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Solve the simultaneous equations: 3x + 2y = 4 and 4x + 5y = 17

Step 1: multiply one or both equations so that the 2 equations have the same coefficient for either x or y (pick easier one) 5(3x + 2y) = 5(4) --> 15x + 10y = 20 AND 2(4x + 5y) = 2(17) --> 8x + 10y = 3...
RD
Answered by Rania D. • Maths tutor
5537 Views

Solve 2sin2θ = 1 + cos2θ for 0° ≤ θ ≤ 180°

sin2θ = 2sinθcosθ (double angle formula for sine)cos2θ = cos 2 θ - sin 2 θ (double angle formula for cosine) = 2cos 2 θ - 1 (utilising the trignometric identity sin 2 θ + cos 2 θ = 1) We substitute these int...
SN
Answered by Samuel N. • Maths tutor
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Solve algebraically the simultaneous equations x^2 + y^2 = 25 and y − 2x = 5 (5 marks)

First consider each equation separately and label them with a number. x 2 + y 2 = 25 (1) y - 2x = 5 (2) This question is difficult as it involves square numbers, unlike a normal simultaneous equation. Hence ...
KS
Answered by Karisma S. • Maths tutor
12845 Views

Factorise y^2 + 27y and simplify w^9/w^4

y 2 + 27y = y(y + 27)To factorise you need to find the common factor between each part of the equation. In this case y is common between the different parts of the equation. Therefore you take y outside of t...
LM
Answered by Lucy M. • Maths tutor
9097 Views

there are 11 sweets in a box four are soft centred and seven hard centred sweets two sweets are selected at random a)calculate the probability that both sweets are hard centred, b) one sweet is soft centred and one sweet is hard centred

a) First sweet you pick will be soft centred in 7 out of 11 cases, so the probability is 7/11when you are picking up a second sweet there are only 6 hard centred left and a total of 10, so the probability is...
AT
Answered by Artem T. • Maths tutor
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