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We have 2 fractions we want to get rid off to make it easier, so what we do is multiply both sides by the things we want to get rid off.In this case we multiply both sides by (2x-3)(x+1)This gives us x(x+...

So u=2+lnx, therefore du/dx=1/x , we can work out the new upper and new lower limit by substitute in e and 1 into 2+lnx , and we get 2+ln(e)=3 , 2+ln(1)=2Rearrange the differential we get dx=xdu , substit...

x²-5x+4 <0First we ignore the inequality and try to solve the equation x²-5x+4=0, which we do via factorising (x-4)(x-1)=0. x = 4 or x=1We draw the graph using our solution, going...

2x² - 2y = 282y - 4 = 12x
y - 2 = 6xy = 6x + 2
Substitute y = 6x + 2
2x² - 2(6x + 2) = 282x² - 12x - 4 = 282(x² - 6x - 2) = 28x² - 6x ...

Call the two numbers x and y. The constraint is that x + y =100, and we need to maximise A=xy.
Rearrange the constraint to y = 100 - x, and substitute into the product equation.
A = x(100-x) = ...