If $(x + y, 1)$ $=$ $(3, y - x)$ , then $x$ $=$ , $y$ $=$

2, 1

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

-1, -2

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

1, 2

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

2, -1

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

Open in App

Solution

The correct option is A 1, 2
We have, $(x + y, 1) = (3, y - x)$
$⟹ x + y = 3$ and $y - x = 1$
$\Rightarrow x + y = 3$ and $- x + y = 1$
Adding both equations, we get
$2 y = 4$
$\Rightarrow y = 2$
Put the value of $y$ in $x + y = 3$ , we get
$x + 2 = 3$
$\Rightarrow x = 1$
So, $C$ is the correct option.

Suggest Corrections

Similar questions

\frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + . . . . . ., then \frac{d y}{d x} =______________________________.

If y = 1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \frac{x^{4}}{4!} - \frac{x^{5}}{5!} + . . . ., then \frac{dy}{dx} =________________________.

x = \frac{1}{1^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + . . . .,

y = \frac{1}{1^{2}} + \frac{3}{2^{2}} + \frac{1}{3^{2}} + \frac{3}{4^{2}} + . .

z = \frac{1}{1^{2}} - \frac{1}{2^{2}} + \frac{1}{3^{2}} - \frac{1}{4^{2}} + . . . .,

y = 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \frac{x 4}{4!}_________________, then \frac{d^{2} y}{d x^{2}} =__________________.

Join BYJU'S Learning Program