(x+2y)+(2x-3y)i+4i=5 find the value of x and y .
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Answered by
35
Given (x + 2y) + (2x - 3y)i + 4i = 5.
(x + 2y) + (2x - 3y + 4)i = 5 + 0i
x + 2y = 5 = > Then x = 5 - 2y ----- (1)
2x - 3y + 4 = 0 = > 2(5 - 2y) - 3y + 4 = 0
10 - 4y - 3y + 4 = 0
10 - 7y + 4 = 0
- 7y = -14
y = 2.
Substitute y = 2 in (1), we get
x = 5 - 4
= 1.
The values are x = 1 and y = 2.
Hope this helps!
(x + 2y) + (2x - 3y + 4)i = 5 + 0i
x + 2y = 5 = > Then x = 5 - 2y ----- (1)
2x - 3y + 4 = 0 = > 2(5 - 2y) - 3y + 4 = 0
10 - 4y - 3y + 4 = 0
10 - 7y + 4 = 0
- 7y = -14
y = 2.
Substitute y = 2 in (1), we get
x = 5 - 4
= 1.
The values are x = 1 and y = 2.
Hope this helps!
Answered by
7
(x+2y)+(2x-3y)i+4i=5
5 is real part
real part equation
x+2y=5=x=5-2y
imaginary part
2x-3y+4=0
2 (5-2y)-3y=-4
10-4y-3y=-4
-7y=-14
y=2,and x=1
5 is real part
real part equation
x+2y=5=x=5-2y
imaginary part
2x-3y+4=0
2 (5-2y)-3y=-4
10-4y-3y=-4
-7y=-14
y=2,and x=1
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