There are a few ways of solving these simultaneous equations, but most of them involve substitution. Perhaps the most intuitive substitution to make is the second equation (y=3x+5) into the first one (xy=2), giving x (3x+5) = 2Expanding and bringing all terms to one side, we obtain a quadratic equation:3x2 +5x = 23x2 +5x -2 =0.We notice that we can factorise this quadratic to solve it (alternatively, we could use the quadratic formula or complete the square)(3x-1) (x+2) =0So we have that either x = 1/3, or x= -2.We now consider what y must be when x= 1/3. From the first equation (xy=2), we are looking for y such that y/3=2, so y=6.Now consider what y must be when x=-2. Similarly, we look for y such that -2y=2, so y=-1.Therefore the set of solutions is:x=-2, y=-1and x= 1/3, y=6.
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