Solve for $x$ and $y$ where $x$ and $y$ are real numbers
$2 y + i x = 4 + x - i$

x = - 1; y = - \frac{3}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = - 1; y = \frac{3}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x = 1; y = \frac{3}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = 1; y = - \frac{3}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $x = - 1; y = \frac{3}{2}$
Given, $2 y + i x = 4 + x - i$
First equating real terms, we get
$2 y = 4 + x$
And now equating imaginary terms,
$x = - 1$
Putting this value of $x$ in equation (1), we get
$2 y = 4 + x$
$\Rightarrow 2 y = 4 - 1$
$\Rightarrow 2 y = 3$
$\Rightarrow y = \frac{3}{2}$
Therefore, $x = - 1$ , and $y = \frac{3}{2}$

Suggest Corrections

Similar questions

\frac{2}{x} + \frac{3}{y} = 6 and \frac{1}{x} + \frac{1}{2 y} = 2

x = 1, y = \frac{2}{3}

x = \frac{2}{3}, y = 1

x = 1, y = \frac{3}{2}

x = \frac{3}{2}, y = 1

(\sqrt{32})^{x} \div 2^{y + 1} = 1

8^{y} - 16^{4 - x / 2} = 0

\frac{3}{x + y} + \frac{2}{x - y} = 2 and \frac{9}{x + y} - \frac{4}{x - y} = 1

x = \frac{1}{2}, y = \frac{3}{2}

x = \frac{5}{2}, y = \frac{1}{2}

x = \frac{3}{2}, y = \frac{1}{2}

x = \frac{1}{2}, y = \frac{5}{2}

3 (2 x + y) = 7 x y 3 (x + 3 y) = 11 x y

Solve for $x$ , if :

${(\sqrt{32})}^{x} \div 2^{y + 1} = 1$ and $8^{y} - 16^{4 - \frac{x}{2}} = 0$

Join BYJU'S Learning Program