This page contains various linear equations word problems. Students have to practice all these questions to score better marks in the exam. To solve these questions on linear equations, you need to understand the given details and make an equation using one variable. And finally, solve that equation. For better understanding, you can check our Linear Equations Practice Problems with Solutions.<\/p>\n
Example 1.<\/strong><\/p>\n Solve the following linear equations:<\/p>\n (a) x\/4 – 1\/2 = x\/5 + 2\/3<\/p>\n (b) x – (x + 1)\/3 = 1 – (x – 1)\/4<\/p>\n Solution:<\/strong><\/p>\n (a) x\/4 – 1\/2 = x\/5 + 2\/3<\/p>\n (x – 2) \/ 4 = (3x + 10) \/ 15<\/p>\n Cross multiply the fractions.<\/p>\n 15(x – 2) = 4(3x + 10)<\/p>\n 15x – 30 = 12x + 40<\/p>\n Transferring 12x from R.H.S to L.H.S becomes -12x, -30 from L.H.S to R.H.S becomes + 30.<\/p>\n 15x – 12x = 40 + 30<\/p>\n 3x = 70<\/p>\n Divide both sides of the equation by 3.<\/p>\n 3x\/3 = 70\/3<\/p>\n x = 70\/3<\/p>\n Therefore, the required solution set is x = 70\/3.<\/p>\n (b) x – (x + 1)\/3 = 1 – (x – 1)\/4<\/p>\n [3x – (x + 1)] \/ 3 = [4 – (x – 1)] \/ 4<\/p>\n Cross multiply the fractions.<\/p>\n 4(3x – x – 1) = 3(4 – x + 1)<\/p>\n 4(2x – 1) = 3(5 – x)<\/p>\n 8x – 4 = 15 – 3x<\/p>\n Transferring -3x from R.H.S to L.H.S becomes 3x, -4 from L.H.S to R.H.S becomes +4.<\/p>\n 8x + 3x = 15 + 4<\/p>\n 11x = 19<\/p>\n x = 19\/11<\/p>\n Therefore, the required solution set is x = 19\/11.<\/p>\n Example 2.<\/strong><\/p>\n Solve the following equations and verify them.<\/p>\n (a) 4(3y + 2) – 5(6y – 1) = 2(y – 8) – 6(7y – 4) + 4y<\/p>\n (b) 2\/(3x + 1) + 4\/(3x + 1) = 5\/3x<\/p>\n Solution:<\/strong><\/p>\n (a) 4(3y + 2) – 5(6y – 1) = 2(y – 8) – 6(7y – 4) + 4y<\/p>\n 12y + 8 – 30y + 5 = 2y – 16 – 42y + 24 + 4y<\/p>\n 13 – 18y = 8 – 36y<\/p>\n Transferring -36y from R.H.S to L.H.S becomes 36y, 13 from L.H.S to R.H.s becomes -13.<\/p>\n -18y + 36y = 8 – 13<\/p>\n 18y = -5<\/p>\n Divide both sides of the equation by 18.<\/p>\n 18y\/18 = -5\/18<\/p>\n y = -5\/18<\/p>\n Verification:<\/p>\n L.H.S = 4(3y + 2) – 5(6y – 1)<\/p>\n Substitute y = -5\/18<\/p>\n L.H.S = 4(3(-5\/18) + 2) – 5(6(-5\/18) – 1)<\/p>\n = -60\/18 + 8 + 150\/18 + 5<\/p>\n = 90\/18 + 13<\/p>\n = 5 + 13<\/p>\n = 18<\/p>\n R.H.S = 2(y – 8) – 6(7y – 4) + 4y<\/p>\n Substitute y = -5\/18<\/p>\n R.H.S = 2(-5\/18 – 8) – 6(7(-5\/18) – 4) + 4(-5\/18)<\/p>\n = -10\/18 – 16 + 210\/18 + 24 -20\/18<\/p>\n = 180\/18 + 8<\/p>\n = 10 + 18<\/p>\n = 18<\/p>\n L.H.S = R.H.S<\/p>\n Hence, proved.<\/p>\n (b) 2\/(3x + 1) + 4\/(3x + 1) = 5\/3x<\/p>\n 6\/(3x + 1) = 5\/3x<\/p>\n Cross multiply the fractions.<\/p>\n 6 x 3x = 5 x (3x + 1)<\/p>\n 18x = 15x + 5<\/p>\n Transferring 15x from R.H.S to L.H.S becomes -15x.<\/p>\n 18x – 15x = 5<\/p>\n 3x = 5<\/p>\n Divide both sides of the equation by 3<\/p>\n 3x\/3 = 5\/3<\/p>\n x = 5\/3<\/p>\n Verification,<\/p>\n L.H.S = 2\/(3x + 1) + 4\/(3x + 1)<\/p>\n Put x = 5\/3<\/p>\n L.H.S = 2\/(3(5\/3) + 1)) + 4\/(3(5\/3) + 1))<\/p>\n = 6\/(15\/3 + 1)<\/p>\n = 6\/((15 + 3)\/3)<\/p>\n = 6\/(18\/3)<\/p>\n = 6\/6<\/p>\n = 1<\/p>\n R.H.S = 5\/3x<\/p>\n Put x = 5\/3<\/p>\n R.H.S = 5\/(3 x 5\/3)<\/p>\n = 5\/(15\/3)<\/p>\n = 5\/5<\/p>\n = 1<\/p>\n L.H.S = R.H.S<\/p>\n Hence proved.<\/p>\n Example 3.<\/strong><\/p>\n The sum of the digits of a 2 digit number is 8. The number obtained by interchanging the digits exceeds the original number by 18.<\/p>\n Solution:<\/strong><\/p>\n Let the 2 digit number at ones place is x, then<\/p>\n Tens place of the number = 8 – x<\/p>\n Original number = 10(8 – x) + x<\/p>\n = 80 – 10x + x<\/p>\n = 80 – 9x<\/p>\n By interchanging the digits<\/p>\n New number = 10x + 1(8 – x)<\/p>\n = 10x + 8 – x<\/p>\n = 9x + 8<\/p>\n According to question,<\/p>\n New number – Original number = 18<\/p>\n 9x + 8 – (80 – 9x) = 18<\/p>\n 9x + 8 – 80 + 9x = 18<\/p>\n 18x – 72 = 18<\/p>\n 18x = 18 + 72<\/p>\n 18x = 90<\/p>\n x = 90\/18<\/p>\n x = 5<\/p>\n Hence, the digits at ones place is 5.<\/p>\n The digits at tens place = (8-5) = 3.<\/p>\n So, the original number is 35 and the new number is 53.<\/p>\n Example 4.<\/strong><\/p>\n The mother\u2019s age is three times her son\u2019s age. Four years ago, she was 4 times her son\u2019s age. Find their present ages.<\/p>\n Solution:<\/strong><\/p>\n Let the present age of son be x, then<\/p>\n Present mother’s age = 3x<\/p>\n Four years ago,<\/p>\n Son age = (x – 4)<\/p>\n Mothers age = (3x – 4)<\/p>\n According to the question,<\/p>\n Mothers age = 4 x son’s age<\/p>\n 3x – 4 = 4(x – 4)<\/p>\n 3x – 4 = 4x – 16<\/p>\n -4 + 16 = 4x – 3x<\/p>\n x = 12<\/p>\n Hence, the son’s present age is 12 years, mothers present age is 12 x 3 = 36 years.<\/p>\n Example 5.<\/strong><\/p>\n If a rectangle has a width of 8 inches and has a perimeter of 36 inches, then what is the length?<\/p>\n Solution:<\/strong><\/p>\n Given that,<\/p>\n Rectangle width w = 8 inches<\/p>\n Rectangle perimeter P = 36 inches<\/p>\n Perimeter of a rectangle = 2(length + width)<\/p>\n 36 = 2(length + 8)<\/p>\n 36 = 2length + 16<\/p>\n 36 – 16 = 2length<\/p>\n 20 = 2length<\/p>\n length = 20\/2<\/p>\n length = 10<\/p>\n Hence, the rectangle length is 10 inches.<\/p>\n Example 6.<\/strong><\/p>\n A bicycle and a bicycle helmet cost 240 dollars. How much did each cost, if the bicycle cost 5 times as much as the helmet?<\/p>\n Solution:<\/strong><\/p>\n Let the cost of the bicycle helmet be x, then<\/p>\n Cost of the bicycle = 5x<\/p>\n According to the question,<\/p>\n x + 5x = 240<\/p>\n 6x = 240<\/p>\n x = 240\/6<\/p>\n x = 40<\/p>\n Hence, the cost of a bicycle helmet is 40 dollars, the cost of the bicycle is 40 x 5 = 200 dollars.<\/p>\n Example 7.<\/strong><\/p>\n The total cost for tuition plus room and board at State University is 2,584 dollars. Tuition costs 704 dollars more than room and board. What is the tuition fee?<\/p>\n Solution:<\/strong><\/p>\n Let the room and board cost be x, then<\/p>\n Tuition cost = 704 + x<\/p>\n According to the question,<\/p>\n 704 + x + x = 2584<\/p>\n 704 + 2x = 2584<\/p>\n 2x = 2584 – 704<\/p>\n 2x = 1880<\/p>\n x = 1880\/2<\/p>\n x = 940<\/p>\n Hence, the tuition fee is 940 + 704 = 1664.<\/p>\n Example 8.<\/strong><\/p>\n A sum of $2700 is to be given in the form of 63 prizes. If the prize is of either $100 or $25, find the number of prizes of each type.<\/p>\n Solution:<\/strong><\/p>\n Let x be the number of $100 and y be the number of $25.<\/p>\n x + y = 63<\/p>\n 100x + 25y = 2700<\/p>\n Multiply both sides of the equation x + y = 63 by 25.<\/p>\n 25(x + y) = 63 x 25<\/p>\n 25x + 25y = 1575<\/p>\n Subtract 100x + 25y = 2700 from 25x + 25y = 1575<\/p>\n 100x + 25y – (25x + 25y) = 2700 – 1575<\/p>\n 100x + 25y – 25x – 25y = 1125<\/p>\n 75x = 1125<\/p>\n x = 1125\/75<\/p>\n x = 15<\/p>\n Substitute x = 15 in x + y = 63<\/p>\n 15 + y = 63<\/p>\n y = 63 – 15<\/p>\n y = 48.<\/p>\n Hence 15 number of $100, 43 number of $25 prizes.<\/p>\n Example 9.<\/strong><\/p>\n Of 240 stamps that harry and his sister collected, Harry collected 3 times as many as his sisters. How many did each collect?<\/p>\n Solution:<\/strong><\/p>\n Let x be the number of stamps collected by harry’s sister, then<\/p>\n Number of stamps collected by harry = 3x<\/p>\n According to the question,<\/p>\n x + 3x = 240<\/p>\n 4x = 240<\/p>\n x = 240\/4<\/p>\n x = 60<\/p>\n Hence, the number of stamps collected by harry is 3 x 60 = 180 and his sister is 60.<\/p>\n Example 10.<\/strong><\/p>\n A lab technician cuts a 12 inch piece off tubing into two pieces in such a way that one piece is 2 times longer than the other.<\/p>\n Solution:<\/strong><\/p>\n Let, the smaller piece length is, longer piece length is y.<\/p>\n y = 2x<\/p>\n According to the question,<\/p>\n x + y = 12<\/p>\n x + 2x = 12<\/p>\n 3x = 12<\/p>\n x = 12\/3<\/p>\n x = 4<\/p>\n y = 2 x 4 = 8<\/p>\n Hence, the length of shorter piece is 4 inches, longer piece is 8 inches.<\/p>\n","protected":false},"excerpt":{"rendered":" This page contains various linear equations word problems. Students have to practice all these questions to score better marks in the exam. To solve these questions on linear equations, you need to understand the given details and make an equation using one variable. And finally, solve that equation. For better understanding, you can check our … Read more<\/a><\/p>\n","protected":false},"author":1,"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:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/35207"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=35207"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/35207\/revisions"}],"predecessor-version":[{"id":138479,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/35207\/revisions\/138479"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=35207"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=35207"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=35207"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}