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Expand the following brackets: (2x + 1)(x + 3)

The FOIL technique (First, Outside, Inside, Last) is very useful when expanding out brackets. Using this method we have:2x * x = 2x 2 (First) 2x * 3 = 6x (Outside) 1 * x = x (Inside) 1 * 3 = 3 (Last)So overa...
BA
Answered by Ben A. • Maths tutor
4374 Views

Solve the following system of simultaneous equations: y=x^2 - 5x + 4 ; x+2y=19

Sub y=x^2 - 5x + 4 into x+2y=19x+2(x^2 -5x + 4) = 19x + 2x^2 -10x + 8 =192x^2 - 9x -11 = 0 (2x -11)(x+1) = 0x= 5.5x=-1
SR
Answered by Sulamaan R. • Maths tutor
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differentiate with respect to x. i). x^(1/2) ln (3x),

From this we can see that equation has 2 parts therefore we should look to using the product rule which is used to differiantiate a two functions multiplied together so (fg)'=f'g+fg'. In this question the di...
JD
Answered by Jesse D. • Maths tutor
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Solve the quadratic equation, x^2 - 4x -5 = 0

Solve x 2 - 4x - 5 = 0 A very full explanation would be: Start by listing the pairs of numbers that can be multiplied together to get 1 * -5 = -5 (i.e. the coefficient of the first term multiplied by the las...
RH
Answered by Ryan H. • Maths tutor
5631 Views

Solve for y and x, where x=y+1 and y=2x+3

Simultaneous equations: x=y+1 y=2x+3 First solve for x. y=2x+3 so first equation becomes: x=(2x+3)+1 x-2x-3-1=0 -1x-4=0x+4=0 x=-4 Solve for y: y=2x+3 y=2(-4)+3 y=-8+3y=-5
TS
Answered by Tom S. • Maths tutor
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