Solve for x and y :
$x + y = 17; x^{2} + y^{2} = 169$

x = 9, y = 12

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

x = 5, y = 12

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

x = 4, y = 12

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

x = 3, y = 12

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

Open in App

Solution

The correct option is B $x = 5, y = 12$
Given, $x + y = 17$ .....(i)
$x^{2} + y^{2} = 169$ .....(ii)
We write the equation (i) as
$x = 17 - y$ ....(iii)
On substituting $x = 17 - y$ in equation (ii). we get,
$(17 - y)^{2} + y^{2} = 169$
$\Rightarrow 17^{2} + y^{2} - 34 y + y^{2} = 169$
$\Rightarrow 289 + y^{2} - 34 y + y^{2} = 169$
$\Rightarrow 2 y^{2} - 34 y + 289 - 169 = 0$
$\Rightarrow 2 y^{2} - 34 y + 120 = 0$
$\Rightarrow y^{2} - 17 y + 60 = 0$
$\Rightarrow (y - 12) (y - 5) = 0$
$∴ y = 12$ and $y = 5$
On substituting $y = 12$ and $y = 5$ in equation (iii). we get,
$x = 17 - 12 = 5$ and $x = 17 - 5 = 12$
$∴ x = 12, y = 4$ or $x = 5, y = 12$

Suggest Corrections

Similar questions

\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}

\frac{1}{2 x} + \frac{1}{3 y} = 2 \frac{1}{3 x} + \frac{1}{2 y} = \frac{13}{6}

Which of the following pair of algebraic expression will add up to give 12xy - 7?

\frac{x^{2}}{y} + \frac{y^{2}}{x} = 18, x + y = 12

Which expression matches the statement
"Subtract the sum of $5 x$ and $3 x$ from $12 y$ ”?

Join BYJU'S Learning Program