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Solve for x and y: 2x +5y + 5= 0 , 2y + 31= 5x

For a question like this you should aim to eliminate either x OR y from one equation in order to deduce the value of the other. 1) 2x +5y + 5= 0 , 2) 2y + 31= 5x Rearrange equation 2) so that 2y +31= 5x --&g...
TD
Answered by Tutor114325 D. • Maths tutor
5767 Views

A cuboid has length x cm. The width of the cuboid is 4 cm less than its length. The height of the cuboid is half of its length. The surface area of the cuboid is 90 cm^2 . Show that 2x^2 − 6x − 45 = 0

Take each side of the cuboid as an algebraic expression and multiply each by 2 to account for both sides of the shape. For example, (x)(x-4), which could be expanded to x 2 -4x, and then multiplied by 2 to r...
TM
Answered by Tom M. • Maths tutor
12405 Views

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
5582 Views

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
12972 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
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