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 2, b1= 3, c1 = -5 and a2 = k, b2 = -6, c2 = -8 Now, for the given system of equations to have a unique solution, we must have: -4 Hence, -4 5. Find the value of k for which the system of equations has a unique solution: x ky = 2, 3x + 2y + 5=0. Sol: The given system of equations are x - ky 2 = 0 3x + 2y + 5 = 0 This system of equations is of the form: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 1, b1= -k, c1 = -2 and a2 = 3, b2 = 2, c2 = 5 Now, for the given system of equations to have a unique solution, we must have: - Hence, - . Page 5 Exercise 3D 1. Show that the following system of equations has a unique solution: 3x + 5y = 12, 5x + 3y = 4. Also, find the solution of the given system of equations. Sol: The given system of equations is: 3x + 5y = 12 5x + 3y = 4 These equations are of the forms: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 3, b1= 5, c1 = -12 and a2 = 5, b2 = 3, c2 = -4 For a unique solution, we must have: , i.e., Hence, the given system of equations has a unique solution. Again, the given equations are: On multiplying (i) by 3 and (ii) by 5, we get: 9x + 15y = 36 On subtracting (iii) from (iv), we get: 16x = -16 x = -1 On substituting x = -1 in (i), we get: 3(-1) + 5y = 12 5y = (12 + 3) = 15 y = 3 Hence, x = -1 and y = 3 is the required solution. 2. Show that the following system of equations has a unique solution: 2x - 3y = 17, 4x + y = 13. Also, find the solution of the given system of equations. Sol: The given system of equations is: 2x - 3y - 4x + y - The given equations are of the form a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 2, b1= -3, c1 = -17 and a2 = 4, b2 = 1, c2 = -13 Now, = = and = = -3 Since, , therefore the system of equations has unique solution. Using cross multiplication method, we have = = = = = = = = x = , y = x = 4, y = -3 Hence, x = 4 and y = -3. 3. Show that the following system of equations has a unique solution: + = 3, x 2y = 2. Also, find the solution of the given system of equations. Sol: The given system of equations is: + = 3 = 3 2x + 3y = 18 2x + 3y 1 and x 2y = 2 x 2y These equations are of the forms: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 2, b1= 3, c1 = -18 and a2 = 1, b2 = -2, c2 = -2 For a unique solution, we must have: , i.e., Hence, the given system of equations has a unique solution. Again, the given equations are: 2x + 3y x 2y On multiplying (i) by 2 and (ii) by 3, we get: 4x + 6y 3x - 6y On adding (v) from (vi), we get: 7x = 42 x = 6 On substituting x = 6 in (iii), we get: 2(6) + 3y = 18 3y = (18 - 12) = 6 y = 2 Hence, x = 6 and y = 2 is the required solution. 4. Find the value of k for which the system of equations has a unique solution: 2x + 3y = 5, kx - 6y = 8. Sol: The given system of equations are 2x + 3y 5 = 0 kx - 6y - 8 = 0 This system is of the form: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 2, b1= 3, c1 = -5 and a2 = k, b2 = -6, c2 = -8 Now, for the given system of equations to have a unique solution, we must have: -4 Hence, -4 5. Find the value of k for which the system of equations has a unique solution: x ky = 2, 3x + 2y + 5=0. Sol: The given system of equations are x - ky 2 = 0 3x + 2y + 5 = 0 This system of equations is of the form: a1x+b1y+c1 = 0 and a2x+b2y+c2 = 0 where, a1 = 1, b1= -k, c1 = -2 and a2 = 3, b2 = 2, c2 = 5 Now, for the given system of equations to have a unique solution, we must have: - Hence, - . 6. Find the value of k for which the system of equations has a unique solution: 5x 7y = 5, 2x + ky = 1. Sol: The given system of equations are 5x - 7y 2x + ky - This system is of the form: a1x+b1y+c1 = 0 a2x+b2y+c2 = 0 where, a1 = 5, b1= -7, c1 = -5 and a2 = 2, b2 = k, c2 = -1 Now, for the given system of equations to have a unique solution, we must have: - Hence, - . 7. Find the value of k for which the system of equations has a unique solution: 4x + ky + 8=0, x + y + 1 = 0. Sol: The given system of equations are 4x + ky + 8 = 0 x + y + 1 = 0 This system is of the form: a1x+b1y+c1 = 0 a2x+b2y+c2 = 0 where, a1 = 4, b1= k, c1 = 8 and a2 = 1, b2 = 1, c2 = 1 For the given system of equations to have a unique solution, we must have: Hence, 8. Find the value of k for which the system of equations has a unique solution: 4x - 5y = k, 2x - 3y = 12. Sol:Read More
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FAQs on RS Aggarwal Solutions: Exercise 3D - Linear Equations in two variables - Extra Documents, Videos & Tests for Class 10
| 1. What are linear equations in two variables? |  |
Ans. Linear equations in two variables are algebraic equations that involve two variables, usually denoted as x and y, and have a degree of one. These equations can be represented graphically as straight lines on a coordinate plane.
| 2. How do you solve linear equations in two variables? | |
Ans. To solve linear equations in two variables, we use various methods such as the substitution method, elimination method, and graphical method. The substitution method involves solving one equation for one variable and substituting it into the other equation. The elimination method involves adding or subtracting the equations to eliminate one variable. The graphical method involves plotting the equations on a coordinate plane and finding the point of intersection.
| 3. What is the importance of solving linear equations in two variables? | |
Ans. Solving linear equations in two variables is important in various real-life situations. It helps in finding the relationship between two variables, determining the point of intersection between two lines, solving optimization problems, and analyzing systems of equations. These skills are crucial in fields such as physics, engineering, economics, and data analysis.
| 4. Can linear equations in two variables have more than one solution? | |
Ans. Yes, linear equations in two variables can have more than one solution. When the two lines representing the equations are parallel, they do not intersect and have no solution. When the two lines are coincident, they overlap and have infinitely many solutions. In other cases, the lines intersect at a single point, which represents a unique solution.
| 5. How can linear equations in two variables be represented graphically? | |
Ans. Linear equations in two variables can be represented graphically by plotting the equations on a coordinate plane. Each equation is represented by a straight line, and the point of intersection between the lines represents the solution to the system of equations. The slope-intercept form (y = mx + c) or the standard form (Ax + By = C) can be used to determine the slope and y-intercept or the x-intercept of the lines.
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