Solving linear equations and rearranging formulae – Year 9 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. Add 9 to both sides: \(4x=28\). Divide by 4: \(x=\)7. Check: \(4\times7-9=19\). (1 mark)
   2. Divide both sides by 5 first: \(x+2=8\). Subtract 2: \(x=\)6. Expanding first, \(5x+10=40\), gives the same answer. (1 mark)
   3. Subtract \(4x\) from both sides to collect the x terms on the left: \(3x+3=24\). Subtract 3: \(3x=21\) (1 mark for collecting correctly). Divide by 3: \(x=\)7 (1 mark).

   Check (c): the left side is \(7\times7+3=52\) and the right side is \(4\times7+24=52\), so both sides balance. Moving the smaller x term keeps the coefficient positive and avoids sign errors. Check (b): \(5\times(6+2)=40\).

2. Answer to Question 2 (4 marks)

   1. Multiply both sides by 3 to clear the fraction: \(x+4=21\) (1 mark). Subtract 4: \(x=\)17 (1 mark). Check: \((17+4)\div3=7\).
   2. Add 1 to both sides: \(\dfrac{x}{2}+4=x\) (1 mark). Multiply every term by 2: \(x+8=2x\). Subtract x from both sides: \(x=\)8 (1 mark).

   Check (b): the left side is \(8\div2+3=7\) and the right side is \(8-1=7\). When multiplying an equation by a number, every term on both sides must be multiplied, not just the fraction; forgetting to double the 4 would give \(x+4=2x\) and the wrong answer \(x=4\).

3. Answer to Question 3 (4 marks)

   1. The perimeter is twice the length plus twice the width: \(2(3x-1)+2(x+2)=42\). (1 mark)
   2. Expand: \(6x-2+2x+4=42\), which simplifies to \(8x+2=42\) (1 mark). Subtract 2: \(8x=40\). Divide by 8: \(x=\)5 (1 mark).
   3. Substitute \(x=5\): length \(=3\times5-1=\)14 cm and width \(=5+2=\)7 cm. (1 mark)

   Check: \(2\times14+2\times7=28+14=42\) cm, the perimeter in the question. The rectangle's area would be \(14\times7=98\) cm². A common error is using only one length and one width, \(4x+1=42\), which gives \(x=10.25\).

4. Answer to Question 4 (4 marks)

   1. Undo the operations on x in reverse order. Add 3 to both sides: \(y+3=5x\) (1 mark). Divide both sides by 5: \(x=\dfrac{y+3}{5}\) (1 mark).
   2. In \(V=lwh\), h is multiplied by both l and w, so divide both sides by their product lw (1 mark): \(h=\dfrac{V}{lw}\) (1 mark).

   Check (a) with a number: if \(x=2\), then \(y=10-3=7\), and the new formula gives \(\dfrac{7+3}{5}=2\). Check (b): a cuboid 5 by 4 by 3 has volume 60, and \(\dfrac{60}{5\times4}=3\), the height. Writing \(x=\dfrac{y}{5}+3\) in (a) would be wrong because the steps were undone in the wrong order.

5. Answer to Question 5 (4 marks)

   1. If a tickets are for adults, the other \(30-a\) are for children. Adults pay \(9a\) and children pay \(6(30-a)\), so \(9a+6(30-a)=219\). (1 mark)
   2. Expand: \(9a+180-6a=219\), which simplifies to \(3a+180=219\) (1 mark). Subtract 180: \(3a=39\). Divide by 3: \(a=\)13 (1 mark).
   3. Subtract the adult tickets from the total: \(30-13=\)17 child tickets. (1 mark)

   Check the money: \(9\times13+6\times17=117+102=219\) pounds, and \(13+17=30\) tickets. Using one letter and \(30-a\) avoids needing two separate equations.

Total : 20 marks