Solve the following equations by the substitution method
$11 x - 8 y = 27, 3 x + 5 y = - 7$

x = 0, y = 1

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

x = 0, y = - 5

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

x = 2, y = 3

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

x = 1, y = - 2

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

Open in App

Solution

The correct option is D $x = 1, y = - 2$
The equations are
$\Rightarrow 11 x - 8 y = 27.... e q (1)$
$\Rightarrow x = \frac{27 + 8}{11}$
$\Rightarrow 3 x + 5 y = - 7.... e q (2)$
Substitute the value of x in eq(2 )
$\Rightarrow \frac{3 (27 + 8 y)}{11} + 5 y = - 7$
$\Rightarrow 81 + 24 y + 55 y = - 77 \Rightarrow 79 y = - 158 \Rightarrow y = - 2$
Substitute the value of y in eq (2 )
$\Rightarrow 3 x + 5 (- 2) = - 7 \Rightarrow 3 x = 3 \Rightarrow x = 1$
$\Rightarrow x = 1, y = - 2$

Suggest Corrections

Similar questions

\frac{1}{x} + \frac{2}{y} = 9; \frac{2}{x} + \frac{1}{y} = 12 (x \neq 0, y \neq 0)

\frac{15}{x} + \frac{2}{y} = 17, \frac{1}{x} + \frac{1}{y} = \frac{36}{5}, x \neq 0, y \neq 0

\frac{1}{x} + \frac{1}{y} = 4

\frac{2}{x} + \frac{3}{y} = 7, x \neq 0, y \neq 0

Question 1 (iii)
Solve the following pair of linear equations by the elimination method and the substitution method:
3x - 5y - 4 = 0 and 9x = 2y + 7

\sqrt{2} x + \sqrt{3} y = 0, \sqrt{3} y - \sqrt{8} y = 0

3 x - 5 y + 1 = 0; x - y + 1 = 0

0.2 x + 0.3 y = 13, 0.4 x + 0.5 y = 2.3

(P . S)

Join BYJU'S Learning Program