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If we have two equations that look like this:
a1 * x + b1 * y = c1
and
a2 * x + b2 * y = c2
where x,y are variables and a1,a2,b1,b2 are coeficients.
then we solve it using the fol...

Expand: x² + 2x + 3x +6 = 2x+4
Take (2x+4) from both sides: x²+3x+2 = 0
Facorise: (x+2)(x+1) = 0
So solutions are x=-2 and x=-1

Using "Integration by parts" or "reverse chain rule" .
Recall formula for intergration by parts: "∫f'(x) g(x) dx = f(x)g(x) - ∫f(x)g'(x)dx"
Then set f'(x) = 1, g(x)...

Use the chain rule: f'(x) = v(du/dx) +u(dv/dx).

Let u = 2x, du/dx = 2, v = lnx, dv/dx = 1/x

Using this information: f'(x) = 2lnx + 2x/x

This simplifies to f'(x...

First factorise the equation into 2 sets of brackets: You can do this by looking at the constant in the quadratic, 8. Try and find 2 numbers that multiply together to make 8 (such as 4 and 2), and that...