\n<\/h3>\n
- \n
- Linear Inequalities in Two Variables by Graphical Method.\n<\/li>\n
- Graphical Solution of Simultaneous Linear Inequalities in Two Variables.\n<\/li>\n
- Linear Programming\n<\/li>\n<\/ul>\n
\u00a0
\n\u00a0<\/p>\nGRAPHICAL SOLUTION OF INEQUALIIES IN TWO VARIABLES<\/p>\n<\/h3>\n
A straight line has the general equation ax+by+c=0, where a,b and c are real numbers. <\/p>\n
\u00a0
\n\u00a0<\/p>\nThe line ax + by + c =0 partitions the x-y plane into two regions<\/p>\n<\/h4>\n
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1031_Week4SS1Se1.png\")
<\/p>\n\u00a0Worked Examples <\/strong>
\n\t1) Show the region representing 2x + y + 1 > 0 <\/p>\nSolution
\n<\/h4>\n2x + y + 1 > 0
\nSteps <\/p>\n- \n
- make y the subject of the inequality\n<\/li>\n
- convert the inequality into a line equation\n<\/li>\n
- obtain x and y co ordinates of the line\n<\/li>\n
- draw the line and shade the required by the inequality 2x + y + 1 > 0 y > – 2x -1\n<\/li>\n<\/ol>\n
When x = 0 , y = -2(0) -1
\n y = -1 (0, 1)
\nWhen y = 0, 0 > -2x – 1
\n1 > -2x
\nx = – \u00bd (-\u00bd ,0) <\/p>\n\u00a0
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\n\u00a02) Show the region represented by x – 2y + 3 \u2264 0
\nSolution <\/strong>x- 2y + 3 \u2264 0
\n2y = – 3 \u2013 x \u2192 y = -3\/-2 – x\/-2 \u2192 y = 3\/2 + x\/2 or y = 3 + x
\n\t2
\nWhen x = 0
\n y = 3 + 0 = 3 (0, 3\/2)
\n2 2
\nWhen y = 0
\n0 = 3 + x
\n\t 2
\n x = -3 (-3, 0) <\/p>\n\u00a0
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\n\u00a0<\/p>\nEvaluation
\n<\/h4>\nShow the region which represents the following inequality a) 2x \u2013 3y + 1 \u2264 0 b) x \u2013 4y + 7 \u22650 <\/p>\n
\u00a0
\n\u00a0<\/p>\nSIMULTANEOUS INEQUALITIES
\n<\/h3>\nThe set of simultaneous inequalities in two variables can be found from the intersection of the areas representing the inequalities. <\/p>\n
\u00a0
\n\u00a0<\/p>\nWorked Examples
\n<\/h4>\n1) Show graphically the region R which satisfies the set of inequalities 2x + 2y \u2264 2, x + 2y\u2264 16, x \u2265 0, y \u2265 0. <\/p>\n
\u00a0Solution <\/strong>2x + 2y \u2264 2
\n2y \u2264 2 \u2013 2x y \u2264 2 \u2013 2x 2 y \u2264 2 – 2x 2 2 y \u2264 1 – x
\nWhen x = 0, y = 1 \u2013 0 =1 (0, 1 ) When y =0, 0 = 1 \u2013 x
\n 1= -x point (1 , 0).
\nX + 2y \u2264 16
\n2y \u2264 16 \u2013 x, y \u2264 16 \u2013 x
\n\t2
\nWhen x = 0
\n y = 16 \u2013 0 = 16 = 8 (0,8)
\n 2 2 <\/p>\n\u00a0When y = 0, 0 = 16 \u2013 x
\n\t2 16 \u2013 x = 0 x = – 16
\n x = 16 (16, 0).
\n x \u2265 0, x = 0 , y \u2265 0, y = 0 <\/p>\n\u00a0
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\n\u00a02) Show graphically, the region which satisfies the set of inequalities 4x + y \u2264 15, 8x \u2013 y \u2265 9, x \u2265 0, y \u22650. <\/p>\n\u00a0<\/p>\n
Solution
\n<\/h4>\n4x + y \u2265 15, y \u2264 15 \u2013 4x
\nWhen x = 0
\n. y = 15 \u2013 4(0)
\n. y = 15 (0, 15)
\nWhen y = 0
\n0 = 15 \u2013 4x
\n-15 = -4
\n\t4 -4
\nX = 15\/4 (3\u00be, 0)
\n8x \u2013 y \u2265 9
\n. \u2013y \u2265 9 \u2013 8x, y \u2265 8x \u2013 9
\nWhen x = 0 , y = 8(0) \u2013 9
\n . y = -9 (0, -9)
\n When y = 0, 0 = 8x \u2013 9 9 = 8x, x = 9\/8 = 1\u215b (1\u215b, 0 )
\n. x \u2265 0 or x = 0, y \u2265 0 or x = 0 <\/p>\n\u00a0
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\n\u00a0<\/p>\nEvaluation
\n<\/h4>\nShow the regions which represent the set of solution of
\n1) 2y \u2264 x + 8, x + 2y + 4 \u2265 0, x \u2264 2y + 12 2) y \u2265 0, x + 2y \u2264 4, -x + 2y \u2264 11, -2x + 5y \u2264 10 <\/p>\n\u00a0<\/p>\n
LINEAR PROGRAMMING
\n<\/h3>\nThe linear function z = ax + by is called the objective function while the given set of the inequalities are called the constraint linear programming attempts to maximize or minimize an objective function under the set of given constraints. Example 1 <\/strong>
\n\tA caterer can make two types of of drinks A and B. A litre of A contains 2gramme of orange juice and 3gramme of pineapple juice. A litre of B contains 4gramme of orange juice and 5gramme of pineapple juice.
\nThere are not more than 16gramme of orange juice and 21gramme of pineapple juice.
\nThe caterer can make a profit of 1ok on 1gramme of A and 15k on 1gramme of B. Assuming that the caterer makes x <\/em>litres of A and y<\/em> litres of B. (a) Write all the inequalities connecting x <\/em>and y<\/em>. <\/p>\n- \n
- Show by shading the required region satisfying the inequalities in (a\ufd3f\n<\/li>\n
- Find the quantity of each type of drink a caterer must make if she is to maximize profit.\n<\/li>\n<\/ol>\n
\u00a0
\n\u00a0<\/p>\nSolution
\n<\/h4>\n- \n
- 2x+4y \u226416, 3x + 5y\u2264 21, x\u22650, y \u2265 0\n<\/li>\n
- 2x + 4y \u226416, 4y = 16 – 2x, y = 16-2x\n\t\t<\/li>\n<\/ol>\n
4
\nWhen x = 0
\n.y = 16 \u2013 2(0) = 16 = 4 (0,4)
\n 4
\nWhen y = 0 , 0 = 16 \u2013 2x<\/p>\n- \n
![\"\"\/](\"https:\/\/ecolebooks.com\/nigeria\/wp-content\/uploads\/9jalessonsimages\/100223_1031_Week4SS1Se2.png\")
x = 8 (8, 0)\n<\/li>\n- 5\n<\/li>\n<\/ol>\n
y = 21-3x
\n\t 5 <\/p>\n\u00a0When x =0 , y = 21- 0 = 21 (0, 21\/5)
\n 5 5 \u00a0\u00a0\u00a0\u00a0 <\/p>\n\u00a0When y = 0, 0 = 21 \u2013 3x
\n\t 5
\n21 \u2013 3x = 0
\n 3x = -21, x = -21 = 7 (7,0)
\n 3 <\/p>\n\u00a0
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\n\u00a0(c) 7 litres of A and none of B. <\/p>\n\u00a0Example 2<\/strong>.
\nA fashion designer makes two types of dresses X and Y by making use of two types of materials P and Q. The quantity of material used for each unit of dress in m2<\/sup>, and the profit on each dress in N are as shown in the following table. <\/p>\n\n\n\n
\n \u00a0<\/td>\n P <\/td>\n Q <\/td>\n Profit <\/td>\n<\/tr>\n \n X <\/td>\n 3 <\/td>\n 2 <\/td>\n 2 <\/td>\n<\/tr>\n \n Y <\/td>\n 4 <\/td>\n 5 <\/td>\n 3 <\/td>\n<\/tr>\n \n Quantity available <\/td>\n 18 <\/td>\n 19 <\/td>\n \u00a0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n \u00a0<\/p>\n
- \n
- Assuming that the designer makes x<\/em> unit of X and y <\/em>and units of Y. write down the four inequalities connecting x <\/em>and y.<\/em>\n\t\t<\/li>\n
- Find how many of each type of dresses the fashion designer should make in order to maximize Profit.\n<\/li>\n<\/ol>\n
Solution
\n<\/h4>\nThe quantity of material P used in making x<\/em> units of dress X and y<\/em> units of dress Y is 3x<\/em>+ 4y<\/em>, since the quantity of material P available is 18m2.<\/sup>
\n\t3x<\/em>+ 4y<\/em> \u2264 18
\nSimilarly for material Q
\n2x + 5y \u2264 19
\nAlso, x \u22650, y \u22650
\n3x + 4y = 18, 4x = 18 -3x
\n y = 18 \u2013 3x
\n 4 <\/p>\n\u00a0When x = 0, y = 18 \u2013 3(0) = 18 = 4\u00bd <\/p>\n
- \n
- 4 (0, 4\u00bd)\n<\/li>\n<\/ol>\n
When y = 0, 18 \u2013 3x = 0
\n3x = -18
\n. x = -18 = 6 (6, 0)
\n 3
\n 2x + 5y = 19 \u2192 5y = 19 -2x, \u2192 y = 19 \u2013 2x
\n\t 5
\nWhen x = 0, y = 19 \u2013 2(0) = 19 (0, 19\/5) <\/p>\n- \n
- 5\n<\/li>\n<\/ol>\n
When y= 0, 19 \u2013 2x = 0
\n 2x = – 19
\n x = -19 = 19
\n\t 2 2 (19\/2, 0) <\/p>\n\u00a0
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\n\u00a0Let z be the profit, then z = 2x + 3y at the point C(2, 3) z = 2(2) + 3(3 ) z = 4+ 9 = 13
\nHence the fashion designer should make 2 dresses of type X and 3 dresses of type Y in order to make a maximum profit of N13.00 <\/p>\n\u00a0<\/p>\n
Evaluation
\n<\/h4>\nA petty trader sells two types of detergents A and B. a dm3<\/sup> of A contains 2gm of Omo detergent and 5gm of Surf detergent. A dm3<\/sup> of B contains 3gm of omo detergent and 2gm of surf detergent. Altogether she has at most 26g of omo detergent and 32g of surf detergent, the trader makes a profit of 2k per gm on A and 1k per gm on B. If the trader sells x dm3<\/sup> of A and y dm3<\/sup> of B 1. \u00a0\u00a0\u00a0\u00a0Write down all the inequalities connecting x and y. <\/p>\n
- \n
- Indicate by shading the region R satisfying all the inequalities in (a)\n<\/li>\n
- Determine the values of x and y which maximises the traders profit.\n<\/li>\n<\/ol>\n
\u00a0<\/p>\n
Solution
\n<\/h4>\n\n\n\n
\n \u00a0<\/td>\n Omo <\/td>\n Surf <\/td>\n Profit <\/td>\n<\/tr>\n \n A <\/td>\n 2 <\/td>\n 5 <\/td>\n 2 <\/td>\n<\/tr>\n \n B <\/td>\n 3 <\/td>\n 2 <\/td>\n 1 <\/td>\n<\/tr>\n \n Total <\/td>\n 26 <\/td>\n 32 <\/td>\n \u00a0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n \u00a0x\u2265 0, y \u2265 0, 2x + 3y \u2264 26, 5x + 2y \u2264 32 2x + 3y = 26,
\n3y = 26 – 2x, y = 26 \u2013 2x
\n\t 3
\nWhen x = 0 , y = 26 \u2013 2(0) = 26 = 8\u2154 (0, 8\u2154)
\n 3 3
\nWhen y = 0, 26 \u2013 2x = 0 <\/p>\n\u00a0– 2x = – 26, x = – 26 = 13 (13, 0) – 2
\n5x + 2y \u2264 32
\n2y = 32 \u2013 5x, y = 32 \u2013 5x
\n\t 2
\nWhen x = 0, y = 32 – 5(0) = 32\/2 = 16 (0, 16) When y = 0, 32 \u2013 5x = 0
\n-5x = -32, x = -32\/-5 = 32 (32\/5, 0) 5 <\/p>\n\u00a0
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\n\u00a0The corner points are A(0, 8.6) , B(4, 6), C(6.4,0) , D(0,0)
\nProfit Z = 2x + y
\nAt A, Z = 2(0 + 8.6) = 8.6
\nAt B, Z = 2(4 + 6) = 14
\nAt C, Z = 2(6.4 + 0)= 12.8 At D, Z = 2(0 + 0) = 0
\nHence, the trader should sell 4 of detergent A and 6 of detergent B to make a profit of 14k. <\/p>\n\u00a0<\/p>\n
Evaluation
\n<\/h4>\n1) Show graphically the region represented by the inequalities (a) y \u2265 4x2<\/sup> + 11x \u2013 3 (b) y \u2265 6x2<\/sup> \u2013 x \u2013 2 2) Show graphically the region R which satisfies the set of inequalities: 2x + 3y \u2264 26, x + 2y \u2264 16, x \u2265 0, y \u2264 0. <\/p>\n
\u00a0<\/p>\n
General Evaluation
\n<\/h4>\n1. show the region R which satisfies the following simultaneous inequalities y + x \u2264 3, y+ x \u2265 1, y – x \u2264 1, x \u2265 0, y \u2265 0. 2. \u00a0\u00a0\u00a0\u00a0show the region R which satisfies simultaneously 2x + y \u2264 7, 3x \u2013 4y \u2265 – 6, x \u2265 0, y \u2265 0. <\/p>\n
- \n
- 3x2<\/sup> + 7x \u2013 3 = 0 \u00a0\u00a0\u00a0\u00a0 solve using formula method\n<\/li>\n
- Using completing the square and formula method solve 3x2<\/sup> \u2013 12x + 10 = 0\n<\/li>\n
- Solve the following exponential equations (a) 22x<\/sup> – 6(2x<\/sup>) + 8 = 0 (b) 22x+1<\/sup> – 5 (2x<\/sup>) + 2 = 0\n<\/li>\n
- Janet buys p sweet and q marbles. The sweets cost \u20a65 each and the marbles cost \u20a66 each. Janet has \u20a690. She wants to share the sweets with her friends, so she needs at least 5sweets, she needs more than 4 marbles to be able to join in the game. (a) Write down three inequalities connecting p and q (b) Draw the graph to show their inequalities (c) What is the highest number of sweets she can buy? (d) What is the highest number of marbles she can buy?\n<\/li>\n<\/ol>\n
\u00a0<\/p>\n
Reading Assignment : F\/maths Project 1 pg 113 \u2013 119 Exercise 8c Q1, 16 and 17<\/p>\n<\/h4>\n
\u00a0 WEEKEND ASSIGNMENT <\/strong>1) Find the range of x for which \u25022x – 1\u2502> 3
\n (a) 1< x < 3\/2 b) -3\/2 < x < -1 c) -3\/2 < x < 1 d) x > 3\/2 and x < -1 <\/p>\n- \n
- \nFind the range of the value that satisfies the inequality x2<\/sup> + 3x \u2013 18 < 0\n<\/div>\n
- \n
- -3 < x < 6 (b)-3 > x <6 (c)-6 >x >3 (d)-6 >x < 3 (e)-6 < x <3\n<\/li>\n<\/ol>\n<\/li>\n
- \nFind the range of values of x for which 2x2<\/sup> \u2013 5x + 2 \u2265 0\n<\/div>\n
- \n
- -2 < x < -\u00bd (b) \u00bd < x < 2 (c) x < -\u00bd or x \u2265 -2 (d) x \u2264 \u00bd or x \u2265 2\n<\/li>\n<\/ol>\n<\/li>\n
- \nFind the range of values of y which satisfies the inequality 2y \u2013 1 < 3 and 2 \u2013 y \u2264 5\n<\/div>\n
- \n
- \u2013 3 \u2264 y \u2264 1 (b) \u2013 2 \u2264y \u2264 3 (c) -3\u2264 y \u2264 4 (d) -3 \u2264 y \u2264 2\n<\/li>\n<\/ol>\n<\/li>\n
- Find the range of values of x for which 1\/x + 3 < 2x is satisfy (a) \u2013 3 < x < 5\/2 (b) x < -3 and x > -5\/2 (c) x < 1 and x < \u00bd\n<\/li>\n<\/ol>\n
\u00a0<\/p>\n
THEORY
\n<\/h3>\n- \n
- Illustrate graphically the set P of all points ( x, y) which satisfy simultaneously the following inequalities:\n<\/li>\n<\/ol>\n
2y \u2264 x + 8, x + 2y + 4 \u2265 0, 3x \u2264 2y + 12. Using your diagram, calculate on the set P the maximum values of (i) x (ii) y (iii) 12x + 5y <\/p>\n
- \n
- Determine the values of x satisfying |x + 3| \u2265 8\n<\/li>\n<\/ol>\n
\u00a0<\/p>\n","protected":false},"excerpt":{"rendered":"
\u00a0 WEEK FOUR \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0 \u00a0\u00a0\u00a0\u00a0DATE\u2026\u2026\u2026\u2026\u2026 TOPIC: LINEAR INEQUALITY…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,188],"tags":[],"class_list":["post-2170","post","type-post","status-publish","format-standard","hentry","category-posts","category-second-term-ss1-further-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2170","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=2170"}],"version-history":[{"count":2,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2170\/revisions"}],"predecessor-version":[{"id":2185,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/2170\/revisions\/2185"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=2170"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=2170"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=2170"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- Determine the values of x satisfying |x + 3| \u2265 8\n<\/li>\n<\/ol>\n
- Illustrate graphically the set P of all points ( x, y) which satisfy simultaneously the following inequalities:\n<\/li>\n<\/ol>\n
- Using completing the square and formula method solve 3x2<\/sup> \u2013 12x + 10 = 0\n<\/li>\n
- 3x2<\/sup> + 7x \u2013 3 = 0 \u00a0\u00a0\u00a0\u00a0 solve using formula method\n<\/li>\n
- 5\n<\/li>\n<\/ol>\n
- Find how many of each type of dresses the fashion designer should make in order to maximize Profit.\n<\/li>\n<\/ol>\n
- Assuming that the designer makes x<\/em> unit of X and y <\/em>and units of Y. write down the four inequalities connecting x <\/em>and y.<\/em>\n\t\t<\/li>\n