MHB Find x and y if x, y are members of real numbers and: (x+i)(3-iy)=1+13i

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To solve for x and y in the equation (x+i)(3-iy)=1+13i, the initial expansion leads to two equations: 3x + y = 1 and -xy + 3 = 13. From the first equation, y can be expressed as y = 1 - 3x. Substituting this into the second equation results in a quadratic equation, which factors to (3x + 5)(x - 2) = 0. The solutions yield the pairs (x, y) = (-5/3, 6) and (2, -5), confirming the calculations.
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Find x and y if x, y are members of real numbers and: (x+i)(3-iy)=1+13i

I first expanded it to give: 3x-yix+3i+y=1+13i
Then I equaled 3x+y=1 and -yx+3=13
But afterwards I do other steps and get the wrong answer.
 
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Your approach and posted work are ok.

To continue,

$$3x+y=1\implies y=1-3x$$

$$3-xy=13\implies xy=-10$$

$$x(1-3x)=-10$$

$$(3x+5)(x-2)=0$$

$$(x,y)=\left(-\frac53,6\right),(2,-5)$$

Does that match your results?
 

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