Topic. Worked solutions 1 Progress checks. 1 Algebra: Topic 1 Revision of the basics. 1 (a) 3(2x + 5) = 6x + 15

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1 1 Algebra: Topic 1 Revision of the basics Topic Worked solutions 1 Progress checks 1 (a) 3(x + 5) = 6x + 15 (b) (x + 6) = x 1 (c) 8(x + 3x + 4) = 8x + 4x + 3 (d) 6(3x 7x + 9) = 18x 4x + 54 = 4x + 54 (e) 3(5x 9) = 15x + 7 (f) 4(x + x + ) = 4x 4x 8 (g) (3x + 5) = 3x 5 (h) (6x + 8) = 6x 8 (i) (x 4x + 8) = x + 4x 8 (j) 4(5 x) = 4x 0 (k) 7( 4x + x ) = 7x + 8x 14 (a) (3x 4) + (x 3) = 6x 8 + x 6 = 8x 14 (b) 5(4 x) (x 4) = 0 10x x + 4 = 4 11x (c) 1(x 8) 4(x 8) = 1x 96 4x + 3 = 8x 64 (d) 4(x 3y) + 6(x + y) = 8x 1y + 6x + 1y = 14x (e) x(x + 4) x(x + 6) = x + 8x x 6x = x + x (f) 4x(3x + 1) 3x(x 4) = 1x + 4x 3x + 1x = 9x + 16x (g) x (x 6) + x 3 + x = x 3 1x + x 3 + x = 3x 3 10x (h) 4x(x + y) y(x + y) = 4x + 4xy xy y = 4x + 3xy y 3 (a) (x + 5)(x + ) = x + x + 5x + 10 = x + 7x + 10 (b) (x 7)(x + 1) = x + x 7x 7 = x 6x 7 (c) (x + 4)(x 4) = x 4x + 4x 16 = x 16 (d) (3x + 1)(5x + 3) = 15x + 9x + 5x + 3 = 15x + 14x + 3 (e) (4x 1)(x 5) = 8x 0x x + 5 = 8x x + 5 (f) (5x 1)(5x + 1) = 5x + 5x 5x 1 = 5x 1 (g) (x 8)(x + 4) = x + 8x 8x 3 = x 3 1

2 1 Algebra: Topic 1 Revision of the basics (h) (3a + b)(4a + b) = 1a + 6ab + 4ab + b = 1a + 10ab + b (i) (5x y)(4x + y) = 0x + 5xy 4xy y = 0x + xy y (j) (6x 5y)(x 3y) = 6x 18xy 5xy + 15y = 6x 3xy + 15y 4 (a) (x + ) + (y + 1) = 0 x + 4x y + y + 1 = 0 x + y + 4x + y + 5 = 0 (b) (x + 5) + (y + 3) = 0 x + 10x y + 6y + 9 = 0 x + y + 10x + 6y + 34 = 0 (c) (x 7) + (y + ) = 0 x 14x y + 4y + 4 = 0 x + y 14x + 4y + 53 = 0 (d) (x 4) + (y + 6) = 0 x 8x y + 1y + 36 = 0 x + y 8x + 1y + 5 = 0 (e) (x 6) + (y + 7) = 1 x 1x y + 14y + 49 = 1 x + y 1x + 14y + 73 = 0 (f) (x + ) + (y 5) + 7 = 0 x + 4x y 10y = 0 x + y + 4x 10y + 36 = 0 (g) (x 1) + (y 9) 10 = 0 x x y 18y = 0 x + y x 18y + 7 = 0 (h) (x + 3) + (y 8) 7 = 0 x + 6x y 16y = 0 x + y + 6x 16y + 46 = 0 (i) (x + 1) + (y 1) = 18 x + x y y = 0 x + y + x y 16 = 0

3 Worked solutions 5 (a) (b) (c) (d) 4x y x y = 4x 1x y 3 4x y = 3xy 16a 3 b c 4a b = 3 abc 4x y 4 8x y = 3x 5 y Divide top and bottom by xy. Divide top and bottom by 4xy. Divide top and bottom by 8a b. Divide top and bottom by 8xy 4. 6 (a) (b) (c) (d) (e) (f) (g) 15abc = 3ab 5c 4x 3 4x y = x y 6pq 3 r 3pqr = q 10x 4 y 3 x y = 5x y 45a 3 bc 9ab c = 5a c b (x + 7)(x + 5) (x + 7) = (x + 5)(x + 3) (x + 3) (x + 3)(x + 5) (x + 5)(x + 3) = 1 Divide top and bottom by 5c. Divide top and bottom by 4x. Divide top and bottom by 3pqr. Divide top and bottom by x y. Divide top and bottom by 9abc. Divide top and bottom by (x + 5). Divide top and bottom by (x + 3)(x + 5). 7 (a) 1x y + 8xy = 4xy(3x + y) (b) 4a b + ab = ab(a + 1) (c) 4x y + 6x = 6x(4xy + 1) (d) 5a 3 b c 5 + 5a b 3 = 5a b (5ac 5 + b) 8 (a) (b) (c) (d) (e) (f) x (x 1) x(x 1) = x x y 3 x y = y 15x 3 y 3 5x 3 y = 3y 5(x 4) 10(x ) = x 4 (x ) (x + 1)(x ) (x 5)(x + 1) = x x 5 x 3 (x 3)(x 1) = 1 (x 1) Divide top and bottom by x(x 1). Divide top and bottom by xy. Divide top and bottom by 5x 3 y. Divide top and bottom by 5. Divide top and bottom by (x + 1). Divide top and bottom by (x 3). 3

4 1 Algebra: Topic 1 Revision of the basics Add 7 to both sides of the equation. Adding 7 will eliminate the 7 on the lefthand side. 9 x 7 = 7 x = x + 7 = 3 x = 10 Subtract 7 from both sides of the equation. Subtracting 7 will eliminate the +7 on the left-hand side of the equation. Remove the denominator by multiplying both sides by x 5 = 1 4x = 60 Remove the multiplier of x by dividing both sides by 4. x = 15 Add 1 to both sides. 1 x 5 1 = 7 x 5 = 8 x = 40 Multiply both sides by 5. Add 1 to both sides. 13 x 3 1 = 5 x 3 = 6 Multiply both sides by 3. x = 18 Divide both sides by. x = 9 Multiply out the bracket. Divide both sides by. 14 (x + 1) = 18 4x + = 18 4x = 16 x = 4 Subtract from both sides. Multiply both sides by to remove the in the denominator x = x = Add 6 to both sides. x = 4 16 (a) x = 4 + x = 4 + x = x 1 = x x = 1 4

5 Worked solutions (b) 4(x 7) = 3(x 10) 4x 8 = 6x 30 8 = x 30 = x x = 1 (c) 5(6x 3) = 6(x 1) 30x 15 = 1x 6 18x 15 = 6 18x = 9 TAKE NOTE Be careful here. Students often see the numbers 18 and 9 and automatically think that x is which is incorrect. x = 1 (d) 1 (x 1) = x x 1 = 6x = 5x = 5x x = 13 5 x =.6 17 (a) x 5 4 = 4x x 5 = 16x 5 = 15x x = 5 15 x = 1 3 (b) 6x 1 = 3(x 4) + 7 6x 1 = 3x x 1 = 5 3x = 4 x = 4 3 5

6 1 Algebra: Topic 1 Revision of the basics Multiply both sides by 8 because the denominators 4 and divide exactly into (a) x 4 + x = 15 x + 4x = 10 6x = 10 Divide both sides by 6. Multiply both sides by 1 because the denominators 3 and 4 divide exactly into 1. Divide both sides by 7. (b) x = 0 x 3 + x 4 = 49 4x + 3x = 588 7x = 588 x = 84 Collect the terms in x. Multiply both sides by 9 because the denominators 9 and 3 divide exactly into 9. (c) x 9 + x 3 = 4 x + 6x = 378 Collect the terms in x together. 7x = 378 Divide both sides by 7. x = 54 and 3 are factors of 6 so we multiply both sides by (x 3) x 3 + x = 3 + 6(x + 1) 3 = 18 3(x 3) + (x + 1) = 18 3x 9 + x + = 18 5x 7 = 18 5x = 5 x = 5 6

7 Worked solutions 0 1 x x 5 = 3 10x 10x 5 = 30 5x x = 30 3x = 30 x = (x 1) = 1 (x ) 3 3(x 1) = 4(x ) 3x 3 = 4x 8 Multiply both sides by 10. This step is usually not written down. Divide both sides by 3. Multiply both sides by 1. Multiply out the brackets. Subtract 3x from both sides. 3 = x 8 Add 8 to both sides. x = 5 x 3 x 4 = 5 4x 3 1x 4 = 60 8x 3x = 60 5x = 60 x = 1 Multiply both sides by 1. This step is usually not written down. Divide both sides by 5. 3 (a) A = πr A π = r r = A π (b) A = 4πr A 4π = r r = A 4π 7

8 1 Algebra: Topic 1 Revision of the basics (c) V = 4 3 πr3 3V = 4πr 3 3V 4π = r3 r = 3 3V 4π Subtract at from both sides. 4 (a) v = u + at (u) u = v at Subtract u from both sides. (b) v = u + at (a) v u = at Divide both sides by t. a = v u t Divide both sides by a. (c) v = as (s) s = v a Subtract u from both sides. (d) v = u + as (a) v u = as Divide both sides by s. a = v u s Subtract as from both sides. (e) v = u + as (u) v as = u Square root both sides. u = v as Subtract ut from both sides. (f) s = ut + 1 at (a) s ut = 1 at Multiply both sides by. Divide both sides by t. (s ut) = at a = (s ut) t Subtract mx from both sides. (g) y = mx + c (c) y mx = c c = y mx Always put the subject of the equation on the left-hand side of the equation. 8

9 Worked solutions (h) y = mx + c (m) y c = mx m = y c x (i) s = 1 (u + v)t (t) s = (u + v)t Subtract c from both sides. Divide both sides by x. Multiply both sides by. Divide both sides by (u + v). t = s u + v (j) E = 1 mv (v) Multiply both sides by. E = mv E m = v Divide both sides by m. Square root both sides. v = E m (k) V = πr l (l) Divide both sides by πr. l = V πr (l) V = πr l (r) V πl = r Divide both sides by πl. Square root both sides. r = V πl 9

10 1 Algebra: Topic 1 Revision of the basics 5 (a) x + y = 5... (1) 5x + y = () Multiplying equation (1) by we obtain x + y = 10 Subtracting the above equation from equation () we obtain At GCSE level the simultaneous equations you solved usually resulted in whole number (i.e. integer) answers. At this level you can often get fractions so do not automatically assume you have done something wrong if you get fractions. 5x + y = 11 Subtracting x + y = 10 3x = 1 x = 1 3 Substituting x = 1 into equation (1) we have y = 5 Hence, y = 4 3 >>> TIP Always specifically say what your answers are. The examiner should not have to wade through your working to find what your answers are. Substituting x = 1 3 and y = 4 3 into LHS of equation () we obtain 5( 1 3) + (4 3) = = = RHS Hence the solutions are x = 1 3 and y = 4 3 Both sides of the equation are equal so the values of x and y satisfy the second equation. (b) x 3y = 5 (1) >>> TIP Notice that the terms in y have opposite sign. It is easier to make these terms the same in value but opposite in sign so that the two equations can be added together in order to eliminate the term in y. It is easier to add the equations so this is why we have chosen to eliminate y rather than x. 5x + y = 16 () Multiplying equation (1) by and equation () by 3 we obtain 4x 6y = 10 15x + 6y = 48 Adding 19x = 38 x = Substituting x = into equation (1) we have () 3y = 5 4 3y = 5 10

11 Worked solutions 3y = 9 y = 3 Substituting x = and y = 3 into LHS of equation () we obtain 5() + (3) = = RHS Both sides of the equation are equal so the values of x and y satisfy the second equation. Hence the solutions are x = and y = 3 6 (a) 3x 5 = x 1 x 5 = 1 The y-values are equated and the resulting equation solved. x = 4 x = Substituting x = into the equation y = 3x 5 we obtain y = 3() 5 = 1 Checking by substituting x = and y = 1 into y = x 1 we obtain 1 = 1 1 = 1 Both sides of the equation are equal, showing that the values of x and y satisfy the second equation. Hence solutions are x = and y = 1. (b) From the equation x + 3y = 8 we have 3y = x + 8 Notice that 3y appears in both equations so it is best to substitute the value of 3y into the second equation in order to eliminate y. 5x + 3y = 11 The 3y is replaced by x + 8. Hence, 5x + ( x + 8) = 11 3x + 8 = 11 3x = 3 x = 1 Substituting x = 1 into the equation x + 3y = 8 we obtain (1) + 3y = 8 3y = 6 y = 11

12 1 Algebra: Topic 1 Revision of the basics Checking by substituting x = 1 and y = into 5x + 3y = 11 we obtain 5(1) + 3() = 11 Both sides of the equation are equal so the values of x and y satisfy the second equation. 11 = 11 Hence solutions are x = 1 and y = = = = = (3 3 )( 3 + ) ( 3 )( 3 + ) = = = ( 6 ) = ( 3 ) = = 8 3 Subtract 11 from both sides. Add 5 to both sides. Divide both sides by 15. Test yourself 1 (a) x + 11 = 5 x = 14 x = 7 (b) 3x 5 = 10 3x = 15 x = 5 (c) 15x = 60 x = 4 Divide both sides by 7. Divide both sides by 3. (d) x 4 = 8 x = 3 Multiply both sides by 4. Multiply both sides by 5. (e) 4x 5 = 0 4x = 100 x = 5 Divide both sides by 4. 1

13 Worked solutions (f) x 3 = 6 x = 18 x = 9 Multiply both sides by 3. (g) 5 x = 7 5 = 7 + x Add x to both sides to make x positive. Subtract 7 from both sides. = x x = (h) x 7 9 = 3 x 7 = 1 x = 84 Add 9 to both sides. Multiply both sides by 7. (a) 35x 3 y = 5x 7xy (b) 15ab3 c 3ab = 5b c Divide top and bottom by 3ab. (x 4)(x 7) (x 7) (c) = (x 1)(x 4) (x 1) (x + 3) (d) (x + 3) = (x 6)(x + 3) (x 6) 3 (a) 4(x 3) + 5(x + 1) = 8x x + 5 = 18x 7 (b) (x + 4) = x 8 (c) (x 5) = x + 5 (d) 4(x 6) (5x 4) = 8x 4 5x + 4 = 3x 0 (e) 3(5x 9) 4(x 6) = 15x 7 8x + 4 = 7x 3 (f) 4(x 7) + 5x 9 = 8x 8 + 5x 9 = 13x 37 (g) (3x + 4x ) = 3x 4x + (h) x(x 4x + 8) = x 3 4x + 8x Divide top and bottom by 7xy. Divide top and bottom by (x 4). Divide top and bottom by (x + 3). (i) 3a(a + b) + b(a + b) = 3a + 3ab + ab + b = 3a + 5ab + b (j) 4a(a b) 3b(a b) = 8a 4ab 6ab + 3b = 8a 10ab + 3b (k) 5x (x 3) + 3x(x + 4) = 5x 3 15x + 3x + 1x = 5x 3 1x + 1x (l) x(x 1) (x 3x) = x x x + 3x = x + x 13

14 1 Algebra: Topic 1 Revision of the basics 4 (a) (x 5)(x + 3) = x + 3x 5x 15 = x x 15 (b) (4x 1)(x 5) = 4x 0x x + 5 = 4x 1x + 5 (c) (x 7)(3x + 5) = 6x + 10x 1x 35 = 6x 11x 35 (d) (9x 1)(9x + 1) = 81x + 9x 9x 1 = 81x 1 (e) (4a b)(a + 4b) = 8a + 16ab ab 4b = 8a + 14ab 4b (f) (5y 1)(y + 5) = 10y + 5y y 5 = 10y + 3y 5 5 (a) V = nrt p (b) n = pv RT (c) T = pv nr (d) p = nrt V 6 f = E h 7 m = y c x 8 (a) λ = c f (b) V = n c (c) T = Q mc (d) V = 1000n c (e) c = 1000n V (f) h = E f 9 (a) = = = 5 3 (b) = ( + 5 )(3 5) (3 + 5)(3 5) = =

15 Worked solutions 10 (a) 5 = 5 = 5 1 (b) = = = (c) = = = 7 (d) ( 5 ) = 4 5 = x + 4y =... (1) 3x y = 9... () Multiplying equation () by 4 gives 1x 4y = 36 7x + 4y = Adding these two equations to eliminate y, we obtain 19x = 38 x = Substituting x = into equation (1) we obtain 7x + 4y = 7() + 4y = y = 4y = 1 y = 3 Checking by substituting x = and y = 3 into equation () we obtain 3x y = 9 3() ( 3) = = 9 9 = 9 Both sides of the equation are equal showing that the values of x and y satisfy the second equation. Hence the solutions are x = and y = 3. 15

16 1 Algebra: Topic Manipulation of algebraic expressions Topic Worked solutions Progress check 1 (a) (x + 1) = x + x + 1 (b) (x + 11) = x + x + 11 (c) (x + 13) = x + 6x (d) (x 6) = x 1x + 36 (e) (x 11) = x x + 11 (f) (x + 7) = x + 14x + 49 (a) (x + 3) = x + 6x + 9 (b) (x + 4) = x + 8x + 16 (c) (x + 1) = x + x + 1 (d) (x + 6) = x + 1x + 36 (e) (x + 8) = x + 16x + 64 (f) (x + 5) = x + 10x + 5 (g) (x 4) = x 8x + 16 (h) (x 5) = x 10x + 5 (i) (x 9) = x 18x + 81 (j) (x 7) = x 14x + 49 (k) (x 10) = x 0x (l) (x + 1) = x + 4x (a) x + 4x + 8 = (x + ) = (x + ) + 4 (b) x + x + 6 = (x + 1) = (x + 1) + 5 (c) x 6x + 4 = (x 3) = (x 3) 5 (d) x x 10 = (x 1) 1 10 = (x 1) 11 (e) x 10x = (x 5) 5 = (x 5) 7 (f) x 8x + 4 = (x 4) = (x 4) 1 (g) x 6x + 1 = (x 3) = (x 3)

17 Worked solutions 4 Factorise each of the following expressions. (a) x + 3x + = (x + 1)(x + ) (b) x + 6x + 8 = (x + 4)(x + ) (c) x + 10x + 1 = (x + 3)(x + 7) (d) x + 3x 4 = (x + 4)(x 1) (e) x x 3 = (x 3)(x + 1) (f) x 3x + = (x 1)(x ) (g) x 4x 5 = (x 5)(x + 1) (h) x + 5x 14 = (x + 7)(x ) (i) x 5x + 4 = (x 1)(x 4) (j) x + 3x 10 = (x + 5)(x ) 5 Factorise each of the following expressions. (a) x x 3 = (x 3)(x + 1) (b) x + 9x + 4 = (x + 1)(x + 4) (c) 3x + 4x + 1 = (3x + 1)(x + 1) (d) 5x + 19x 4 = (5x 1)(x + 4) (e) 5x 7x + = (5x )(x 1) (f) 4x 3x 1 = (4x + 1)(x 1) (g) 3x + 8x + 5 = (3x + 5)(x + 1) (h) x + 3x 14 = (x + 7)(x ) (i) 4x 1x + 0 = (4x 5)(x 4) (j) x 3x 10 = (x 5)(x + ) 6 (a) x + 3x + = (x + 1)(x + ) (b) x + 6x + 5 = (x + 5)(x + 1) (c) x + 11x + 4 = (x + 3)(x + 8) (d) x + 10x + 9 = (x + 1)(x + 9) (e) x + 8x + 15 = (x + 5)(x + 3) (f) x x + 1 = (x 1)(x 1) (g) x + 5x 6 = (x + 6)(x 1) (h) x + 4x 1 = (x + 7)(x 3) (i) x 5x + 6 = (x 3)(x ) (j) x + 7x 30 = (x + 10)(x 3) (k) x + x 15 = (x 3)(x + 5) 7 (a) x + x 1 = (x 1)(x + 1) (b) x + 13x + 6 = (x + 1)(x + 6) (c) 4x 3x 1 = (4x + 1)(x 1) (d) 3x + 19x 14 = (3x )(x + 7) (e) 5x + 18x 8 = (5x )(x + 4) (f) 8x + 30x 7 = (4x 3)(x + 9) (g) 1x + 8x 5 = (6x 1)(x + 5) (h) 1x 7x + 1 = (4x 1)(3x 1) 8 (a) x 1 = (x + 1)(x 1) (b) 4x 5 = (x + 5)(x 5) (c) 4c b = (c + b)(c b) (d) 16x 49 = (4x + 7)(4x 7) (e) p q = (p + q)(p q) (f) 5x y = (5x + y)(5x y) (g) x y = (x + y)(x y) (h) y 100 = (y + 10)(y 10) (i) 4a 1 = (a + 1)(a 1) (k) c 5 = (c + 5)(c 5) 9 (a) x = 1 or (b) x = 4 or (c) x = 3 or 7 (d) x = 4 or 1 (e) x = 3 or 1 (f) x = 1 or (g) x = 5 or 1 (h) x = 7 or (i) x = 1 or 4 (j) x = 5 or 17

18 1 Algebra: Topic Manipulation of algebraic expressions 10 (a) x 10x + 1 = (x 7)(x 3) = 0 so x = 7 or 3 (b) a a 4 = (a + 6)(a 7) = 0 so a = 6 or 7 (c) 3x + 11x 4 = (3x 1)(x + 4) = 0 so x = 1 or (a) x + 4x + 1 = 0 (x + ) = 0 (x + ) = 3 x + = ± 3 x = 3 or 3 x = 0.7 or 3.73 ( d.p.) (b) x + 4x 5 = 0 (x + x 5 ) = 0 Dividing both sides by we obtain. x + x 5 = 0 (x + 1) 1 5 = 0 (x + 1) = 3.5 x + 1 = ± 3.5 x = or x = 0.87 or.87 ( d.p.) 1 x + 8x 1 = 0 Completing the square we obtain (x + 4) 16 1 = 0 (x + 4) 8 = 0 (x + 4) = 8 x + 4 = ± 8 x = ± 8 4 x = 8 4 or 8 4 x = 1.9 or 9.9 ( d.p.) 18

19 Worked solutions 13 (a) Comparing the equation 3x 4x + 6 = 0, with ax + bx + c = 0 gives a = 3, b = 4 and c = 6. b 4ac = ( 4) 4(3)(6) = 16 7 = 56 As b 4ac < 0, there are no real roots. (b) Comparing the equation 3x + 6x + = 0, with ax + bx + c = 0 gives a = 3, b = 6 and c =. >>> TIP Be careful here. It is easy to substitute the values for part (a) into this equation. Substituting these values into the quadratic equation formula gives: x = 6 ± (6) 4(3)() (3) = 6 ± = 6 ± 1 6 = or = 0.4 or 1.58 ( d.p.). 19

20 1 Algebra: Topic Manipulation of algebraic expressions One way to do this is to consider the graph of the function. The minimum point would occur at (, 8). The least value is the y-value (i.e. 8) and the value of x for which this occurs is the x-coordinate(i.e. ). Another way is to look at the function and spot that the bracket squared will only be zero or positive no matter what the value of x is. Whatever the value of the bracket squared is, it will be added to the 8. The smallest value of the whole function would be if the bracket squared was zero and this would occur if x =. In this case nothing would be added to the 8 so the least value of x + 4x + 1 would be 8. Test yourself 1 x 3x = (x 3 ) = (x 3 ) (a) x 8x + 1 = (x 4) = (x 4) 4 Hence a = 4 and b = 4. (b) x 8x +1 = 0 (x 4) 4 = 0 (x 4) = 4 Taking the square root of both sides gives (x 4) = ± Hence, x 4 = or x 4 = x = + 4 or x = + 4 x = 6 or x = 3 (a) x + 4x + 1 = (x + ) = (x + ) + 8 Hence a = and b = 8. (b) The least value of (x + ) + 8 is 8 and this occurs when x =. 4 (i) x + 8x 9 = (x + 4) 16 9 = (x + 4) 5 Hence a = 4 and b = 5. (ii) The minimum point on the curve y = x + 8x 9 will be at ( 4, 5) Hence the least value is 5 and this occurs when x = 4. 5 (a) x + x + 1 = (x + 1)(x + 1) (b) x + 5x + 6 = (x + 3)(x + ) (c) x + 3x + 1 = (x + 1)(x + 1) (d) 3x + 10x + 3 = (3x + 1)(x + 3) (e) x x = (x )(x + 1) (f) x + 3x 4 = (x + 4)(x 1) (g) x + x 1 = (x 3)(x + 4) (h) x 6x + 5 = (x 1)(x 5) (i) x x 35 = (x 7)(x + 5) 0

21 Worked solutions 6 (a) 3x + 5x + = (3x + )(x + 1) (b) 4x + 5x + 1 = (4x + 1)(x + 1) (c) 5x + 1x + 4 = (5x + 1)(x + 4) (d) 0x + 17x + 3 = (4x + 1)(5x + 3) (e) 3x + 11x 4 = (3x 1)(x + 4) (f) 5x 34x 7 = (5x + 1)(x 7) (g) 7x 31x + 1 = (x 4)(7x 3) (h) 6x 5x + 1 = (x 1)(3x 1) (i) 4x + 19x 30 = (4x 5)(x + 6) (j) 8x 49x + 6 = (8x 1)(x 6) (k) 1x 31x + 7 = (3x 7)(4x 1) (l) 9x + 89x 10 = (9x 1)(x + 10) Remember that the coefficient of x needs to be 1 before you complete the square. Here it is necessary to take a 5 out of the square bracket as a factor. 7 5x 0x + 10 = 5(x 4x + ) = 5[(x ) 4 + ] = 5(x ) 10 Giving a = 5, b = and c = y = x + 4 and y = x 7x + 0 x + 4 = x 7x + 0 If the line and the curve touch then the resulting equation will have a repeated root. x 8x + 16 = 0 (x 4)(x 4) = 0 (x 4) = 0 There is only one solution to the quadratic which means the straight line and curve touch at only one point. There is just one solution to this equation which proves that the straight line and curve touch. Solving gives x = 4 When x = 4, y = = 8 The x-coordinate is substituted in the equation of the line to find the corresponding y-coordinate. Hence the coordinates of the point of contact are (4, 8). 1

22 1 Algebra: Topic 3 The remainder and factor theorems and solving cubic equations Topic Worked solutions 3 Progress checks 1 f(x) = x 3 7x 6 f(3) = (3) 3 7(3) 6 = = 0 As f(3) = 0 then (x 3) is a factor of the function. g(x) = x 3 7x + 3x + 1 g(1) = (1) 3 7(1) + 3(1) + 1 = = 1 Hence, remainder = 1. 3 f(x) = x 3 4x + x + 8 f( ) = ( ) 3 4( ) + ( ) + 8 = = 18 Hence, remainder = f(x) = x 3 x + 6 f() = () 3 () + 6 = = 6 Hence, remainder = 6. 5 Let f(x) = x 3 + 4x + x 6 If (x 1) is a factor of x 3 + 4x + x 6 then f(1) = 0. Now, f(1) = (1) 3 + 4(1) + (1) 6 = = 0 As f(1) = 0, (x 1) is a factor of x 3 + 4x + x 6.

23 Worked solutions 6 (a) (x + )(x + x +1) = x 3 + x + x + x + x + = x 3 + 3x + 3x + (b) (x 4)(x 3x +1) = x 3 3x + x 4x + 1x 4 = x 3 7x + 13x 4 (c) (x 1)(x + x +1) = x 3 + x + x x x 1= x 3 + x + x 1 (d) (x + 1)(x + 4)(x + 5) = (x + 1)(x + 9x + 0) = x 3 + 9x + 0x + x + 9x + 0 = x x + 9x + 0 (e) (x 1)(x + )(x + 1) = (x 1)(x + 3x + ) = x 3 + 3x + x x 3x = x 3 + x x (f) (x + 3) (x ) = (x + 3)(x + 3)(x ) = (x + 3)(x + x 6) = x 3 + x 6x + 3x + 3x 18 = x 3 + 4x 3x 18 (g) (x 1) (x 3) = (x 1)(x 1)(x 3) = (x 1)(x 4x + 3) = x 3 4x + 3x x + 4x 3 = x 3 5x + 7x 3 7 f() = () 3 4() + () + 1 = = 5 Hence, remainder = 5. 8 Let f(x) = x 3 x + ax + 6 f( 1) = ( 1) 3 ( 1) + a( 1) + 6 = 1 a + 6 = a + 3 Now as (x 1) is a factor, f( 1) = 0 Hence a + 3 = 0 Solving, we have a = 3. 9 (i) f() = () 3 5() + 7() = = 0 (ii) x is a factor of the function. 3

24 1 Algebra: Topic 3 The remainder and factor theorems and solving cubic equations 4 Test yourself 1 Let f(x) = 4x 3 + 3x 3x + 1 f( 1) = 4( 1) 3 + 3( 1) 3( 1) + 1 = 3 Remainder = 3 (a) Let f(x) = x 3 + 6x + ax + 6. f( ) = ( ) 3 + 6( ) + a( ) + 6 = a As x + is a factor, f( ) = 0 Hence, a = 0 So a = 11 (b) x 3 + 6x + 11x + 6 = (x + )(ax + bx + c) Equating coefficients of x 3 gives a = 1 Equating coefficients of x gives b + a = 6 and since a = 1 this gives b = 4 Equating constant terms gives c = 6 so c = 3. x 3 + 6x + 11x + 6 = (x + )(x + 4x + 3) Now (x + )(x + 4x + 3) = 0 So (x + )(x + 1)(x + 3) = 0 Solving gives x =, 1 or 3. 3 (a) (i) f( ) = ( ) 3 ( ) 4( ) + 4 = 0 (ii) As there is no remainder, (x + ) is a factor of x 4x + 4. (b) x 3 x 4x + 4 = (x + )(ax + bx + c) Equating coefficients of x 3 gives a = 1. Equating coefficients of x gives 1 = b + a so 1 = b + Hence b = 3. Equating constant terms gives 4 = c so c =. Substituting these values in for a, b and c gives x 3 x 4x + 4 = (x + )(x 3x + ) = (x + )(x )(x 1) Now f(x) = 0 so (x + )(x )(x 1) = 0 Solving gives x =, or 1

25 1 Algebra: Topic 4 Problem solving and inequalities Topic Worked solutions 4 Progress check From the question her mother is three times older. Both will have aged 1 years. 1 Let Amy s age = x years so her mother s age = 3x years. In 1 years time Amy will be x + 1 years and her mother will be 3x + 1 years. Then Amy will be half the age of her mother so x + 1 = 3x + 1 Multiplying both sides by we obtain x + 4 = 3x + 1 Solving gives x = 1 years. Amy s age is 1 years and her mother s age is 36. The difference of the two numbers is 1 (with x being the larger number) so we can write this as x y = 1. The product of the numbers is 7 and this can be written as xy = 7 From the first equation we can write x = ( y + 1) and substituting this into the second equation for x we obtain y( y + 1) = 7 y + y 7 = 0 Factorising, we obtain ( y + 9)( y 8) = 0. Solving, we obtain y = 9 or 8. As the questions says that x and y are both positive, y = 8. Now xy = 7 so 8x = 7 and hence x = 9. Hence x = 9 and y = 8. 5

26 3 Let the integers be x, x + 1, x +. The square of the largest integer minus the square of the smallest integer is equal to 64 can be written as (x + ) x = 64 So we have x + 4x + 4 x = 64 4x + 4 = 64 4x = 60 x = 15 Hence the integers are 15, 16, 17 4 (a) x 5 (b) x 10 (c) x > 1 (d) x > 4 (e) x 50 5 (a) 1 x 8 (b) x 5 (c) 1 < x 8 (d) 4 < x < 4 (e) 4 x < 10 6 (a) 3, 4, 5, 6, 7, 8, 9 10 (b) 4, 3,, 1, 0 (c) 1,, 3, 4, 5, 6, 7 (d) 16, 17, 18 (e), 1, 0, 1, (f) 16, 17, 18, 19 (g) 4, 3,, 1, 0, 1,, 3, 4 (h) 3, 4, 5, 6, 7, 8, 9 (i) 0, 1,, 3, 4, 5 (j) 1, 13, 14, 15, 16, 17, 18 7 (a) x 3 and x 8 (b) 3 < x < 7 (c) 4 < x < 14 (d) 4 x < 5 (e) x < 5 and x 8 (f) x 1 and x > 8 14 x 18 9 (a) 4x > 3 x (b) 5x > 3 5x > 5 x > 1 (x + 1) > 8 x x + > 8 x 3x > 6 x > (c) (5x 3) 4(x 3) 10x 6 4x 1 6x 6 x 1 (d) 4 x < 3x + 7 (e) 4 4x < 7 4x < 3 x > x 4 x 9 4x 4 4x 5 x 5 4 (f) 5 x < 3(x ) 5 x < 3x 6 5 4x < 6 4x < 11 x >

27 Worked solutions 10 x 6x + 8 > 0 As the curve has a positive coefficient, it will be -shaped, cutting the x-axis at x = 4 and x =. Sketching the curve for y = x 6x + 8 gives the following: y y = x 6x x We want the part of the graph which is above the x-axis. The range of values for which this occurs are x < and x > 4. >>> TIP If you do not cancel fractions you may lose marks x < x + 7 3x < x + 6 4x < 6 x > 6 4 x 3 The inequality sign is reversed because both sides have been divided by a negative quantity (i.e. 4). 7

28 1 Algebra: Topic 4 Problem solving and inequalities 1 5x + 7x 6 0 Considering the case where 5x + 7x 6 = 0 Factorising gives (5x 3)(x + ) = 0 Giving x = 3 5 or (these are the intercepts on the x-axis) As the curve y = 5x + 7x 6 has a positive coefficient of x the curve will be -shaped. Sketching the curve for y = 5x + 7x 6 gives the following: y y = 5x + 7x 6 O 3 5 x We want the part of the graph which is below or on the x-axis. Meaning that x lies between and 3 inclusive, which can be written 5 mathematically as x

29 Worked solutions If the line and the curve touch then the resulting equation will have a repeated root. 13 y = x + 4 and y = x 7x + 0 x + 4 = x 7x + 0 x 8x + 16 = 0 (x 4)(x 4) = 0 (x 4) = 0 There is only one solution to the quadratic which means the straight line and curve touch at only one point. There is just one solution to this equation which proves that the straight line and curve touch. Solving gives x = 4 Putting x = 4 into the equation of the straight line y = = 8 Hence, the coordinates of the point of contact are (4, 8). An alternative method for proving that the curve and straight line touch at one point is to find the discriminant and show that it equals zero. For example, the equation x 8x + 16 = 0 has discriminant b 4ac = ( 8) 4(1)(16) = = 0. This shows there are two real equal roots showing the curve and straight line touch at a single point. 14 Substituting y = 5x + 13 into x + y = 13 for y gives x + (5x + 13) = 13 x + 5x + 65x + 65x +169 = 13 6x + 130x = 0 Dividing through by 13 we obtain x + 10x + 1 = 0 (x + 4)(x + 3) = 0 Hence, x = or 3 Substituting these two values into the linear equation y = 5x + 13 to find the corresponding y-coordinates we obtain When x =, y = 5( ) + 13 = 3 When x = 3, y = 5( 3) + 13 = Hence, the two points are (, 3) and ( 3, ). 9

30 1 Algebra: Topic 4 Problem solving and inequalities 15 Solving the two equations y = 3x + 6 and y = x x + 1 simultaneously by equating the y-values, we obtain 3x + 6 = x x + 1 Note there is no point in trying to factorise this as the question asks that the answer be given to two decimal places. You have to solve this quadratic equation by either completing the square or using the formula. Here we will use the formula. 0 = x 5x 5 Comparing the equation above, with ax + bx + c = 0 gives a = 1, b = 5 and c = 5. Substituting these values into the quadratic equation formula x = b ± b 4ac a gives: x = 5 ± ( 5) 4(1)( 5) (1) x = 5 ± = 5 ± 45 = or = 5 45 = 5.85 or 0.85 ( d.p.) Test yourself 1 (a) 3x > 7 3x > 9 Add to both sides. Divide both sides by 3. x > 3 (b) 3(x ) > 9 3x 6 > 9 3x > 15 Multiply out the brackets. Add 6 to both sides. (c) Divide both sides by 3. x > 5 x x 5 1 x 16 Multiply both sides by 7. Add 5 to both sides. 30 Subtract 4x from both sides. Divide both sides by minus 1 and remember to reverse the inequality. (d) 3x 4 < 4x + 6 x 4 < 6 x < 10 x > 10 Add 4 to both sides.

31 Worked solutions Subtract x from both sides. (a) x 4 > x + 6 x 4 > 6 x > 10 Add 4 to both sides. Add 4x to both sides. (b) 4 + x < 6 4x 4 + 5x < 6 Subtract 4 from both sides. Divide both sides by 5. 5x < x < 5 x < 0.4 Multiply out the brackets. (c) x + 9 5(x 3) x + 9 5x 15 Subtract 5x from both sides. Subtract 9 from both sides. 3x x 4 x 8 Divide both sides by 3 remembering to reverse the inequality in the process. 3 Rearranging the inequality we have x 3x 18 > 0 Considering the case where x 3x 18 = 0 Factorising gives (x + 3)(x 6) = 0 Giving x = 3 or 6 (these are the intercepts on the x-axis) y y = x 3x 18 As the curve y = x 3x 18 has a positive coefficient of the curve will be -shaped. Note we need the section of the curve which is above (and not on) the x-axis. Sketching the curve for y = x 3x 18 gives the following: 3 O 6 x Hence, x < 3 and x > 6. 31

32 1 Algebra: Topic 4 Problem solving and inequalities TAKE NOTE Note with practice you may find you do not need to draw the curve, which will save you a bit of time. 4 Factorising x x 15 = 0 gives (x 5)(x + 3) = 0 Hence x = 5 or 3 As the coefficient of x is positive, the graph of x x 15 is -shaped. Now x x This is the region below the x-axis (i.e. where y 0). Hence 3 x 5 5 (a) 5 < x < x 14 Add 1 to each side. Divide each side by. 3 < x 7 (b) 7 < 3x 5 < 4 < 3x < 9 Add 5 to each side. Divide each side by 3. 3 < x < 3 (c) 4(x 3) 3(x ) 4x 1 3x 6 x 1 6 Multiply out the brackets. Subtract 3x from both sides. Add 1 to both sides. x 6 3

33 1 Algebra: Topic 5 The binomial expansion and probability Topic Worked solutions 5 Progress check 1 (a) ( 8 5) = (b) ( 7 3) = (c) ( 1 6 ) = (d) 7 C = n! r!(n r)! = 8! 5!3! = = 56 n! r!(n r)! = 7! 3!4! = = 35 n! r!(n r)! = 1! = 6!6! = 94 n! r!(n r)! = 7!!5! = = 1 (e) (f) 1 C 5 = 4 C = n! r!(n r)! = 1! = 5!7! = 79 n! r!(n r)! = 4!!! = = 6 (a) ( 1 0) = 1 (b) ( 1) = (c) ( 3 ) = 3 (d) ( 10 3 ) = 10 3 (a) ( 5 0) = 1 TAKE NOTE You have to use the binomial theorem here as it is specified in the question. If you found the answer by multiplying out the brackets you would not gain any marks. (b) ( 5 1) = 5 (c) ( 8 5) = 56 (d) ( 10 5 ) = 5 4 The formula is as follows: (a + b) n = a n + na n 1 b + n(n 1)! a n b + n(n 1)(n ) 3! a n 3 b

34 1 Algebra: Topic 5 The binomial expansion and probability Here n = 3, a = 3 and b = x. (3 + x) 3 = (3) (x) + (3)()! = x + 36x + 8x 3 5 The formula is as follows: 3 1 (x) + (3)()(1) 3! 3 0 (x) 3 (a + b) n = a n + ( n 1)a n 1 b + ( n )a n b + + ( n r)a n r b r + + b n Here n = 6, a = x and b = 3 x. Looking at the above it can be seen that the term in x is the third term in the expansion. Substituting in the values for a, b and n we obtain (x + 3 x) 6 = x 6 + ( 6 1) x 5 ( 3 x) + ( 6 ) x 4 ( 3 x) + ( 6 3) x 3 ( 3 x) 3 + Term in x = ( 6 ) x 4 ( 3 x) To find the coefficients we will expand Pascal s triangle The last line of Pascal s triangle shows the line we need as we need the second number in the line to be a 6 which is the power to which the bracket is to be raised As ( 6 ) = 15, we have term in x = 15x 4 ( 3 x) = 135x 6 Obtaining the formula and following the pattern in the terms gives: (a + b) n = a n + ( 1)a n n 1 b + ( )a n n b + ( 3)a n n 3 b 3 + (a + b) 5 = a 5 + ( 5 1)a 4 b + ( 5 )a 3 b + ( 5 3)a b 3 + ( 5 4)ab 4 + ( 5 5)b 5 34 The coefficients of this expansion can be found using Pascal s triangle or worked out using a calculator. Putting a = x, b =, and n = 5 into the equation, gives: x (x + x) 5 = x 5 + 5x 4 ( x) + 10x 3 ( x) + 10x ( x) 3 + 5x( x) 4 + ( x) 5 = x x x + 80 x + 80 x x 5

35 Worked solutions Note we have used Pascal s triangle here to determine the coefficients in the expansion which are You could have alternatively used the formula to find these. 7 (a) (a + b) 6 = a 6 + 6a 5 b + 15a 4 b + 0a 3 b 3 +15a b 4 + 6ab 5 + b 6 (1 + x) 6 = 1 + 6x + 15x + 0x 3 +15x 4 + 6x 5 + x 6 (b) (1.0) 6 = ( ) 6 Hence we substitute a = 1 and b = 0.0 into the expansion from part (a). ( ) 6 = (1) 5 (0.0) + 15(1) 4 (0.0) + 0(1) 3 (0.0) (1) (0.0) (1)(0.0) 5 + (0.0) 6 = 1.16 (4 d.p.) 8 (i) P(X = r) = ( n r)p r (1 p) n r p = 0.4, n = 10 and r = 5. P(X = 5) = ( 10 5 )0.4 5 (1 0.4) 5 P(X = 5) = ( 10 5 )0.4 5 (0.6) 5 = = 0.01 (3 s.f. ) We need to find the probability of each of the above and then add the two probabilities together. (ii) Less than female turtles means 0 or 1 turtle. P(X = 0) = ( 10 0 )0.4 0 (1 0.4) 10 0 = ( 10 0 )0.4 0 (0.6) 10 = = (3 s.f.) P(X = 1) = ( 10 1 )0.4 1 (1 0.4) 10 1 = ( 10 1 )0.4 1 (0.6) 9 = = (3 s.f.) Now, P(X = 0 or 1) = P(X = 0) + P(X = 1) = = Probability of less than female turtles = (3 s.f.) 35

36 1 Algebra: Topic 5 The binomial expansion and probability You could use the binomial formula but here we can use the AND law and the fact that the probability of throwing one six is (a) P(X = 3) = ( 1 6) 3 = 1 16 = (b) P (X = 0) = ( 5 6) 3 = = Note the probability of not getting a six = = 5 6. (c) Note that at least sixes means or 3 sixes. We first find the probability of sixes and then add this to the probability of obtaining 3 sixes already found in part (a). P(X = ) = ( 3 )( 1 6) (1 1 6) 3 = ( 3 )( 1 6) ( 5 6) 1 = Hence, probability at least sixes = P( sixes) + P(3 sixes) = = = (3 s.f.) Test yourself 1 (a + b) n = a n + ( 1)a n n 1 b + ( )a n n b + ( n = r) n! r!(n r)! The term in x is given by: ( n )a n b Here a =, b = 3x and n = 5 So the term in x is Hence, the coefficient of x is 70. 5!!(5 )! ()3 (3x) = x = 70x 36

37 Worked solutions n(n (1 + x) n 1)x = 1 + nx + +! n(n 1)(n )x3 + 3! In this case we substitute 3x for x and 6 for n. Hence (1 + 3x) 6 = 1 + (6)(3x) + (6)(5)(3x) 1 = x + 135x + 540x 3 + (6)(5)(4)(3x) P(X = r) = ( n r)p r (1 p) n r p = 0.5 and n = 0. P(X = 4) = ( 0 4 )0.5 4 (1 0.5) 0 4 = ( 0 4 )0.5 4 (0.75) 16 = (4 s.f.) 4 P(X = 15) = ( 0 15) (1 0.8) 5 = (4 s.f.) 5 P(X = r) = ( n r)p r (1 p) n r Now p = 0.1, n = 10, r = 1, so we have P(X = 1) = ( 10 1 )0.1 1 (1 0.1) 10 1 P(X = 1) = ( 10 1 )0.1 1 (0.88) 9 = (4 s.f.) 37

38 Coordinate geometry: Topic 6 Coordinate geometry of straight lines Topic Worked solutions 6 Progress checks 1 (a) Negative (b) Zero (c) Positive (d) Negative (e) Positive (f) Infinite (g) Negative (h) Zero (a) Gradient = y y 1 x x 1 = = 6 = 3 (b) Gradient = y y 1 = 1 0 x x = 1 8 = 3 (c) Gradient = y y = x x 1 5 ( ) = 3 3 = 1 (d) Gradient = y y = x x 1 5 ( 1) = 9 4 = 9 4 (e) Gradient = y y 1 = 6 ( 3) x x = 9 1 = 9 (f) Gradient = y y 1 = ( ) x x = 4 3 (g) Gradient = y y 1 5 ( 4) = x x = 1 11 = (a) ( x + x 1, y + y 1 ) = ( 1 + 3, + 8 ) = (, 5) (b) ( x + x 1, y + y 1 ) = ( 0 + 4, + 1 ) = (, 3 ) (c) ( x + x 1, y + y 1 ) = ( + 0, 5 + ( 5) ) = ( 1, 0) (d) ( x + x 1, y + y ( ) ) = (, 4 + ( 6) ) = ( 5, 1) 38

39 (e) ( x 1 + x, y 1 + y ) = ( ) = ( (f) ( x 1 + x, y 1 + y (g) ( x 1 + x, y 1 + y 10 + ( 3), ( 4), 4 + 6, ) = ( 8 + ( 5) ) = ( 7 ), 6 ) = ( 7 ), 1 ) = ( 3 ), 3 4 (a) (x x 1 ) + (y y 1 ) = (5 1) + (9 5) = 3 = 5.66 (b) (x x 1 ) + (y y 1 ) = (6 3) + (9 4) = 34 = 5.83 (c) (x x 1 ) + (y y 1 ) = (6 1) + (1 0) = 169 = 13.0 (d) (x x 1 ) + (y y 1 ) = ( ( 3)) + (6 ) = 41 = 6.40 (e) (x x 1 ) + (y y 1 ) = (0 ( 5)) + (4 0) = 41 = 6.40 (f) (x x 1 ) + (y y 1 ) = (0 ( 1)) + (10 5) = 169 = 13.0 (g) (x x 1 ) + (y y 1 ) = ( 7 ( 6)) + ( ( 3)) = 6 = (a) A to B = ( 3 1) C to D = ( 3 1) Hence AB = CD (b) A to B = ( 5 1) C to D = ( 5 1) Hence AB = CD (c) A to B = ( 1) 6 C to D = ( 1) 6 Hence AB = CD (d) A to B = ( 6 1) C to D = ( 6 1) Hence AB = CD Worked solutions 39

40 Coordinate geometry: Topic 6 Coordinate geometry of straight lines (e) A to B = ( 4 ) C to D = ( 4 ) Hence AB = CD (f) A to B = ( 6 1) C to D = ( 1 6) Hence AB = CD (g) A to B = ( 1 ) C to D = ( 1) Hence AB = CD 6 (a) (x x 1 ) + (y y 1 ) = (6 1) + (7 4) = 34 = 5.83 (b) (x x 1 ) + (y y 1 ) = (5 0) + (17 5) = 169 = 13.0 (c) (x x 1 ) + (y y 1 ) = ( 3 [ 1]) + (0 [ 5]) = 9 = 5.39 (d) (x x 1 ) + (y y 1 ) = ( 4) + (4 [ 1]) = 9 = (a) m = 3, c = (b) m =, c = 3 (c) m = 3, c = 1 (d) m = 4 3, c = 3 (e) m = 1, c = 3 (f) m = 1, c = The mid-point of a line joining the points (x 1, y 1 ) and (x, y ) is given by: ( x 1 + x, y + y 1 ) Mid-point of AB = ( 5 + 1, ) = (, 8) 40

41 Worked solutions 9 (a) Mid-point of PQ, M = ( 4 + 4, ) = (0, 4) (b) P to Q = ( 8 ) PQ = 8 + PQ = PQ = 68 PQ = 4 17 PQ = (a) Length = (x x 1 ) + (y y 1 ) = (6 1) + (1 ( )) = = 34 = 5.83 (b) Length = (x x 1 ) + (y y 1 ) = (0 [ 4]) + ( 3 0) = = 5 = 5 (c) Length = (x x 1 ) + (y y 1 ) = (4 0) + (7 8) = = 17 = 4.1 Remember that the midpoint of the line joining the points (x 1, y 1 ) and (x, y ) is ( x 1 + x, y 1 + y ). 11 Let the coordinates of B be (x, y). The x-coordinate of the mid-point = x + The x-coordinate is 4, so 4 = x + Solving gives x = 6 The y-coordinate of the mid-point = y + 3 The y-coordinate is 4, so 4 = y + 3 Solving gives y = 5 Hence, coordinates of B are (6, 5) 1 Gradient of line y y 1 = k x x = k 5 Gradient of AB = 4 5, so k 5 =

42 Coordinate geometry: Topic 6 Coordinate geometry of straight lines Solving, gives k = 6 13 (a) Gradient = 4 (b) Gradient = 3 (c) Gradient = 1 4 (d) Gradient = 3 14 (a) As point B lies on the line, its coordinates will satisfy the equation of the line. Hence, 5() (k) = Solving gives k = 4 (b) Mid-point = ( x 1 + x, y 1 + y ) = ( 4 +, ) = (3, 6.5) 15 4x 3y = c Substituting the coordinates ( 3, ) into this equation gives 4( 3) 3() = c Solving gives c = 18. Hence required equation of line is 4x 3y = (a) Gradient of AB = = 1 Gradient of BC = = 1 (b) Product of the gradients = (1)( 1) = 1. As m 1 m = 1, AB and BC are perpendicular to each other. Both gradients are found using the formula: Gradient = y y 1 x x 1 17 (a) Gradient of PQ = = 3 For equation of line PQ, y y 1 = m(x x 1 ) where m = 3 and (x, y 1 1 ) = (0, 6) y 6 = 3 (x 0) y 1 = 3x y = 3x + 1 4

43 Worked solutions (b) Mid-point of PQ = ( x + x 1, y + y 1 ) = ( 0 + 4, ) = (, 3) Gradient of line perpendicular to PQ is given by m ( 3 ) = 1 18 (a) y = 6x 5 m = 3 Equation of line perpendicular to PQ, y y 1 = m(x x 1 ) where m = 3 and (x, y 1 1 ) = (, 3) y 3 = (x ) 3 3y 9 = x 4 y = 6(1) 5 3y = x + 5 y = 1 which is the y-value of the point so the point lies on the line. (b) y = 6x 5 y = 6(0) 5 y = 5 which is the y-value of the point so the point lies on the line. (c) y = 6x 5 y = 6() 5 y = 1 5 y = 7 as this is not the y-coordinate of the point, the point does not lie on the line. (d) y = 6x 5 y = 6( 1 ) 5 y = 3 5 y = This is the y-value of the point, so the point lies on the line. (e) y = 6x 5 y = 6( 1) 5 y = 11 This is not the y-value of the point, so the point does not lie on the line. 43

44 Coordinate geometry: Topic 6 Coordinate geometry of straight lines 19 (a) y = 4x 1 y = 4(0) 1 y = 1 (b) y = 3x + 5 y = 3(0) + 5 y = 5 (c) 4x y = 0 4(0) y = 0 y = 0 y = 0 (d) 5y x = 5y 0 = y = 5 (e) x + y 1 = 0 (0) + y 1 = y 1 = 0 y = 1 (f) y x 3 = 0 y 0 3 = 0 y = 3 0 (a) y = 4x + 0 = 4x + = 4x x = 1 (b) y = 3x = 3x = 3x x = 5 (c) 3x + y = 1 3x +(0) = 1 x = 4 (d) 5y x = 9 5(0) x = 9 x = 9 (e) 5x 7y = 5 5x 7(0) = 5 5x = 5 x = 5 (f) x y + 7 = 0 x = 0 x = 7 (g) 5x 3y 10 = 0 5x 3(0) 10 = 0 x = 44

45 Worked solutions 1 7x + y = (1) x y = 4... () Multiplying equation () by gives x y = 8 7x + y = 19 Adding the above two equations gives 9x = 7 x = 3 Substituting x = 3 into equation (1) we obtain 7(3) + y = y = 19 y = y = 1 Checking by substituting x = 3 and y = 1 into equation () LHS = 3 ( 1) = 4 RHS = 4 So LHS = RHS Point of intersection is (3, 1). 7y = 5x 7... (1) 4y = 3x () Multiplying equation (1) by 3 and equation () by 5 gives the following 1y = 15x 81 0y = 15x 80 Note that 81 ( 80) = 1. Be very careful when subtracting negative numbers. Subtracting these two equations gives y = 1 Substituting y = 1 into equation (1) gives 7 = 5x 7 0 = 5x x = 4 45

46 Coordinate geometry: Topic 6 Coordinate geometry of straight lines Checking in equation () LHS = 4( 1) = 4 RHS = 3(4) 16 = 4 So LHS = RHS Point of intersection is (4, 1). 3 Rearranging each equation so that it is the form y = mx + c. y 3x + 4 = 0 so y = 3x 4. This line has a gradient of 3. 4y 1x + 1 = 0 so y = 3x 1. This line also has a gradient of 3. 4 Both lines have the same gradient and are therefore parallel so they will not intersect. 4 4x + y = 1 3x + y = 7 Multiplying the first equation by gives (1) () gives 8x + y =... (1) 3x + y = 7... () 5x = 5 Hence, x = 1 Substituting x = 1 into equation (1) gives 8 + y = y = 10 y = 5 Checking by substituting x = 1 and y = 5 into equation () gives LHS = 3( 1) + (5) = = 7 RHS = 7 So LHS = RHS Coordinates of A are ( 1, 5) 46

47 Worked solutions Test yourself 1 (a) Gradient of AB = y y 1 = 1 0 x x = 1 3 Gradient of CD = y y = x x 1 ( 1) = 1 3 As the gradients of AB and CD are the same the two lines are parallel. (b) Gradient of AB = 1 3 and AB passes through point A (1, 0) so equation of AB is: y y 1 = m(x x 1 ) y 0 = 1 3 (x 1) 3y = x 1 Rearranging this equation so that it is in the form asked for by the question gives: x 3y 1 = 0 (a) Gradient of AB = y y 1 = 1 4 x x 1 k ( 7) = 5 k + 7 But gradient of AB = 1 so The equation is multiplied through by the common denominator, (k + 7). 5 k + 7 = 1 5 = 1(k + 7) 10 = k 7 Giving k = 3. (b) The product of the gradients of perpendicular lines is 1. Hence, m( 1 = 1 ) Hence gradient of BC = Equation of BC is: y y 1 = m(x x 1 ) where m = and (x 1, y 1 ) = (3, 1). y ( 1) = (x 3) y + 1 = x 6 x y 7 = 0 47

48 Coordinate geometry: Topic 6 Coordinate geometry of straight lines 3 (a) y 6 B (1, 6) ( 3, ) A 1 C (6, 1) x Gradient of AB = y y 1 6 = x x 1 1 ( 3) = 4 4 = 1 Gradient of BC = y y 1 x x 1 = = 5 5 = 1 TAKE NOTE This formula needs to be remembered. If you forget it you can plot the two points on a sketch graph and form a triangle and use Pythagoras' theorem to find the length of the hypotenuse. Product of gradients = (1)( 1) = 1 proving that the two lines are perpendicular to each other. (b) Length = (x x 1 ) + (y y 1 ) Putting the coordinates A ( 3, ) and B (1, 6) into the formula gives AB = (1 [ 3]) + (6 ) = = 3 units Using the coordinates B (1, 6) and C (6, 1) in the formula gives BC = (6 1) + (1 6) = = 50 units (c) Tan AĈB = AB BC = 3 50 = 16 5 = 4 5 = The equation of the line parallel to 3x + y = 5, will be of the form 3x + y = c. Substituting the coordinates (3, 4) into the equation, we obtain 3(3) + (4) = c Giving c = 17. Hence, the equation of the line is 3x + y = 17 48

49 Worked solutions 5 (a) Gradient of PQ = y y 1 0 = x x 1 0 ( 4) = 4 = 1 Equation of line PQ with gradient 1 and passing through the point Q (0, ) is y y 1 = m(x x 1 ) y = 1 (x 0) y 4 = x Hence, equation of line is y x = 4 (b) Mid-point of PQ = ( x 1 + x, y 1 + y ) = ( 4 + 0, 0 + ) = (, 1) Gradient of PQ = 1 Gradient of line perpendicular to PQ, m is given by (m)( 1 ) = 1 ( i.e. the products of the gradients of two lines are 1 if they are perpendicular). Hence, m = Equation of required line through mid-point is: y y 1 = m(x x 1 ) y 1 = (x [ ]) y 1 = x 4 >>> TIP For questions involving coordinate geometry, it is always worthwhile spending a little time sketching a graph showing the positions of the coordinates. It is then easier to see the shape formed when certain points are joined up with lines. You can also check the signs of any gradients you have found numerically. 6 (a) y = x 3 R (, 6) y Q (3, 5) 1 P (1, 0) Gradient of PQ = y y 1 x x 1 = = 5 x 49

50 Coordinate geometry: Topic 6 Coordinate geometry of straight lines (b) Equation of straight line PQ which passes through (1, 0) and has gradient 5 is given by: y y 1 = m(x x 1 ) where m = 5 and (x y ) = (1, 0) 1, 1 y 0 = 5 (x 1) Multiply both sides by to remove the denominator in the fraction. y = 5 (x 1) y = 5(x 1) y = 5x 5 y 5x = 5 You can now form the parallelogram and it is possible to see from the sketch the rough position of point S. If you give it some thought you can get the coordinates of point S using the diagram. (c) R (, 6) y 6 Q (3, 5) S 1 P (1, 0) x As lines RS and PQ are opposite sides of a parallelogram, they are parallel and therefore have the same gradient. Hence, gradient of RS = 5. Equation of straight line RS which passes through (, 6) and has gradient 5 is given by: y y 1 = m(x x 1 ) where m = 5 and (x y ) = (, 6) 1, 1 y 6 = 5 (x [ ]) y 6 = 5 (x + ) y 1 = 5(x + ) y 1 = 5x + 10 y 5x = 50

51 Worked solutions (d) To find the coordinates of the point S, the equations for lines RS and SP are solved simultaneously. y 5x =... (1) 5y + x = 1... () Multiplying equation () by 5 gives 5y + 5x = 5 y 5x = Adding the two above equations gives 7y = 7 So, y = 1 Substituting y = 1 into equation () and solving gives 5 + x = 1 So, x = 4 Hence the coordinates of point S are ( 4, 1) 7 (a) Because there are lots of points in this question, it is worth spending a little time doing a sketch showing their positions. y D (5, ) x 3 4 C (9, 4) 5 6 (, 5) A 7 B (6, 7) 51

52 Coordinate geometry: Topic 6 Coordinate geometry of straight lines Gradient of AB = y y 1 7 ( 5) = x x 1 6 = 4 = 1 (b) Gradient of DC = y y 1 4 ( ) = x x = 4 = 1 Gradient of AB = gradient of DC, so lines are parallel The two lines at right angles which have the vector as the hypotenuse are the same for both triangles so the lengths of the hypotenuse will be the same. (c) The vector to go from A to B is ( 4 ). The vector to go from D to C is ( 4 ). These vectors are the same so the lines are the same length. (d) The vector to go from A to D is ( 3 3). The vector to go from B to C is ( 3 3). These vectors are the same so the lines are both parallel and the same length. Hence, this and the answer to parts (b) and (c) prove that ABCD is a parallelogram. 8 (a) y 6 5 B (1, 5) C (4, 5) 4 3 M A (4, ) x Mid-point, M = ( x 1 + x, y 1 + y ) = ( 4 + 1, + 5 ) = ( 5, 7 ) (b) Gradient of AB = y y 1 x x 1 = = 3 ( 3) = 1 (c) Gradient of MC = y y 1 = 5 7 x x = (3 ) ( 3) = 1 Product of gradients of lines AB and MC = ( 1)(1) = 1 As the product of two perpendicular lines is 1 so AB and MC are perpendicular. 5

53 Worked solutions (d) Equation of line MC with gradient 1 and passing through the point C (4, 5) is y y 1 = m(x x 1 ) y 5 = 1(x 4) y 5 = x 4 Hence, equation of MC is y = x (a) x + 3y = 5 3y = x + 5 y = 3 x Gradient of line = 3 (b) Equation of line with gradient and passing through the 3 point R (3, 3) is y y 1 = m(x x 1 ) y 3 = (x 3) 3 3y 9 = x + 6 Hence, equation of line is 3y = x + 15 Now, the equation of the y-axis is x = 0. Solving this simultaneously with the equation of the line, we have 3y = so y = 5 Hence, S is the point (0, 5) 10 (a) 4x + 5y = 10 5y = 4x + 10 y = 4 5 x + Hence, gradient of AB =

54 Coordinate geometry: Topic 6 Coordinate geometry of straight lines (b) Substituting the coordinates of C into the equation of the line we obtain LHS = 4( 5) + 5(6) = = 10 RHS = 10 As LHS = RHS, point C lies on the line. (c) As lines are perpendicular, m 4 5 = 1, so m = 5 4 Gradient of line at right-angles to AB = 5 4 Equation of line with gradient 5 and passing through the 4 point C ( 5, 6) is y y 1 = m(x x 1 ) y 6 = 5 (x [ 5]) 4 4y 4 = 5(x + 5) 4y 4 = 5x + 5 Hence, equation of line is 4y = 5x

55 Coordinate geometry: Topic 7 Coordinate geometry of the circle Topic Worked solutions 17 Progress check Note that the radius of a circle cannot be zero as it is a length. 1 (a) x + y = 1 This equation is in the form x + y = r Hence r = 1 = 1 (b) x + y = 9 This equation is in the form x + y = r Hence r = 9 = 3 (c) x + y = 5 This equation is in the form x + y = r Hence r = 5 = 5 (d) x + y 4 = 0 x + y = 4 This equation is in the form x + y = r Hence r = 4 = (e) x + y 49 = 0 x + y = 49 This equation is in the form x + y = r Hence r = 49 = 7 (f) 4x + 4y = 16 Dividing both sides by 4 we obtain x + y = 4 This equation is in the form x + y = r Hence r = 4 = 55

56 Coordinate geometry: Topic 7 Coordinate geometry of the circle (g) 8x + 8y = 7 Dividing both sides by 8 we obtain x + y = 9 This equation is in the form x + y = r Hence r = 9 = 3 (h) 3(x + y ) 7 = 0 Dividing both sides by 3 we obtain x + y 9 = 0 x + y = 9 This equation is in the form x + y = r Hence r = 9 = 3 (i) y = 16 x x + y = 16 This equation is in the form x + y = r Hence r = 16 = 4 (j) x + y 5 = 0 x + y = 5 This equation is in the form x + y = r Hence r = 5 (k) x + y 50 = 0 x + y = 50 This equation is in the form x + y = r Hence r = 50 = 5 = 5 (a) x + y = 9 (b) x + y = 16 3 (a) (x 3) + (y + 1) = 9 x 6x y + y + 1 = 9 x + y 6x + y + 1 = 0 (b) (x ) + (y + 4) = 16 x 4x y + 8y + 16 = 16 x + y 4x + 8y + 4 = 0 (c) (x 1) + (y 3) = 1 x x y 6y + 9 = 1 x + y x 6y + 9 = 0 (d) (x + 4) + (y 5) = 5 x + 8x y 10y + 5 = 5 x + y + 8x 10y + 16 = 0 (e) (x + 5) + (y 1) = 9 x + 10x y y + 1 = 9 x + y + 10x y + 17 = 0 (f) (x 6) + (y + 7) = 49 x 1x y + 14y + 49 = 49 x + y 1x + 14y + 36 = 0 (g) (x 5) + (y 4) = 16 x 10x y 8y + 16 = 16 x + y 10x 8y + 5 = 0 (h) x + (y 1) = 4 x + y y + 1 = 4 x + y y 3 = 0 (c) x + y = 81 (d) x + y = 6 (e) x + y = 1 (f) x + y = 45 (g) x + y = 18 56

57 Worked solutions Test yourself 1 (a) Comparing the equation x + y 8x 6y = 0 with the equation x + y + gx + fy + c = 0 we can see g = 4, f = 3 and c = 0. Centre A has coordinates ( g, f) = (4, 3) Radius = g + f c = ( 4) + ( 3) 0 = 5 = 5 (a) Comparing the equation x + y 4x + 6y 3 = 0 with the equation x + y + gx + fy + c = 0 we can see g =, f = 3 and c = 3. Centre A has coordinates ( g, f) = (, 3) Radius = g + f c = ( ) + (3) ( 3) = 16 = 4 (b) If point P (, 1) lies on the circle its coordinates will satisfy the equation of the circle. x + y 4x + 6x 3 = 0 x + y 4x + 6y 3 = () + (1) 4() + 6(1) 3 = = 0 Both sides of the equation equal zero so P (, 1) lies on the circle. 3 (a) Equation of the circle is: (x a) + (y b) = r (x ) + (y 3) = 5 x 4x y 6y + 9 = 5 x + y 4x 6y 1 = 0 (b) Gradient of line joining the centre A (, 3) to P (5, 7) = = 4 3 AP is a radius of the circle and will make an angle of 90 to the tangent at point P. Gradient of tangent = 3 4 Equation of tangent is y 7 = 3 (x 5) 4 4y 8 = 3x y + 3x 43 = 0 57

58 Coordinate geometry: Topic 7 Coordinate geometry of the circle 4 (a) x + y 4x + 8y + 4 = 0 Completing the squares for x and y gives (x ) + (y + 4) (x ) + (y + 4) 16 = 0 (x ) + (y + 4) = 16 Centre of circle is at (, 4) (b) If the point P (6, 4) lies on the circle, the coordinates will satisfy the equation of the circle. Hence, (x ) + (y + 4) = (6 ) + ( 4 + 4) = = 16 This is the same as the RHS of the equation so the point lies on the circle. 5 (a) Centre of circle is at the mid-point of the diameter AB. Mid-point of line joining A(1, 4) and B(9, 10) is ( , ) = (5, 3) (b) Distance between the points (1, 4) and (5, 3) is given by The distance from the midpoint to the circumference is the radius of the circle. r = (x x 1 ) + (y y 1 ) r = (5 1) + (3 [ 4]) r = r = 65 (c) The equation of a circle having centre (a, b) and radius r is given by (x a) + (y b) = r For this circle, centre is (5, 3) and radius is 65. (x 5) + (y 3) = 65 Make sure you give the equation of the circle in the format asked for in the question (i.e. x + y + ax + by + c = 0 in this case). Multiplying out the brackets we obtain x 10x y 6y + 9 = 65 x + y 10x 6y 31 = 0 58