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Solve this pair of simultaneous equations: 3x + 2y = 4 and 2x + y = 3

2x + y = 3   therefore  y = 3 - 2x

Substitute y = 3 - 2x in to the first equation:

3x + 2(3 - 2x) = 4

3x + 6 - 4x = 4

-x = -2   therefore x = 2

Substitute x = 2 in to ei...

Answered by Tim C. Maths tutor
5442 Views

Solve x² - 3x - 2 = 0

To complete the square for the equation ax² + bx + c = 0 there are three steps.. 1. In order to complete the square, we need to take the first two parts of the quadtratic equation, and put them in the for...

Answered by Natalia M. Maths tutor
9221 Views

Use the equation 2(x+4) + 6(x-1) = 14, to find x.

Firstly, you need to get rid of any brackets. You do this by expanding them. This becomes 2x + 8 + 6x - 6 = 14.

The next step is to get all the x's on one side, and all the numbers on the other. Yo...

Answered by Ursula R. Maths tutor
3042 Views

Find dy/dx for y = x^3*e^x*cos(x)

In this problem, we see that y is a product of 3 functions of x. That means that in order to find dy/dx we need to use the product rule. The product rule tells us that in this case we should differentiate...

Answered by Lyudmil N. Maths tutor
9519 Views

Solve x² ≥ | 5x - 6 | (Question from AQA Core 3 June 2016)

To solve this question we can break it down in to a multiple small stages.

The first thing to tackle is the use of modulus around 5x - 6. This effectively means we have two equations to solve:

Answered by James H. Maths tutor
6217 Views

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