Solving Equations and Inequalities

Solving Equations and Inequalities 59 minutes 53 marks Page 1 of 21

Q1. (a) Solve a 2 = 9 Answer a =... (b) Solve = 5 Answer b =... (c) Solve 2c 3 = 11 Answer c =... (2) (Total 4 marks) Q2. In the magic square, the rows, columns and diagonals add to the same number. 10 w x 5 y 9 6 11 4 Work out the values of w, x and y.......... Answer w =... x =... y =... (Total 3 marks) Page 2 of 21

Q3. (a) Solve 8t 5 = 19 Answer t =... (2) (b) Nails cost 3 pence each. Screws cost 5 pence each. Write down an expression for the cost of x nails and y screws. Answer... pence (2) (c) Expand and simplify 4(2w + 3) 5(3w + 7) Answer... (2) (Total 6 marks) Q4. (a) Simplify a + a + a Answer... (b) Simplify 8b + 3 2b + 7 Answer... (2) (c) Solve the equation = 6 Answer x =... (Total 4 marks) Page 3 of 21

Q5. Complete this table. Expression Value 2x 8 5x 2x + 3y 5 y 3x y............ (Total 5 marks) Q6. (a) Solve the equation = 7... Answer w =... (b) Solve the equation 7x 3 = 60... Answer x =... (2) (c) Write down one integer which satisfies the inequality 7y > 63 Answer... (Total 4 marks) Page 4 of 21

Q7. (a) Solve the equation = 4 Answer x =... (b) Solve the equation 8w 5 = 3w + 1 Answer w =... (3) (c) Simplify y + 2 y y Answer... (d) Factorise 15t + 25 Answer... (e) Factorise z + 8z 2 Answer... (Total 7 marks) Q8. ABCD is a quadrilateral. Not drawn accurately Page 5 of 21

(a) Write down an expression for the perimeter of the quadrilateral in terms of x and y. Simplify your answer. Answer... (2) (b) When x = 4 cm, the perimeter of the quadrilateral is 68 cm. Find the value of y. Answer...cm (3) (Total 5 marks) Q9. (a) Solve the equation Answer x =... (3) (b) Solve the inequality 3x + 8 < 29 Answer... (2) (Total 5 marks) Page 6 of 21

Q10. (a) Explain why the sum of the angles in any quadrilateral is 360. (2) (b) A quadrilateral has one right angle. The other angles are and Not drawn accurately (i) Write down an equation in terms of x. Answer... (ii) Solve your equation and find the size of the largest angle in the quadrilateral................... Answer =... degrees Largest angle =... degrees (3) (Total 6 marks) Page 7 of 21

Q11. Parveen is using trial and improvement to find a solution to the equation x 3 + 7x = 30 This table shows her first two trials. x x 3 + 7x Comment 2 22 Too small 3 48 Too big Continue the table to find a solution to the equation. Give your answer to 1 decimal place. Answer... (Total 3 marks) Q12. Use trial and improvement to find a solution to the equation x 3 x = 21 Page 8 of 21

Give your answer to one decimal place. You must show your working............................... Answer x =... (Total 4 marks) Page 9 of 21

. (a) 11 (b) 15 (c) (2c = 11 + 3 = 14) Their 14 2 7 [4] M2. w = 3 x = 8 y = 7 each B3 [3] M3. (a) 8t = 19 + 5 3 (b) 3x + 5y oe Allow 3 x + 5 y not x3 + y5 For one term correct Do not ignore fw B2 (c) 8w + 12 15w 35 Allow one term incorrect 7w 23 [6] Page 10 of 21

M4. (a) 3a (b) 6b + 10 Allow 3 a and a 3; not a3 for 6b or + 10 fw eg, 16b; deduct 1 mark B2 (c) 12 [4] M5. (5x) = 20 (5 8) 3 (y =) 1 or x = 4 seen Allow 5 8 3 or 3y = 3 3 (their) 4 (their) 1 13 ft (their) x or x = 4 and (their) y ft [5] M6. (a) (w =) 63 (b) 7x = 63 (x = ) 9 63 7 9 embedded (c) One integer > 9 Any whole number greater than 9 [4] Page 11 of 21

(c) y + 2y 2 or y(1 + 2y) M7. (a) 80 (b) 8w 3w or 1 + 5 Or better 5w = 6 or 1.2 oe ft Allow y + 2 y 2 B0 fw eg, y + 2y 2 = 3y 2 (d) 5 (3t + 5) Allow 5 (3t + 5) or (3t + 5) 5 Condone missing final bracket (e) z (z + 8) Allow z (z + 8) or (z + 8) z Condone missing final bracket [7] M8. (a) 2x + 3y + 4x + 2y + 3x + 5x + y 14x + 6y 14x or 6y seen (b) Their 14 4 + their 6 y = 68 Their 6y = their 12 (y = ) 2 ft Their answer for (a) with 2 terms ft [5] Page 12 of 21

M9. (a) 23 2x = 15 4.6 0.4x = 3 gets allow one error 23 15 = 2x 1.6 = 0.4x 4 f.t. if awarded. ft (b) 3x < 21 3x = 21 gets iff recovered x < 7 Must have inequality in answer. Accept.. [5] 0. (a) Complete explanation eg, Quadrilateral can be divided into 2 triangles and 2 180 Use of (n 2) 180 with n = 4 or Using (external angles) = 360 eg, (internal angles + external angles) = 4 180 (internal angles) = 4 180 360 B for partial explanation B0 for 2 180 only B2 (b) (i) 3x 12 + x 6 + 2x + 90 = 360 or better eg, 6x + 72 = 360 B0 for 3x 12 + x 6 + 2x + 90 = 180 Page 13 of 21

(ii) 6x = 288 or 6x = 360 72 or x = (Their 288) 6 ft for 6x = 108 or 6x = 180 72 or (Their 108) 6 x = 48 ft for x = 18 132 3 (Their x) 12 for 35 x 63 SC1 48 with no working or using T & I SC2 (48 and)132 with no working or using T & I ft [6] 1. Trial between 2 and 3 Trial between 2.3 and 2.4 inclusive that bracket the answer Trial at 2.35 or 2.36 or 2.37 and 2.4 stated as answer In this question final answer on its own will not get any marks Working must be seen. All trials must be correctly evaluated either rounded or truncated to a degree of accuracy that allows comparison. D [3] Page 14 of 21

2. Trial above 2.8796 2 gives 6, 3 gives 24 Trial below 2.8796 2.9 gives 21.489, all values to at least l dp rounded or truncated Testing a value that justifies 2.9 as answer 2.5 13.125, 2.6 14.976, 2.7 16.983 2.8 19.152, 2.85 gives 20.299 D x = 2.9 Dep on any M mark [4] Page 15 of 21

E1. This question is drawn from our specimen paper produced in advance of live examinations. As such, the question was not used in a live examination and therefore no Examiner's Remarks exist. E2. This question is drawn from our specimen paper produced in advance of live examinations. As such, the question was not used in a live examination and therefore no Examiner's Remarks exist. E3. Whilst it was pleasing in part (a) to see many correct answers, it was disappointing to see the number who used trial and error or failed to show any method. Correct embedded answers often led to candidates offering 19 or 24 as their answer. Many candidates did not understand the term expression with answers of 8 or 15 being common. Some candidates who correctly wrote 3x and 5y either failed to combine them with an addition sign or having combined them went on to offer, for example 15xy. In part (c) there were many poor responses and those candidates who knew the correct procedure often made errors with signs when expanding the second bracket. Fully correct answers were rare. E4. Half the candidates appeared to know that a + a + a simplified to 3a, in part (a). The rest gave varying responses including a 3. Half the candidates appeared to know what to do, in part (b), but not all could do it accurately. Some obtained the correct answer but lost a mark by trying to simplify 6b + 10 further, for example by giving their answer as 16b. About five percent of candidates did not attempt this question. Part (c) was reasonably well done with about sixty percent of the candidates gaining the mark. As expected a common incorrect response was 3. About eight percent of candidates did not attempt this part of the question. Additional Examiner s Commentary (b) There was a mark here for combining 8b and -2b to obtain 6b. There was a mark here for combining 3 and +7 to obtain 10 Candidates who are challenged to combine algebraic expressions should be encouraged to at least attempt the numbers, as half marks can be achieved without attempting any algebra. Page 16 of 21

E5. (6 5; 20 3; 2 2; 29 1; 32 0; 13 0; 10 dna). Most candidates found this a difficult question with about ten percent making no attempt. About sixty percent of candidates managed to score the first mark for finding the value of 5x and about half of these made some further progress by finding the value of y or by finding the value of 3x y for their values of x and y. Very few candidates attempted to show any method. Additional Examiner s Commentary With practice at this type of question candidates may be able to make further progress. Realising that the first row of entries in the table is equivalent to 2x = 8 is important here. Candidates should be encouraged to use the working lines provided to show their working perhaps 2x = 8 x = 4? The (their) in the mark scheme indicates that candidates who, having determined an incorrect value for y, continue with the question using this incorrect value, will be rewarded for their positive achievement. E6. Parts (a) and (b) were well answered. In part (c) there was evidence of a lack of understanding as fortuitously answers of 63, 70 or 763 were correct and common responses. Candidates who appeared to understand would usually give 10 or 11 as their answer. 9 was a very common error. Candidates should be discouraged from embedding solutions in their expressions without identifying their answer. E7. The majority of candidates answered part (a) of this question correctly although many gave their solution as 5. Nearly seventy percent obtained the correct solution, in part (b), nearly always from an algebraic approach although some used Trial and Improvement. Many of those who attempted a valid algebraic approach made slips usually with signs during the initial rearrangement. Obtaining 1.2 from 5w = 6 defeated some with 5/6 occurring fairly frequently. Nearly fifteen percent of candidates scored zero not encouraging for what is very nearly a banker topic. Page 17 of 21

Only about a third of the candidates managed to obtain the mark for part (c) and not all of these obtained the simplest answer as y + 2 y 2 was also allowed. Clearly applying the correct hierarchy of operations to an algebraic expression is not well understood. A fairly common incorrect response was 2y 3. Candidates at Higher tier should expect routine questions similar to those in part (d) and about three-quarters of the candidates managed to answer it correctly. However, some gave their answer as 3t + 5 suggesting an incomplete understanding of the concept involved. Many candidates clearly did not appreciate the meaning of the term factorise. Part (e) saw a similar level of performance to that in part (c). Here incorrect responses included trying to factorise into two brackets and the completely misconceived 8z 3. Additional Examiner s Commentary (b) There was a mark here for combining 8b and -2b to obtain 6b There was a mark here for combining 3 and +7 to obtain 10 Candidates who are challenged to combine algebraic expressions should be encouraged to at least attempt the numbers, as half marks can be achieved without attempting any algebra. E8. Most candidates knew how to find the perimeter, in part (a), but made errors in the simplification. The correct answer of 14x + 6y was quite often simplified to 20xy or 7x + 3y. Despite the question being deliberately structured, few candidates used their answer to part (a) in part (b) and very few used an algebraic method. Most candidates totalled the x values to 56, found 68 56 and then gave the correct answer of 2.68 4 = 17 was often seen. E9. Higher Tier Part (a) was well done by the majority, but many candidates are unable to deal with firstly the division by 5 and secondly the negative value for x. Answers of ±19 and 4 were common. Part (b) was very badly done. Too many candidates replace the inequality with an equals sign then fail to recover this in the answer. Others misinterpreted the question and gave answers of 6, 5, 4,. presumably assuming x is an integer. Many candidates introduced the sign. This was condoned this year but may not be in future. Intermediate Tier In (a) the correct answer was seen quite often but it seems to have been found mainly by trial and improvement. A correct algebraic solution was rarely seen. Of the candidates who tried to multiply both sides by 5, many made errors [usually sign errors] in the manipulation of the x term and/or the numbers Page 18 of 21

Part (b) was not well answered for a straightforward inequality. Many decided to convert to an equation and solved 3x + 8 = 29 to give an answer of x = 7 and consequently scored zero. Others just gave a list of integers which they thought fitted the inequality. E10. Intermediate Tier Part (a) was not well done with only about 20% of candidates gaining any marks. Most of these described the splitting into triangles method but some successfully remembered 180(n 2). There were many incorrect explanations with reference to angles on a straight line, opposite angles adding up to 180 and 4 90 0 = 360 0, being fairly popular. Part (b)(i) was very badly done with many candidates confusing an equation with an expression. Only about 20% of candidates gained any marks for part (b)(ii), with less than half of these able to give a fully correct answer. The rare candidate who managed to start with the correct equation was often let down by poor rearrangement skills, with, for example, 6x 18 = 270often leading to 6x = 252. Most candidates who scored marks did so by division by 6 and/or following through accurately from their answer for x to the largest angle. Higher Tier In part (a) the descriptions were often incorrect, a common misconception being that the opposite angles of a quadrilateral add up to 180 or that a quadrilateral was simply a square that had been stretched a bit so the four 90 angles became a little bit less or more but it all balanced out in the end. The two triangles method was the most successful but many correctly quoted (n 2) 180 0 and then substituted n = 4. Those candidates who tried to argue the exterior angle case often came to grief. 62% of candidates scored zero on this part. Part (b) was also disappointing. In (b)(i) an equation was required but many candidates only gave an expression. In part (b)(ii) the working was often very unclear and this led to errors in arithmetic, which might otherwise have been avoided. Credit was given for correct method and also follow through was allowed for the largest angle, but only for realistic answers (i.e., it had to be an obtuse angle). Only 43% of candidates scored all 3 marks in part (b)(ii). E11. Higher Tier This was well done by all candidates, virtually all of whom scored at least 2 marks. The common errors were not testing a 2 decimal place value to establish which 1 decimal place value was nearest to the root or finding a value of x such that the answer to x 3 + 7x was within 1 decimal place of 30. Page 19 of 21

Intermediate Tier Many candidates knew how to approach this problem and scored the first two marks for establishing a sandwich in the range 2.3 x 2.4 but then failed to score the final mark for carrying out a trial at 2.35 or 2.36 or 2.37. Others had problems deciding what the answer was since numbers like 30.02 were seen quite often on the answer line. E12. Intermediate Tier Most candidates scored at least two marks with many scoring three but it was rare to score four marks as most did not use a value of x that justified 2.9 as the answer rather than 2.8. Most did not do a trial between 2.85 and 2.876 inclusive. Some only did trials above the answer and so could only score half marks. Some gave the final answer as 2.88 which was not to one decimal place. Higher Tier This was almost always at least 3 marks. The main reason for losing the 4th mark is a failure to test a value (2.85, say) that confirms that 2.9 is the correct answer. Page 20 of 21

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