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5 and 7
[tex]x + y = 12[/tex]
[tex]x(y) = 35[/tex]
[tex]y=12-x[/tex]
Substitute x for y
[tex]x(12-x) = 35[/tex]
Expand [tex]x(x-12)[/tex]
[tex]-x^2 +12x = 35[/tex]
Subtract 35 to both sides
[tex]-x^2 + 12x - 35[/tex]
All form of equation [tex]ax^2 + bx + c[/tex] can be solved using quadratic formula [tex]x = \frac{-b±\sqrt{b^2-4ac} }{2a}[/tex]
Thus, a = -1, b = 12 and c = -35
Substitute the value in each variables
[tex]x=\frac{-(12)±\sqrt{12^2-4(-1)(-35)} }{2(-1)}[/tex]
[tex]x=\frac{-12±\sqrt{144-140} }{-2}[/tex]
[tex]x = \frac{-12±\sqrt{4} }{-2}[/tex]
Therefore [tex]x = 5[/tex] or [tex]x = 7[/tex]
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