How do you solve $2 i x - 5 + 3 i = (2 - i) x + i$ $2ix - 5 + 3i = (2 - i)x + i$ ?

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Answer 1 · 2015-12-05T20:06:47

$x = \frac{(- 16 - 11 i)}{5}$ $x=((-16-11i))/(5)$

Given: $2 i x - 5 + 3 i = (2 - i) x + i$ $2ix-5+3i=(2-i)x+i$

Multiply out the brackets

$2 i x - 5 + 3 i = 2 x - i x + i$ $2ix-5+3i=2x-ix+i$

Collecting like terms

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

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

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

$x=((5-2i))/((-2+3i))......................(1)$

Using $(a^2-b^2)=(a-b)(a+b)$
Multiply equation (1) by 1 in the form of $((-2-3i))/((-2-3i))$

$x=((5-2i)(-2-3i))/((-2+3i)(-2-3i)) color(white)(..)=color(white)(..)(-16-11i)/(5)$

How do you solve 2ix - 5 + 3i = (2 - i)x + i2ix−5+3i=(2−i)x+i2ix - 5 + 3i = (2 - i)x + i?