{"id":12291,"date":"2019-06-17T19:12:51","date_gmt":"2019-06-17T13:42:51","guid":{"rendered":"https:\/\/www.cbselabs.com\/?page_id=12291"},"modified":"2021-09-18T15:33:01","modified_gmt":"2021-09-18T10:03:01","slug":"rd-sharma-class-10-solutions-pair-of-linear-equations-in-two-variables","status":"publish","type":"page","link":"https:\/\/www.cbselabs.com\/rd-sharma-class-10-solutions-pair-of-linear-equations-in-two-variables\/","title":{"rendered":"RD Sharma Class 10 Solutions Pair Of Linear Equations In Two Variables"},"content":{"rendered":"

RD Sharma Class 10 Solutions Pair Of Linear Equations In Two Variables<\/h2>\n

RD Sharma Class 10 Solutions Pair Of Linear Equations In Two Variables Exercise 3.1<\/h3>\n

Question 1.<\/strong><\/span>
\nAkhila went to a fair in her village. She wanted to enjoy rides on the Giant Wheel and play Hoopla (a game in which you throw a rig on the items kept in the stall, and if the ring covers any object completely you get it). The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs Rs. 3, and a game of Hoopla costs Rs. 4. If she spend Rs. 20 in the fair, represent this situation algebraically and graphically.
\nSolution:<\/strong><\/span>
\nLet number of rides on the wheel = x
\nand number of play of Hoopla = y
\nAccording to the given conditions x = 2y \u21d2 x – 2y = 0 ….(i)
\nand cost of ride on wheel at the rate of Rs. 3 = 3x
\nand cost on Hoopla = 4y
\nand total cost = Rs. 20
\n3x + 4y = 20 ….(ii)
\nNow we shall solve these linear equations graphically as under
\nWe take three points of each line and join them to get a line in each case the point of intersection will be the solution
\nFrom equation (i)
\nx = 2y<\/p>\n\n\n\n\n
X<\/td>\n4<\/td>\n0<\/td>\n6<\/td>\n<\/tr>\n
y<\/td>\n2<\/td>\n0<\/td>\n3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

y = 2, then x = 2 x 2 = 4
\ny = 0, then x = 2 x 0 = 0
\ny = 3, then x = 2 x 3 = 6
\nNow, we plot these points on the graphs and join them to get a line
\nSimilarly in equation (ii)
\n3x + 4y = 20 \u21d2 3x = 20 – 4y
\n\"RD
\nNow we plot these points and get another line by joining them
\nThese two lines intersect eachother at the point (4, 2)
\nIts solution is (4, 2)
\nWhich is a unique Hence x = 4, y = 2
\n\"Pair<\/p>\n

Question 2.<\/strong><\/span>
\nAftab tells his daughter, \u201cSeven years ago, I w as seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.\u201d Is not this interesting ? Represent this situation algebraically and graphically.
\nSolution:<\/strong><\/span>
\nSeven years ago
\nLet age of Aftab\u2019s daughter = x years
\nand age of Aftab = y years
\nand 3 years later
\nAge of daughter = x + 10 years
\nand age of Aftab = y + 10 years
\nAccording to the conditions,
\ny = 7x \u21d2 7x – y = 0 ……….(i)
\ny + 10 = 3 (x + 10)
\n=> y + 10 = 3x + 30
\n3x – y = 10 – 30 = -20
\n3x – y = -20 ….(ii)
\nEquations are
\n7x – y = 0
\n3 x – y = -20
\nNow we shall solve these linear equations graphically as under
\n7x – y = 0 \u21d2 y = 7x<\/p>\n\n\n\n\n
X<\/td>\n0<\/td>\n1<\/td>\n-1<\/td>\n<\/tr>\n
y<\/td>\n0<\/td>\n7<\/td>\n-7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

If x = 0, y = 7 x 0 = 0
\nIf x = 1, y = 7 x 1=7
\nIf x = -1, y = 7 x (-1) = -7
\nNow plot these points on the graph and join
\nthen
\n3x – y = -20
\ny = 3x + 20<\/p>\n\n\n\n\n
X<\/td>\n-1<\/td>\n-2<\/td>\n-3<\/td>\n<\/tr>\n
y<\/td>\n17<\/td>\n14<\/td>\n11<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

If x = -1, y = 3 x (-1) + 20 = -3 + 20= 17
\nIf x = -2, y = 3 (-2) + 20 = -6 + 20 = 14
\nIf x = -3, y = 3 (-3) + 20 = -9 + 20= 11
\nNow plot the points on the graph and join them we see that lines well meet at a point on producing at (5, 35).
\n\"RD<\/p>\n

Question 3.<\/strong><\/span>
\nThe path of a train A is given by the equation 3x + 4y – 12 = 0 and the path of another train B is given by the equation 6x + 8y – 48 = 0. Represent this situation graphically.
\nSolution:<\/strong><\/span>
\nPath of A train is 3x + 4y – 12 = 0
\nand path of B train is 6x + 8y – 48 = 0
\nGraphically, we shall represent these on the graph as given under 3x + 4y- 12 = 0
\n\"RD
\n\"RD
\n\"RD<\/p>\n

Question 4.<\/strong><\/span>
\nGloria is walking along the path joining (-2, 3) and (2, -2), while Suresh is walking along the path joining (0, 5) and (4, 0). Represent this situation graphically.
\nSolution:<\/strong><\/span>
\nPlot the points (-2, 3) and (2, -2) and join them to get a line
\nand also plot the points (0, 5), (4, 0) and joint them to get another line as shown on the graph
\nWe see that these two lines are parallel to each other
\n\"Learncbse.In<\/p>\n

Question 5.<\/strong><\/span>
\nOn comparing the ratios , and without drawing them, find out whether the lines representing following pairs of linear equations intersect at a point, are parallel or coincide :
\n(i) 5x – 4y + 8 = 0
\n7x + 6y – 9 = 0
\n(ii) 9x + 3y +12 = 0
\n18x + 6y + 24 = 0
\n(iii) 6x – 3y +10 = 0
\n2x – y + 9 = 0
\nSolution:<\/strong><\/span>
\n\"Class
\n\"RD
\n\"RD<\/p>\n

Question 6.<\/strong><\/span>
\nGiven the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is :
\n(i) intersecting lines
\n(ii) parallel lines
\n(iii) coincident lines.
\nSolution:<\/strong><\/span>
\n\"RD
\n\"RD<\/p>\n

Question 7.<\/strong><\/span>
\nThe cost of 2kg of apples and 1 kg of grapes on a day was found to be Rs. 160. After a month, the cost of 4kg of apples and 2kg of grapes is Rs. 300. Represent the situation algebraically and geo-metrically.
\nSolution:<\/strong><\/span>
\nLet cost of 1kg of apples = Rs. x
\nand cost of 1kg of grapes = Rs. y
\nNow according to the condition, the system of equation will be
\n2x + y = 160
\n4x + 2y = 300
\nNow 2x + y = 160
\ny = 160 – 2x<\/p>\n\n\n\n\n
X<\/td>\n20<\/td>\n40<\/td>\n60<\/td>\n<\/tr>\n
y<\/td>\n120<\/td>\n80<\/td>\n40<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

If x = 20, then y = 160 – 2 x 20 = 160 – 40 = 120
\nIf x = 40, then y = 160 – 2 x 40 = 160 – 80 = 80
\nIf x = 60, then y = 160 – 2 x 60 = 160 – 120 = 40
\nNow plot the points and join them and 4x + 2y = 300
\n=> 2x + y = 150
\n=> y = 150 – 2x<\/p>\n\n\n\n\n
X<\/td>\n40<\/td>\n50<\/td>\n60<\/td>\n<\/tr>\n
y<\/td>\n70<\/td>\n50<\/td>\n30<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

If x = 40, then y = 150 – 2 x 40 = 150 – 80 = 70
\nIf x = 50, then y = 150 – 2 x 50 = 150 – 100 = 50
\nIf x = 60, then y = 150 – 2 x 60 = 150 – 120 = 30
\nNow plot the points and join them We see that these two lines are parallel
\n\"RD
\nRD Sharma Class 10 Solutions Pair Of Linear Equations In Two Variables Ex 3.1<\/p>\n

1.
\n\"Pair-Of-Linear-Equations-In-two-Variables-RD-Sharma-Class-10-Solutions-Ex-3.1-Q-1\"
\n2.
\n\"Pair-Of-Linear-Equations-In-two-Variables-RD-Sharma-Class-10-Solutions-Ex-3.1-Q-2\"
\n\"Pair-Of-Linear-Equations-In-two-Variables-RD-Sharma-Class-10-Solutions-Ex-3.1-Q-2-i\"
\n\"Pair-Of-Linear-Equations-In-two-Variables-RD-Sharma-Class-10-Solutions-Ex-3.1-Q-2-ii\"
\n3.
\n\"Pair-Of-Linear-Equations-In-two-Variables-RD-Sharma-Class-10-Solutions-Ex-3.1-Q-3\"
\n\"Pair-Of-Linear-Equations-In-two-Variables-RD-Sharma-Class-10-Solutions-Ex-3.1-Q-3-i\"
\n\"Pair-Of-Linear-Equations-In-two-Variables-RD-Sharma-Class-10-Solutions-Ex-3.2-Q-3-ii\"
\n4.
\n\"Pair-Of-Linear-Equations-In-two-Variables-RD-Sharma-Class-10-Solutions-Ex-3.1-Q-4\"
\n5.
\n\"Pair-Of-Linear-Equations-In-two-Variables-RD-Sharma-Class-10-Solutions-Ex-3.1-Q-5-i\"
\n\"Pair-Of-Linear-Equations-In-two-Variables-RD-Sharma-Class-10-Solutions-Ex-3.1-Q-5-ii\"
\n\"Pair-Of-Linear-Equations-In-two-Variables-RD-Sharma-Class-10-Solutions-Ex-3.1-Q-5-iii\"
\n6.
\n\"Pair-Of-Linear-Equations-In-two-Variables-RD-Sharma-Class-10-Solutions-Ex-3.1-Q-6-i\"
\n\"Pair-Of-Linear-Equations-In-two-Variables-RD-Sharma-Class-10-Solutions-Ex-3.1-Q-6-ii\"
\n\"Pair-Of-Linear-Equations-In-two-Variables-RD-Sharma-Class-10-Solutions-Ex-3.1-Q-6-iii\"
\n7.
\n\"Pair-Of-Linear-Equations-In-two-Variables-RD-Sharma-Class-10-Solutions-Ex-3.1-Q-7\"
\n\"Pair-Of-Linear-Equations-In-two-Variables-RD-Sharma-Class-10-Solutions-Ex-3.1-Q-7-i\"
\n\"Pair-Of-Linear-Equations-In-two-Variables-RD-Sharma-Class-10-Solutions-Ex-3.1-Q-7-ii\"<\/p>\n

RD Sharma Class 10 Solutions<\/a><\/h4>\n