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Yr 11 MCAT Algebra Practice 2

Yr 11 MCAT Algebra Practice 2. 6. Write an expression for the area of the following shape… 7. If the above area is 450cm 2 what is the value of a? 8. Factorise fully… a) a 2 + 22a + 40 = b) 15ab + 20a = c) a 2 – 100 = 9. Solve the following… a) 13a – 16 = 10a + 2

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Yr 11 MCAT Algebra Practice 2

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  1. Yr 11 MCAT Algebra Practice 2 • 6. Write an expression for the area of the following shape… • 7. If the above area is 450cm2 what is the value of a? • 8. Factorise fully… • a) a2 + 22a + 40 = • b) 15ab + 20a = • c) a2 – 100 = • 9. Solve the following… • a) 13a – 16 = 10a + 2 • b) 5a4 = 80 • c) (a – 4)(a + 9) = 0 • d) a2 + 14a + 40 = 0 10. Simplify fully… a2 + 8a – 20 = a + 10 11. Three consecutive numbers beginning with a are added together. Write a simplified expression for the sum of these three numbers. 12. If these three numbers total up to 171. What is the value of the first number (a). 13. Simplify fully… a + a = 3 5 3a2 ÷ 9a = 4b 20b3 • 1. Expand and Simplify… • (3a + 8)(a – 1) = • 2. (2a2 x 3a3)2 = 36ak • What is the value of k? • 3. Rearrange to make x the subject of… • y = 3x – 6 • 4. Simplify fully… • a) 8ab – 3ab + 5c3 + c3 = • b) 24a10b6 = • 30a4b8 • c) 8a2 x 2a3 x 3a = • 5. Simultaneously solve for a & b. • 6a + 5b = 108 • 3a + 5b = 84 6a 3a

  2. Yr 11 MCAT Algebra Practice 2 • 6. Write an expression for the area of the following shape… • 7. If the above area is 450cm2 what is the value of a? • 8. Factorise fully… • a) a2 + 22a + 40 = • b) 15ab + 20a = • c) a2 – 100 = • 9. Solve the following… • a) 13a – 16 = 10a + 2 • b) 5a4 = 80 • c) (a – 4)(a + 9) = 0 • d) a2 + 14a + 40 = 0 10. Simplify fully… a2 + 8a – 20 = a + 10 11. Three consecutive numbers beginning with a are added together. Write a simplified expression for the sum of these three numbers. 12. If these three numbers total up to 171. What is the value of the first number (a). 13. Simplify fully… a + a = 3 5 3a2 ÷ 9a = 4b 20b3 • 1. Expand and Simplify… • (3a + 8)(a – 1) = • 2. (2a2 x 3a3)2 = 36ak • What is the value of k? • 3. Rearrange to make x the subject of… • y = 3x – 6 • 4. Simplify fully… • a) 8ab – 3ab + 5c3 + c3 = • b) 24a10b6 = • 30a4b8 • c) 8a2 x 2a3 x 3a = • 5. Simultaneously solve for a & b. • 6a + 5b = 108 • 3a + 5b = 84 (a + 10)(a – 2) . a + 10 6a 3a2 - 3a + 8a - 8 = 3a2 + 5a - 8 = a – 2 3a Area (rectangle) = base x height Area = 6a x 3a k = 10 (6a5)2 = 36a10 Area = 18a2 1st + 2nd + 3rd = a + a + 1 + a + 2 3x – 6 = y 450 = 18a2 = 3a + 3 3x = y + 6 18a2 = 450 x = y + 6 . 3 a2 = 25 a = 5 a = √25 + 6c3 5ab ( )( ) a a + 2 + 20 3a + 3 = 171 4 a6 . 3a = 168 ( ) 5 a 3b + 4 5 b2 a = 56 ( )( ) + 10 a - 10 a 48 a6 5a + 15 3a = 15 8a . 15 3a = 18 13a – 10a = 2 + 16 a = 6 eqn 1 subtract eqn 2 off eqn 1 a = 4√16 3a2 x 20b3 4b 9a eqn 2 and -2 a = 2 a4 = 16 3a = 24 a = 8 = 60a2b3 36ab 4 -9 a is ___ or ___ substitute a back into eqn 1 6(8) + 5b = 108 = 12/3ab2 5b = 60 a is ___ or ___ -10 -4 (a + 10)(a + 4) = 0 b = 12

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