{"id":2179,"date":"2020-09-08T11:08:40","date_gmt":"2020-09-08T05:38:40","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=2179"},"modified":"2021-07-02T15:03:43","modified_gmt":"2021-07-02T09:33:43","slug":"kseeb-solutions-class-9-maths-chapter-10-ex-10-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-10-ex-10-3\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.3"},"content":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.3<\/strong>\u00a0are part of KSEEB Solutions for Class 9 Maths<\/a>. Here we have given Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Exercise 10.3.<\/p>\n

Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.3<\/h2>\n

Question 1.
\nDraw the graph of each of the following linear equations in two variables :
\ni) x + y = 4
\nii) x – y = 2
\niii) y = 3x
\niv) 3 = 2x + y
\nSolution:
\ni) x + y = 4
\nThis can be written in the form of y = mx + c, then
\ny = 4 – x.<\/p>\n\n\n\n\n
x<\/td>\n0<\/td>\n1<\/td>\n4<\/td>\n-1<\/td>\n<\/tr>\n
y<\/td>\n4<\/td>\n3<\/td>\n0<\/td>\n5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"KSEEB<\/p>\n

ii) x – y = 2
\n-y = 2 – x
\ny = -2 + x
\ny = x – 2<\/p>\n\n\n\n\n
\n

x<\/p>\n<\/td>\n

0<\/td>\n1<\/td>\n2<\/td>\n4<\/td>\n<\/tr>\n
y<\/td>\n-2<\/td>\n-1<\/td>\n0<\/td>\n\n

2<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"KSEEB<\/p>\n

iii) y = 3x<\/p>\n\n\n\n\n
x<\/td>\n0<\/td>\n1<\/td>\n2<\/td>\n-1<\/td>\n<\/tr>\n
y<\/td>\n0<\/td>\n3<\/td>\n6<\/td>\n-3<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"KSEEB<\/p>\n

iv) 3 = 2x + y
\n2x + y = 3
\ny = 3 – 2x<\/p>\n\n\n\n\n
x<\/td>\n0<\/td>\n1<\/td>\n2<\/td>\n-1<\/td>\n<\/tr>\n
y<\/td>\n3<\/td>\n1<\/td>\n-1<\/td>\n5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"KSEEB<\/p>\n

Question 2.
\nGive the equations of two lines passing through (2, 14). How many more such lines are there, and why ?
\nSolution:
\nE.g. i) y = 7x
\nii) x = \\(\\frac{y}{7}\\)
\niii) x + y = 16 .
\nWe may write infinite equation, because Graph of a linear equation in two variables has many solutions.<\/p>\n

Question 3.
\nIf the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
\nSolution:
\n3y = ax + 7
\nIf point (3, 4) then a = ?
\n3y = ax + 7
\n3(4) = a(3) + 7 \u2235 x = 3, y = 4
\n12 = 3a + 7
\n3a + 7 = 12
\n3a = 12 – 7
\n3a = 5
\n\u2234 a = \\(\\frac{5}{3}\\)<\/p>\n

Question 4.
\nThe taxi fare in a city is as follows:
\nFor the first kilometre, the fare is Rs. 8 and for the subsequent distance, it is Rs. 5 per km. Taking the distance covered as x km and the total fare is Rs. y. write a linear equation for this information, and draw its graph.
\nSolution:
\nLet the distance travelled be x km.
\nLet the total distance be y km.
\n\"KSEEB
\nFor the first kilometre, the fare is Rs. 8 + fare for the remaining distance, i.e. (x – 1) km.
\nRs. 5 for 1 km.
\nif it travels (x – 1) km … ? Rs.
\nRs. (x – 1)5
\n\u2234 Total cost = Rs. y
\nFare for first km + fare for subsequent distance means (x-1) km = Rs. y
\n8 + (x – 1)5 = y
\n8 + 5x – 5 = y
\n8 – 5 + 5x = y
\n3 + 5x = y
\n\u2234 y = 3 + 5x linear equation.<\/p>\n\n\n\n\n
x<\/td>\n0<\/td>\n1<\/td>\n-1<\/td>\n-2<\/td>\n<\/tr>\n
y<\/td>\n3<\/td>\n8<\/td>\n-2<\/td>\n-7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"KSEEB<\/p>\n

Question 5.
\nFrom the choices given below, choose the equation whose graphs are given in Fig. 10.6 and Fig. 10. 7.
\n\"KSEEB
\nSolution:
\nSolution for graph of linear equation of Fig. 10.6 :
\n\"KSEEB
\nSum of (x + y) = 0.
\nEquation for this is<\/p>\n

(ii) x + y = 0
\nbecause x = -y we have OR -x = y
\n0 = 0
\n1 = -1
\n-1 = 1
\nSolutions for graph for linear equation in Fig. 10.7 :
\n\"KSEEB
\nx + y = 2 is common in all.
\n\u2234 y = -x + 2
\n\u2234 Ans: (iii) y = -x + 2 equation.<\/p>\n

Question 6.
\nIf the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is
\n(i) 2 units
\n(ii) 0 unit.
\nSolution:
\nLet the work done be, W
\nDistance travelled be D.
\nIf constant force is K, then
\nForce is directly proportional to the distance travelled by the body.
\n\u2234 W \u221d D.
\nW = K \u00d7 D
\nConstant force, K = 5 units.
\nW = 5 \u00d7 D linear equation.
\nIf D = 2, then W = 5D
\nW = 5 \u00d7 2
\nW = 10
\nIf D = 0, then W = 5D
\nW = 5 \u00d7 0
\nW = 0
\n\"KSEEB<\/p>\n

Question 7.
\nYamini and Fatima, two students of Class IX of a school, together contributed Rs. 100 towards the Prime Minister\u2019s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs. x and Rs. y.) Draw the graph of the same.
\nSolution:
\nYamini and Fatima together contributed Rs. 100 towards the Prime Minister\u2019s Relief Fund to help the earthquake victims.
\nIn that contribution of Yamini is R. x Contribution of Fatima is Rs. y Together Rs. 100.
\n\u2234 x + y = 100
\n\u2234 y = 100 – x.<\/p>\n\n\n\n\n
x<\/td>\n20<\/td>\n40<\/td>\n50<\/td>\n60<\/td>\n<\/tr>\n
y<\/td>\n80<\/td>\n60<\/td>\n50<\/td>\n40<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

\"KSEEB
\n\"KSEEB<\/p>\n

Question 8.
\nIn countries like USA and Canada, the temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius :
\nF = \\(\\left(\\frac{9}{5}\\right)\\) C + 32.
\n(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for the y-axis.
\n(ii) If the temperature is 30\u00b0 C, what is the temperature in Fahrenheit?
\n(iii) If the temperature is 95\u00b0 F, what is the temperature in Celsius?
\n(iv) If the temperature is 0\u00b0 C, what is the temperature in Fahrenheit, and if the temperature is 0\u00b0 F, what is the temperature in Celsius?
\n(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
\nSolution:
\nLet Temperature be Celsius C,
\nThe temperature be Fahrenheit F.
\nF = \\(\\left(\\frac{9}{5}\\right)\\) C + 32<\/p>\n\n\n\n\n
\n

C<\/p>\n<\/td>\n

10<\/td>\n30<\/td>\n0<\/td>\n35<\/td>\n<\/tr>\n
F<\/td>\n50<\/td>\n86<\/td>\n32<\/td>\n\n

95<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

In the x-axis Celsius,
\nIn the y-axis Fahrenheit.
\n(i) Graph
\n\"KSEEB
\n(ii) If temperature is 30\u00b0 C, then 86\u00b0 F.
\n(iii) If the temperature is 95\u00b0 F, then 35\u00b0 C.
\n(iv) If temperature is 0\u00b0 C, then 32\u00b0 F.
\nIf 0\u00b0F, then -32\u00b0C
\n(v) Temperature is not equal in Celsius and Fahrenheit.
\nbecause C \u2260 F<\/p>\n

We hope the KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.3 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Exercise 10.3, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.3\u00a0are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Exercise 10.3. Karnataka Board Class 9 Maths Chapter 10 Linear Equations in Two Variables Ex 10.3 …<\/p>\n","protected":false},"author":6,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2179"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=2179"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2179\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=2179"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=2179"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=2179"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}