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Dvinal [7]
1 year ago
5

Find the real numbers x and y. 27+yi= 11-x + 4i

Mathematics
1 answer:
Korvikt [17]1 year ago
8 0

\quad \huge \quad \quad \boxed{ \tt \:Answer }

\qquad \tt \rightarrow \:x = -16

\qquad \tt \rightarrow \: y =4

____________________________________

\large \tt Solution  \: :

\qquad \tt \rightarrow \: 27 + yi = 11 - x + 4i

\qquad \tt \rightarrow \: yi + x = 11 + 4i - 27

\qquad \tt \rightarrow \: x + yi =  - 16 + 4i

By equating both equations we get :

\qquad \tt \rightarrow \: x =  - 16

\qquad \tt \rightarrow \: y = 4

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

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The equations -6x + 3y = 6 and x² + y = 10 intersect at the
eimsori [14]

The values of the letters in coordinates (a, b) and (c,d) are;

<u><em>a = -4</em></u>

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We are given two equations;

-6x + 3y = 6  ---(eq 1)

x² + y = 10 ---(eq 2)

  • We are told that they intersect at coordinates; (a, b) and (c, d).

Let us make y the subject in eq 2 to get;

y = 10 - x²  --(eq 3)

  • Let us put 10 - x² into eq 1 to get;

-6x + 3(10 - x²) = 6

expanding further gives;

-6x + 30 - 3x² = 6

rearranging gives;

3x² + 6x - 24 = 0

Using online quadratic equation <em>solver</em>, we have;

x = -4 and x = 2

Putting x = -4 into eq 3 gives;

y = 10 - (-4)²

y = 10 - 16

y = -6

Putting x = 2 into eq 3 gives;

y = 10 - (2)²

y = 10 - 4

y = 6

  • Thus, the coordinates are; (-4, -6) and (2, 6)

Comparing with (a, b) and (c,d), we have;

a = -4

b = -6

c = 2

d = 6

Read more at; brainly.com/question/15165519

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