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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+...
First of all, these more complex tenses are often very complicated and difficult so do not feel discouraged if you do not understand them initially, but once you do you will be using the plus question par...
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 = 12xy - 2 = 6xy = 6x + 2Substitute y = 6x + 22x2 - 2(6x + 2) = 282x2 - 12x - 4 = 282(x2 - 6x - 2) = 28x2 - 6x ...
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