So, we can write it like this: x+y=3 x*y=12 from the first we know that x+y=3 y=3-x so: x*y=12 can be written as: x*(x-3)=12 so x^2-3x=12 Actually this equation does not have a solution within the real numbers but we can calculate: x^2-3x-12=0 and if we use the quadratic formula, we will have: (remember the quadratic formula is the following, where for us a=1,b=-3 and c=-12) [latex] \frac{-b- \sqrt{b*b-4a*c} }{2a} [/latex] [latex] \frac{3- \sqrt{3*3-4*(-12)} }{2} = \frac{3- \sqrt{9+48} }{2} = \frac{3- \sqrt{57} }{2} [/latex] So this would be the solution, where the other number would be [latex] 3-\frac{3- \sqrt{57} }{2}= \frac{6}{2}-\frac{3- \sqrt{57} }{2}=\frac{3- \sqrt{57} }{2}[/latex] The other solution would be the following one; x=[latex]\frac{3+\sqrt{57} }{2} [/latex], y=[latex]1\frac{1+\sqrt{57} }{2} [/latex]

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