$(2 x + y) + (3 x - 4 y) i = (x - 2) + (4 y - 5) i$ $(2x+y)+(3x-4y)i=(x-2)+(4y-5)i$ Find x and y..?

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Answer 1 · 2018-05-12T16:05:00

$x = - \frac{21}{11}, y = - \frac{1}{11}$ $x=-21/11,y=-1/11$

We have

$(2 x + y) + (3 x - 4 y) i = (x - 2) + (4 y - 5) i$ $(2x+y)+(3x-4y)i=(x-2)+(4y-5)i$

We know two complex numbers $a + b i$ $a+bi$ and $c + d i$ $c+di$ iff $a = c$ $a=c$ and $b = d$ $b=d$ . So, in this case, we must have

$2 x + y = x - 2$ $2x+y=x-2$

$3 x - 4 y = 4 y - 5$ $3x-4y=4y-5$

so

$x = - y - 2$ $x=-y-2$

$3 (- y - 2) - 4 y = - 3 y - 6 - 4 y = - 7 y - 6 = 4 y - 5$ $3(-y-2)-4y=-3y-6-4y=-7y-6=4y-5$

so $11 y = - 1$ $11y=-1$ and $y = - \frac{1}{11}$ $y=-1/11$

Hence, $x = \frac{1}{11} - 2 = \frac{1}{11} - \frac{22}{11} = - \frac{21}{11}$ $x=1/11-2=1/11-22/11=-21/11$

We make a quick check:

$2 (- \frac{21}{11}) - \frac{1}{11} = - \frac{43}{11}$ $2(-21/11)-1/11=-43/11$

$- \frac{21}{11} - 2 = - \frac{21}{11} - \frac{22}{11} = - \frac{43}{11}$ $-21/11-2=-21/11-22/11=-43/11$

So, our values of $x$ $x$ and $y$ $y$ are correct.

(2x+y)+(3x-4y)i=(x-2)+(4y-5)i(2x+y)+(3x−4y)i=(x−2)+(4y−5)i(2x+y)+(3x-4y)i=(x-2)+(4y-5)i Find x and y..?