MATHEMATICS FORM 1 – COORDINATE GEOMETRY msomimaktaba, November 11, 2018August 17, 2024 COORDINATE GEOMETRY IntroductionThe position of points on a line found by using a number line, that is When two number lines one vertical and another one horizontal are considered one kept at 90o and intersecting at their zero marks, The result is called xy – plane or Cartesian plane. The horizontal one is called x – axis and the vertical is called y – axis.Origin is the where the two axes that is x – axis and y axis (intersect) Coordinates of a pointThe position of a point in the xy – plane is given by a pair of in the form of ordered pair. Thus ordered pair is called coordinate. The coordinate of the point is therefore written in the form of (a, b), Where the first number ‘’a’’ is the value in the horizontal axis i.e x – axis b is the value in the y – axisThe value in the x – axis is also referred to as abscissa and y – axis is called ordinate. All distance in the xy – plane are measured the origin.Examples write the coordinates of the following point A, B, C, D, E, Ffigure 10.1Solutions The coordinates of the points are A = (0,5) B = (5,0) C = (0,4) D = (-5,5)Exercise 10.11. a) Write down the coordinates of each of the labeled points in figure 9.2b) State the quadrant in which each of these points F, H, V and I belong figure 10.2 2. Draw axes on a graph paper and plot the points given below. Join in the order given with straight lines forming polygonal figures shape have you drawn in each case.a) (1, 1), (3, 1), (3, 3), (1, 3)b) (-2, 1), (2, 5), (2, -2)c) (3, 1), (5.4, 1) (4.3, 2), (3.3, 2)d) (5.5, 3.4), (6.5, 3.4), (6.8, 4.3), (6.0, 4.9), (5.2, 4.3)e) (1.5, -3), (6, 3), (1.5, 3), (-6, -3)f) (-1, 0), (-2, 2), (0, 1), (2, 2), (1, 0), (2, -2), (0, -1)SLOP/GRADIENT OF A LINE Slope / gradient is the change in the vertical axis to the change in the horizontal axis. Example: Find the gradients of the lines joining (a) A (2, 4) and B (-2, 6)(b ) A (-2, -2) and B (2, -4)(c) A (0, -1) and B (2, 3) Solutiona. Let (x1, y1) be (2, 4) and(x2, y2) be (-2, 6)M=b. let (x1, y1) be (-2, -2)(x2, y2) be (2, -4)M = = m= c. A (0, -1) and B (2, 3)SolutionLet (x1, y1) be (0, -1)(x2, y2) be (2, 3)M = = M = 2 Exercise 10.21.Plot pair of the following points on a graph paper and join them by straight line. For each pair, calculate the gradient of the line and state whether it is positive, negative, zero or undefined.(a) (0,3), (2,5) (b) (5,8), (4,1) (c) (1,5), (4,7) (d) (2,6), (5,3) (e) (1,6), (3,-1) (f) (3,6), (-2,-1) (g) (0,2), (6,2) (h) (2,3), (-1,-3) (i) (2,10), (2,0) (j) (–), , 2) (k) (-2,1), (4,3) (l) (-4,4),(-3,3) (m) (0,0), (-3,4) (n) (99,6), (119,1) (o) (0.64,-1.62, (1.36,-0.62)) EQUATION OF A LINEWe have already discussed how to find the gradient of a line for example the gradient of the line joining points (2, – 4) and (5,0) is given as. Since the two points are collinear. we can find the equation of the line having any point on a line say (x,y) and any point, then from let (x1,y1) = (x,y), and (x2,y2) = (5,0) 4(5-x) = 3x-y 20-4x = -3y ∴y = – In general the equation of a straight line is written as y = mx + c. Where m – Is the slope of the line and c is ordinate of the y. called y- intercept The point on the line (x,y) is called arbitrary pointExample: – find the equation of line passing through the points.(12,-6) and (2, 6)Solution let (x,y)=(12,-6), (x2-y2)= (2,6) 5(y-6)=-6(x-2) 5y-30=-6x+12 5y=-6+12+30 5y=-6x+42 (2) Give that y = – + 6 find the gradient of this lineThe gradient is Example: find the equation of the line passing through the point (4, 6) and having a slope -1/2Solution(x,y) , (4,6) , M = = 2(y-6) = x – 42y – 12 = x – 42y = x + 8X-intercept and y – intercept.X-intercept is the point where a line meets (cuts) the x-axis, at the value of y (ordinate) is equal to zero.That is to say the x-intercept is found by substituting y = 0 in the equation. Therefore for the equation y = mx + c.Y = 0, 0 = mx + cMx + c = 0Mx = -cx= -c/m.Therefore the coordinate of x-intercept is (-c/m, 0).y- Intercept is the point where the line and the y- axis meet. All this point the abscissa is normally equal to zero. The x- intercept is found by setting. x=0i.e y = mx + cx=0y = m(0) + cy = cThe intercept (0,c)The coordinate of the Y-intercept is (0,c)Example : – a line L is passing through the points A(5 – 2) and B(1,4).Findi. The equation of the line in the form of Y= mx +c and ax + by + c = oii. The x and y intercept.Solution: i. (x,y) = (1,4)= -3 + 3x = 8 – 2y2y = 11- 3xY =Y= then -3 + 3x = 8 – 2y-3 + 3x = 8 – 2y-3 + 3x – 8 + 2y = 03x – 11 + 2y = 03x + 2y – 11 = 0(ii) Y – interceptLet x = 03x + 2y – 11 = 0= y = Y – Intercept =0, The coordinate of Y-intercept is (0,11/2)∴ find the x-interceptLet y=0mx + c = y0 = + 11 x =The coordinate of X-intercept is (22/3,0) (iii) If ax + by = 12 goes through points (1,-2) and (4, 1) find the value a and b-solution + let the two collinear point be (x,y),(4,1) and gradient 1 then from x-4 = y-1 x-y=–1+4 x-y=3 if the equation is multiple by 4 both side we have 4(x-y)=4×3 4x-4y=12 compare the equations. x-y=3 and ax+by=12 ax+by=4(3) x-y=3 then a=1 b=-4 The value of a=1 and b=-4 EXERCISE 10.3 3. Find the equations of lines through points a) (2,1) with gradient 2. b) (0,5) with gradient -2 c) (1,-3) with gradient-3 d) (-2, -4) with gradient e) (0, 0 ) with gradient -3 f) (-3 , -3) and y- intercept g) ( 6, 2) and y intercept -2 h) (-1 , -1 ) and y – intercept – i) ( 1 , 2 ) and y- intercept = 2 j) (5 , 5) and y intercept 0 2. Find the equations of the following linesa. (i) y – intercept – 2 , gradient 1b. (ii) y – intercept 7, gradient (iii) y- intercept -16, gradient 4 d. (iv) y -intercept 2, gradient – 10 e. (v) y- intercept 0.4, gradient -0.7 (3) Rewrite the following equations in the form y = mx + c, and then determine the gradient and the y- intercept of each.a) (i) 7x + 4y = 11 (ii) 14x + 3y = 12 c) (iii) 2x = 5 + yd) (iv) 4x + 5y = 40 (v) 8x – ()y = 0 (vi) 6x = 5 – 2y (vii) + = 9Qn. 4. Find the area of the shaded region in the following figure. If the equation line AB is 5x + 6y-60 =0SIMULTANEOUS EQUATIONSAre equations of more than one variable which can be solved at the same time. There are two ways of solving simultaneous equations.1. (i) Elimination method (ii) Substitution method. (iii) Graphical Method The principle of solving equations is that the number of equations should be equal to the number of unknownsExample of simultaneous equation(a) = (b) = These are examples of simultaneous equations with two unknowns. 1. Elimination methodIs the method of omitting one variable and solve the remaining variables.How to eliminatei. Check if there are equal coefficients ii. If there are equal coefficients of same variables in the both equations subtract.iii. If there are equal and opposite to coefficients of same variable in both equations, add.2. If all coefficients are different modify the equationsExample 1. 2x + 3y = 63x + 2y = 4ModificationOmitting: xhere we can now omit x by subtracting. 13y=10Omitting y+ 13x=24=x=Let find the value of y by take one equation2x+ 3y=6 Example 2: Solution 3x=6x =2Let find the value of y by take one equation 6x+y=156 2 + y = 1512 + y = 15y = 15 – 12y = 3∴x =2 and y = 3 Example 03:2x + y = 103x – 2y = 1Solution. by Eliminate 7x= 21x=3Let find the value of y by taken one equation3x-2y=139-2y=1-2y=1-9= y= 4y=4 and x = 3 Solve the following simultaneous equations by elimination method1. x + y = 75x + 12y = 7SolutionModify-7y = -28 y= 4Let find the value of x by taken one equationx + y=7x + 4 =7x=7- 4 x=3Therefore: x=3 and y = 42.x + 8y =192x + 11y = 28 Solution== 5y=10= y= 2Let find the value of x by take one equationx + 8y = 19x + 8 x=19-16 x = 3Therefore: x=3 and y = 2 3. 8x + 5y = 93x + 2y = 4 Solution– = y=5 Let find the value of x by taken one equation8x + 5y = 98x + 5(5) =98x + 25 = 98x = 9 – 25 8x = -16= x = -2Therefore: x = -2 and y = 5 4. 2x– 3y = 715x + y = 9Solution17x = 34 = x = 2Let find the value of y by taken one equation2y – 3y = 72(2) – 3y =7-3y=7-4, -3y=34 – 3y = 7= y=-1 Therefore: x = 2, y = -1 5. 2x + 3y = 82x = 2 + 3ysolution 2x + 3y = 8 2x -3y = 2 6y=6 = y=1Let find the value of x by taken one equation2x + 3y = 82x + 3(1) = 8 2x = 8-3 x=5== x = Therefore: x= and y = 1 6.3x – 4y = 20x + 2y = 5Solution– 2x = 20x= 10Let find the value of y by taken the one equationx + 2y =510+ 2y = 52y = 5 – 10= Therefore: y = and x = 10 7. 6x = 7y + 77y – x = 8 Solution= – 5x= 15= x=3Let find the value of y by taken one equation6x – 7y =76 (3) -7y = 7 Therefore : 8. y = 4x – 716x – 5y = 25 solution -4x + y = -7 16x – 5y = 25 + -4x=-10x = 2.5 or 5/2 Let find the value of y by taken equation oney – 4x = -7y = – 4 (2.5)= -7y = -7 +10y= 3 Therefore x = 2.5 and y = 39. 2x + 7y = 393x + 5y = 31 solution– y= 5Let find the value of x by taken one equation3x +5y = 313x+ 5(5) = 313x=31- 25= x = 2Therefore: x = 2 and y = 510. 15x – 8y = 2917x + 12y = 75Solutionx =3Let find the value of y by taken one equation15x – 8y = 2915(3) – 8y = 29 Therefore: y = 2 and x = 32: SUBSTITUTIONExample: 01 solve6x + y = 153x + y = 9 by substitution method.solution6x + y = 15 ……… (i)3x + y = 9……….. (ii)from equation 1 6x+y =15 y=15-6x……………..(iii)Substitute equation (iii) into (ii)3x + y = 93x + 15 – 6x = 93x – 6x + 15 = 9-3x = 9 – 15-3x=-6= = x=2y = 15 – 6xy = 15 – 6 x 2y= 15 – 12 = 3x= 2 and y = 3 Example 022x + y = 10……………..(i)3x – 2y = 1………………(ii) From (i)2x + y = 102x – 2x + y = 10 – 2xy = 10 – 2x ……………..(iii)Put (iii)into (ii) 3x – 2 (10 – 2x) = 13x – 20 + 4x = 13x + 4x – 20 = 1= = ∴x = 3 y = 10 – 2xy = 10 – 2 (3)y = 10 – 6y = 4∴x = 3 and y = 4 Solve: 3x + 2y = 8………. (i)2x + 3y = 12…… (ii)solution from (i) 3x + 2y = 8 2y=8-3x= Y = ……. (iii) Put (iii) into (ii) 3x + 2y = 83(0) + 2y = 8 2y=8= ∴Y = 4X = 0 and y = 4Exercise 10 .4 1. Solve the simultaneous equations by using elimination method.(i). 2x + y = 5 4x – y = 7(ii) 3x + y = 65x + y = 8(iii) 5x – 2y = 16x + 2y = 8(iv) 8x+5y=40 9x + 5y = 5(v) 7x – 4y = 17 5x – 4y = 11(vi) 0.7x – 0.5y = 2.50.7x – 0.3y = 2.92. Solve the following simultaneous equations by using substitution method(i). 3x – 2y = 2x + y = 8(ii) 5x + y = 23 3x- 2y = 6 (iii) x – 3y = 2 4x + 2y = 36(iv) 7x – y = 14 8x – 2y = 16(v) 7x + y = 14 8x – 2y = 16 3. Solve the following by using any method (i) 3y – x = y + 2x = 6(ii) 8m – n = 38 m – 3n = -1(iii) 5x – 2y = 10-x + 3y = 24 1. 4.Solve the following simultaneous equations by substitution method(i) x – y = -3 2x – y = -5(ii) X – 2y = 6 X + 2y = 23. (iii) 3x – 4y = -112x + 3y = 1 4. (iv) 2x – 3y =32 3x – 4y = 30(v) 5a – 5b = 7 2a – 4b = 25. Solve the following system of simultaneous equations by elimination.6. (i) 10u + 3v – 4 = 06u + 2v – 2 =0(ii)x – y = 1 4x + 3y = (iii) 3x + 3y = 152x + 5y =14(iv) 7x – 3y = 15 5x– 2y = 19(v) x + y = 5 x – y = 16. Solve the following by any method Solving word problems leading to simultaneous equations.1. A fathers age is four times the age of his son. If the sum of their ages is 60 years. Find the age of the son and that of the father.Solution.Let x be the age of the sonLet y be the age of fathery = 4x……………….. (i) x+ y = 60 …………(ii)By substitution method x + y = 6, but y=4x then x + 4x = 60 5x=60 = x = 12By solve value of y you can take one equationy= 4x y = 4 (12) = 48The age of the father is 48 years and that of the son is 12 years 2. 2. The sum of two number is 12 and their difference is 2 find the numberSolutionLet the first number be = xLet the second number be= y=2x + 0 = 14 2x=14 = x = 7Let find the value of y7 + y = 12y = 12 – 7 y = 5The numbers are 5 and 7Example 03. If the numerator of a fraction is decreased by 1 its value become2/3 but if it denominator is increased by 5 its value becomes ½ , what the fraction? solution let the fraction be a/b= …………….(i) + 5 = …………..(ii)from (i) = 3(a-1) =2b 3a – 3 =2b 3a – 2b=3………………(iii) from (ii) Example 4The sum of the digits of a two digit number is 7. If the digits are reversed, the new number is increased by 3. Equal to 4 times the original numberFind the original numberLet the number be x and yx + y = 7……………… (i)The meaning of x, y = 10x + ySimilarly y, x = 10y + xy, x + 3 = 4 (x, y)10y+x+3=4(10x+y) 10y + x + 3 = 40x + 4y6y+3=39x = 13 x – 2y = 1……. (2) = = + 15x + 0 = 15 15x = 15= x = 1Let find the value of yx + y = 71+ y = 7 the number was x ,y is 1,6y = 6The original number is 16Exercise 10.51. The sum of two number is 109 and the difference of the same numbers 29. find the numbers 2. Two number are such that the first number plus three times the second number is 1. And the first minus three times the second is 1/7. Find the two numbers3. The sum of the number of boys and girls in a class is 36. If twice the number of girls exceeds the number of boys by 12, find the number of girls and that of boys in the class.4. Twice the length of a rectangle exceeds three times the width of the rectangle by one centimeter and if one – third of the difference of the length and the width is one centimeter find the dimensions of the rectangle.5. The cost of 4 pencils and five pens together is 6000 shillings while the cost of 6 pencils and 8 pens is 940 shillings, calculate the cost of one pencil and one pe6. Half of Paul’s money plus one – fifth of John’s money is 1400 shilling John’s money is 2650 shillings. How much has each? 7. A farmer buys 3 sheep and 4 goats for shs 290,Another buys sheep and goats from the some market for shs 170.What price did they pay for (a) 1 goat (b) 1 sheepSOLUTIONS. EXERCISE 10.11. (a) Their coordinates are A (2, 7) J (0, 4) F (3, 4) N (4, 1.5) M (6, 1) V (2.5, -2) D (6, -2) K (3, -5) C (-3, -5) I (-2, -2) P (-4.5, -2.5) H (-4, 4) G (-2, 6) L (-3, 1) E (0, 0) (b) F belongs to quadrant I H belongs to quadrant II V belongs to quadrant IV I belong to quadrant III2 (a) Square (b) Triangle (c) Trapezium(d) Parallelogram(e) Octagon EXERCISE 10.21. a) (0, 3), (2, 5)let (x1,y1) be (0,3) (x2,y2) be (2,5) M = 1 it is positive gradient.(b) (5, 8), (4, 1) Let (x1, y1) be (5, 8)(x2, y2) be (4, 1) m= 7 Slope is positive(c) (1, 5), (4, 7)Let (x1, y1) be (1, 5)(x2, y2) be (4, 7) The gradient is positive(d) (2,6), (5,3)Let (x1, y1) = (2,6)(x2, y2) = (5, 3) It has a negative gradient(e) (1,6), (3, -1)Let (x1, y1) be (1, 6)(x2, y2) be (3, -1)The gradient is negative(f) (3, 6), (-2, -1) Let (x1, y1) be (3, 6) (x2, y2) be (-2, -1)m == = The gradient is positive(g) (0,2), (6,2) Let (x1, y1) be (0, 2)(x2, y2) be (6, 2)M ==== 0M =0 The gradient is zero(h) (2,3), (-1,-3)Let (x1, y1) be (2, 3)(x2, y2) be (-1, -3) m = 2 The gradient is positive(i) (2,10), (2,0)Let (x1, y1) be (2, 10)(x2, y2) be (2, 0)M= M is undefined The gradient is undefined(j) (–), , 2) The gradient is positive(k) (-2,1), (4,3)let (x1,y1) be (-2,1) (x2,y2) be (4,3) The gradient is positive(i ) (-4,4), (-3,-3) let (x1,y1) be (-4,-4) (x2,y2) be (-3,-3)M m= -7 The gradient is negative(m) (0,0), (-3,4) let (x1,y1) be (0,0) (x2,y2) be (-3,4)M The gradient is negative(n) (99,6), (119,1) let (x1,y1) be (99,0) (x2,y2) be (119,1) M The gradient is negative(o) (0.64,-1.62), (1.36,-0.62) let (x1,y1) be (0.64,-1.62) (x2,y2) be (1.36,-0.62)M m=0 The gradient is zeroEXERCISE 10.31. a) (2,1) with gradient 2.(x, y), (2, 1) m= 2 2(2-x) = 1(1-y)4-2x = 1-y4 -2x =1-yy=1-4+2xy= -3+2x∴y = 2x – 3.b) (0,5) with gradient -2(x, y), ( 0, 5) m = -2 -2(0 – x) = 1(5-y)0+2x = 5 –yy= 5 -2x∴ y= -2x + 5c) (1,-3) with gradient -3(x, y) , (1, -3) m = – 3-3 – y = 3 – 3x-y = – 3 + 3x + 3 -y = 3x y= -3x d) (-2, -4) with gradient 3/2(x, y) (-2 , -4) m = 3/2 = = 3(-2 – x) = 2(-4 – y)-6 – 3x = -8 – 2y2y = 6 + 3x – 8= + e) (0, 0 ) with gradient -3(x,y) (0,0) m = -3= = -3(0 – x) = 1(0 – y)0 +3x = 0 – yy = 0 – 3xy = -3x f) (-3 , -3) and y- intercept 1/2 Solution (-3, -3)(x, y) y – intercept =1/2Y = mx + c(-3, -3),(x, y)substitute (-3,-3) to y = mx +c g) ( 6, 2) and y intercept -2(6, 2), (0, -2) Gradient = 2/3 Arbitrary(x,y), (6,2)3(2-y) = (6-x) 6-3y = 12 – 2x -3y = 12 – 6-2x -3y = 6- 2x h) (-1 , -1 ) and y – intercept – 1/3(-1, -1) and (0, -1/3) Arbitrary point = (x,y) (x,y), (-1,-1) 2(-1-x) = 3(-1-y) -2-2x=-3-3y 3y=-3+2+2x 3y=-1+2xi) (1,2) and y-intercept = 2(1,2), (0,2) 0 = 2-y y = 2j) (5,5) and y-intercept 0(5,5) (0,0) y = 5-5 + x y=x(2) (i) y – intercept -2 , gradient 1 (0, – 2) gradient 1Arbitrary (x, y)y – 0 = y + 2x = y+2y = – 2 + x(ii) y-intercept 7,gradient 3/4(0,7), (x,y) , gradient 3/4iii) y-intercept -16, gradient 4 (0,-16), (x,y) m = 44x-0 = y +16 -y =-4x +16 y = 4x -16 y = 4x – 16iv) y-intercept 2, gradient is -10 3 (i) 7x + 4y = 11 Alternatively7x +7y = 11 4y = -7x +11 ii) 14x + 3y = 12iii) 2x=5 +yx=0 2x = 5 + y, y=0-5, y =-5 y = 2x -5 y =2x-5 Gradient = 2 y- intercept = (0,2)iv) 4x +5y= 40 x=0 5y= -4x + 40 y = 24x y = 24 (0) y = 0 y = -8x X -3 = 24x y = 24x + 0Gradient = 24y-intercept = (0,0)vi) 6x =5-2yx=0 Gradient = -3 4. x-intercept , y=05x + 6(0) – 60 = 0 5x – 60 = 0 + 60 x= 12 X- intercept = (12,0)Y- intercept, x=0 5(0) + 6y – 60 = 0 5x – 60 = 60 y = 10 y-intercept = (0,10) Area = 60 square unitsEXERCISE 10.4 1. (i) 2x + y = 54x – y = 7 x = 2 2x + y =5 2 (2) + y = 5 4 + y = 5 y = 5-4 y = 1x =2 and y = 1(ii). 3x + y = 6 5x + y =6 x = 1 from 3x + y = 6 3 x 1 + y = 6 3 +y = 6 y = 6-3 y = 3x = 1 and y = 3(iii) 5x -2y = 16 x + 2 = 8x = 4 Let find the value of y by taken equation one 5x -2y = 16 5 x 4 – 2xy =16 20 – 2y = 16 -2y = 16 – 20 y = 2 x = 4 and y = 2(iv) 8x + 5y = 40 9x + 5y = 5 x+0 = 35 x = 35 let find the value of y by taken one equation8 x 35 + 5y =40 280 + 5y = 40 y = -48 x = 35 and y = -48(v). 7x -4y = 17 5x -4y = 11 Let find the value of y taken one equation 7x – 4y = 17 (vi) 0.7x – 0.5y = 2.5 0.7x – 03y = 2.9 0-0.2 = -0.5 0.2y = -10.4 0.2 = -10.4 y = 2Let find the value of x by taken one equation0.7x – 0.5y = 2.5 0.7x – 0.5(2) = 2.5 0.7x = 2.5 + 1 2. (i) 3x -2y = 5 2x +y = 83x – 2y = 5……………(i) 2x + y = 8…………….(ii) y=8 – 2x ……………..(iii)put eqn (iii) into (iv) 3x -2x (8 -2x) = 5 3x – 16 + 4x = 5 3x + 4x -16 =5 7x = 5 + 16 x=3 y= 8-2 x3 y= 8-6 y = 2 x =3 and y = 2(ii) 5x + y =23 3x – 2y = 65x – y = 23…………(i) 3x -2y = 6………….(ii) 5x-y +y = 23 + y. 5x =23 + y 5x – 23 =y…………….(iii)put eqn (iii) in (ii)3x -2 (5x – 23) = 63x -10x + 46 = 6 -7x+ 46 = 6 7x = 6 – 46 Let find the value of y by taken one equation(iii). x -3y=24x + 2y = 36x -3y=2 ……………(i) 4x + 2y = 36 …………(ii) x=2 + 3y ……………..(iii)put eqn (iii) in (ii)4 (2 + 3y) + 2y = 36 8 + 12y +2y = 36 8 + 14y = 36 14y = 36 -8 y = 2Let find the value of x by taken one equationx= 2 +3y x = 2+3(2) x = 2 +6 x = 8y = 2 and x = 8(iv) 7x – y = 14 8x – 2y = 167x – y =11……………(i) 8x – 2y = 16 …………(ii) 7x – 14 = y ………………(iii)put eqn (iii) into eqn (ii)8x – 2x (7x – 14) = 16 8x – 14 + 28 = 16 -6x + 28 = 16 x = 2Let find the value of y by taken one equationy = 7 x2 = 14 y = 14 -14 = 0 y = 0x = 2 and y =o(v) 7x + y = 14 8x – 2y =67x + y =14…………….(i) 8x – 2y = 6…………… (ii) y = 14 – 7x ……………(iii)put eqn (iii) in (ii)8x – 2(14 – 7x) =6 8x – 28 + 14x =6 22x – 28 = 6 x = -90 Let find the value of y by taken one equationx = -90 and y = 273. (i) 3y – x = 4 y + 2x = 6 x = 2 Let find the value of y by taken one equationy +2x = 6 y + 2 x2 = 6 y +4 =6 y = 6-4 y =2x =2 and y = 2(ii) 8m- n = 38 m – 3n = -1 n = 2Let find the value of m by taken one equation m -3 = -1 m-3 X 2 = -1 m-6 = -1 m = -1 + 6 m = 6-1 m = 5n = 2 and m = 5(iii) 5x – 2y = 10 -x + 3y = 24 Let find the value of y by taken one equation5 x 6 – 2y = 1030 – 2y = 10-2y = 10 -30 y = 10x = 6 and y = 104. (i) x -y = -3 2x – y = -5x-y = -3 ………………..(i) 2x – y = -5 …………….(ii) x=-3 + y ……………….(iii)put eqn (iii) into eqn (ii) 2(-3 + y) – y= -5 -6 + 2y – y = -5 y – 6 = -5 y = -5 + 6 y = 1x = -3 + y = -3 +1 x = -2x= -2 and y = 1(ii). x -2y = 6 x + 2y = 2x – 2y = 6 ………………..(i) x + 2y = 2 ……………….(ii) x =6 + 2y ………………..(iii)put eqn 3 into eqn 26 + 2y +2y = 2 6 + 4y = 2 4y = 2 -6 y = -1x = 6 + 2x -1 x = 6 + -2 x= 4x = 4 and y = -1(iii) 3x – 4y = -11 2x + 3y = 166x -4y = -11 ………………(i) 6x + 3y = 16………………(ii) 6x = -11 + 4y …………….(iii)put eqn (iii) into (ii) -11 + 4y + 3y = 16 -11 + 7y = 16 7y = 16 +11(iv) 2x – 3y = 32 3x – 4y = 306x – 9y =96…………..(i) 6x – 8y = 60…………..(ii) 6x = 96 + 9y ………….(iii)put eqn (iii) into eqn (ii)96 + 9y -8y =60 96 + y =60 y = 60 – 96 y= -366x = 96 +9x -36 6x = 96 – 324 x = -38 and y = -36(v)5a -5b = 7 2a – 4b = 210a-5b =7 ………….(i) 10a -4b = 2 ………….(ii) 10a = 7 + 5b ………….(iii)put eqn (iii) into eqn (ii)i7 +5b -4b= 2 7 + b =2 b =2-7 b = -5 10a = 7+5x -5 10a = 7 + -25 10a =7 – 25 5. (i) 10u + 3v – 4 = 0 6u + 2v – 2 = 010u + 3v = 4 6u + 2v = 2 let find the value of u by taken equation 10u + 3v =4 10u + 3 (-2) = 4 10u -6 = 4 10u = 4 +6 10u = 10 v = -2 and u = 1(ii) x – y = 1 let find the value of x by taken one equation(iii) 3x + 3y =15 2x + 5y = 14Let find the value of x by taken one equation3x + 3y =15 (iv) 7x -3y = 15 5x – 2y = 19 y = 58 Let find the value of x taken one equation 7x -3y =15 7x – 3 x 68 = 15 7x – 204 =15(v). x + y =5 x – y =1 y = 2Let find the value of x by taken one equationx + y = 5 x + 2 = 5 x= 5-2 x = 3 x = 3 and y = 22x + 4y = 48………………… (ii)Now you can take equation (i) and (ii) to solve the equation3x-4y =12 2x+ 4y = 48x = -9Let find the value of y by taken one equation3x – 4y =33(-9) -4y = 3-27 – 4y = 3-4y = 3 + 27 y = -7.3 and x = – 9EXERCISE 10.51. Let the numbers be x and y 2y = 80 y = 40 let find the value of x x + y = 109 x + 40 = 109 x = 109 – 40 x = 69 The two numbers are 69 and 402. let the first number be x let the second number be yx + 3y = 1 ……………(i) Let find the value of yx+3y =13. Let the girls number be x and the boys be y x = 16 let find the value of xx + y = 36 16 + y = 36 y = 20The girls are 16 and boys are 204. let the length be x and the width be y Let find the value of y2x -3y = 1 2 x 2 -3y = 1 4 – 3y = 1 y = 1 The length =2cm and width = 1cm5. Let the pencils be x and pen be y x = 50Let the value of x4x + 5y = 600 4 x 50 + 5y = 600 200 +5y = 600 5y =600 -200 y = 80The pencils are 50 and pens are 806. let paul,s money be x let john’s money be y y = 1500Let find the value of x x = 2200 Paul’s money = 2200 shs and John’s money = 1500 shs7. let the price of the sheep be a Let the price of goat be = b 2a = 100 a = 50 let find the value of b3a + 4b = 290 3 x 50 + 4b = 290 150 + 4b = 290 4b = 290 – 150 b = 35The price of sheep = a The price of goat = b They bought 1 goat at 35 shs and 1 sheep at 50shs ALL NOTES FOR ALL SUBJECTS QUICK LINKS:AGRICULTURE O LEVEL PURE MATHEMATICS A LEVELBAM NOTES A LEVELBASIC MATH O LEVELBIOLOGY O/A LEVELBOOK KEEPING O LEVELCHEMISTRY O/A LEVELCIVICS O LEVELCOMPUTER(ICT) O/A LEVELECONOMICS A LEVELENGLISH O/A LEVELCOMMERCE O/A LEVELACCOUNTING A LEVELGENERAL STUDIES NOTESGEOGRAPGY O/A LEVELHISTORY O/A LEVELKISWAHILI O/A LEVELPHYSICS O/A LEVELMOCK EXAMINATION PAPERSNECTA PAST PAPERS Basic Mathematics Study Notes Form 1 Basic Mathematics Study Notes Msomi Maktaba All Notes FORM 1MATHEMATICSPost navigationPrevious postNext postRelated PostsHow to Access And Login Parent Portal Pisd: A Step-By-Step Tutorial February 4, 2024As a parent, we understand how important it is to stay informed and involved in our children’s education. That’s why Parent Portal PISD (Plano Independent School District) is such a valuable tool for parents like us who want an easy way to access information about our kids’ grades, attendance, assignments,… Read More Form 1 Geography Study Notes GEOGRAPHY FORM 1 – MAP WORK November 11, 2018February 13, 2019ALL NOTES FOR ALL SUBJECTS QUICK LINKS: AGRICULTURE O LEVEL PURE MATHEMATICS A LEVEL BAM NOTES A LEVEL BASIC MATH O LEVEL BIOLOGY O/A LEVEL BOOK KEEPING O LEVEL CHEMISTRY O/A LEVEL CIVICS O LEVEL COMPUTER(ICT) O/A LEVEL ECONOMICS A LEVEL ENGLISH O/A LEVEL COMMERCE O/A LEVEL ACCOUNTING A LEVEL… Read More Economics Study Notes Form 6 Economics – ECONOMIC INTEGRATION AND COOPERATION November 13, 2018April 26, 2020ALL NOTES FOR ALL SUBJECTS QUICK LINKS: AGRICULTURE O LEVEL PURE MATHEMATICS A LEVEL BAM NOTES A LEVEL BASIC MATH O LEVEL BIOLOGY O/A LEVEL BOOK KEEPING O LEVEL CHEMISTRY O/A LEVEL CIVICS O LEVEL COMPUTER(ICT) O/A LEVEL ECONOMICS A LEVEL ENGLISH O/A LEVEL COMMERCE O/A LEVEL ACCOUNTING A LEVEL… Read More Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment *Name * Email * Website Save my name, email, and website in this browser for the next time I comment. Δ
How to Access And Login Parent Portal Pisd: A Step-By-Step Tutorial February 4, 2024As a parent, we understand how important it is to stay informed and involved in our children’s education. That’s why Parent Portal PISD (Plano Independent School District) is such a valuable tool for parents like us who want an easy way to access information about our kids’ grades, attendance, assignments,… Read More
Form 1 Geography Study Notes GEOGRAPHY FORM 1 – MAP WORK November 11, 2018February 13, 2019ALL NOTES FOR ALL SUBJECTS QUICK LINKS: AGRICULTURE O LEVEL PURE MATHEMATICS A LEVEL BAM NOTES A LEVEL BASIC MATH O LEVEL BIOLOGY O/A LEVEL BOOK KEEPING O LEVEL CHEMISTRY O/A LEVEL CIVICS O LEVEL COMPUTER(ICT) O/A LEVEL ECONOMICS A LEVEL ENGLISH O/A LEVEL COMMERCE O/A LEVEL ACCOUNTING A LEVEL… Read More
Economics Study Notes Form 6 Economics – ECONOMIC INTEGRATION AND COOPERATION November 13, 2018April 26, 2020ALL NOTES FOR ALL SUBJECTS QUICK LINKS: AGRICULTURE O LEVEL PURE MATHEMATICS A LEVEL BAM NOTES A LEVEL BASIC MATH O LEVEL BIOLOGY O/A LEVEL BOOK KEEPING O LEVEL CHEMISTRY O/A LEVEL CIVICS O LEVEL COMPUTER(ICT) O/A LEVEL ECONOMICS A LEVEL ENGLISH O/A LEVEL COMMERCE O/A LEVEL ACCOUNTING A LEVEL… Read More