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The quadratic equation is x=(-b+-SQRT(b^2-4ac))/2a. In this instance, a = 3, b = 2, and c = 15. Simply putting the numbers in place of the letter counterpart, gives an answer of 1.93 and -2.59 to two deci...

First of all, you would label the equations, so let's call 3x +4y = 18 equation A, and 5x - 2y = 4 equation B. Then you would try to find a way to eliminate one of the variables (either x or y). We can do...

I have labelled the two separate equations A and B so that it is easier to talk about them. There are two ways in which you can do these equations but I am going to explain the method using substitution. ...

Although this may look more complicated than normal, this is just a normal qudratic equation. First by rearranging the equation we can get it into a simpler form, and then we can go about solving it. The ...

Pythagoras' theorem is: a^2+b^2=c^2 (a=short side, b=short side, c=longest side/hypotenuse, ^=squared). Now applying that to this question would mean that a=PQ, b=PR and c=QR. So we can use the figures gi...