Practise Simultaneous equations. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy3 marks
Solve the simultaneous equations \(3x + 4y = 51\) \(5x - 4y = 21\) You must show all your working.[3]
Show the answer and mark scheme
Answer: \(x = 9, \ y = 6\)
M1 for a correct method to eliminate \(x\) or \(y\) (e.g. coefficients of one variable made equal and the equations added or subtracted, allow one arithmetic error)
M1 (dep) for substituting their value into one of the equations or for a correct method to find the second variable
A1 for \(x = 9\) and \(y = 6\)
Worked solution: Add to eliminate \(y\): \(8x = 72\), so \(x = 9\). Substitute into \(3x + 4y = 51\): \(4y = 24\), so \(y = 6\).
Question 2Medium3 marks
Solve the simultaneous equations \(x - 2y = -11\) \(7x - 4y = -27\) You must show all your working.[3]
Show the answer and mark scheme
Answer: \(x = -1, \ y = 5\)
M1 for a correct method to eliminate \(x\) or \(y\) (e.g. coefficients of one variable made equal and the equations added or subtracted, allow one arithmetic error)
M1 (dep) for substituting their value into one of the equations or for a correct method to find the second variable
A1 for \(x = -1\) and \(y = 5\)
Worked solution: Multiply the first equation by \(2\): \(2x - 4y = -22\) and \(7x - 4y = -27\). Subtract to eliminate \(y\): \(5x = -5\), so \(x = -1\). Substitute into \(x - 2y = -11\): \(-2y = -10\), so \(y = 5\).
Question 3Hard6 marks
A rectangle has a perimeter of 34 cm. The diagonal of the rectangle is 13 cm.
(a) Work out the area of the rectangle.[4]
(b) A second rectangle also has a perimeter of 34 cm. Its area is 10 cm2 more than the area of the first rectangle. Work out the exact length of its diagonal.[2]
Show the answer and mark scheme
(a)Answer: 60 cm²
P1 for \(a + b = 17\)
P1 for \(a^2 + b^2 = 169\) (Pythagoras)
P1 for a method to find \(ab\), e.g. \((a + b)^2 - (a^2 + b^2) = 2ab\), or substituting \(b = 17 - a\) to get \(a^2 - 17a + 60 = 0\) (so the sides are 5 and 12)
A1 for 60
Worked solution: Let the sides be \(a\) and \(b\). \(2(a + b) = 34\) so \(a + b = 17\), and \(a^2 + b^2 = 13^2 = 169\). \((a + b)^2 = a^2 + 2ab + b^2\), so \(289 = 169 + 2ab\) and \(ab = 60\). (The sides are 5 cm and 12 cm.) Area = 60 cm2.
(b)Answer: \(\sqrt{149}\) cm
M1 for \(d^2 = 17^2 - 2 \times 70\) (= 149), or finding the sides 7 and 10
A1 for \(\sqrt{149}\)
Worked solution: Now \(a + b = 17\) and \(ab = 70\), so \(a^2 + b^2 = 17^2 - 2 \times 70 = 149\). (The sides are 7 cm and 10 cm.) Diagonal \(= \sqrt{149}\) cm.