Solve the following equations by substitution method.
$x + 3 y = 10; 2 x + y = 5$

x = 1, y = 3

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

x = 2, y = 3

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

x = 1, y = 1

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

None of these

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

Open in App

Solution

The correct option is D $x = 1, y = 3$
$x + 3 y = 10$ ..... $(1)$ ;
$2 x + y = 5$ ........ $(2)$

$y = 5 - 2 x$ ......from $(2)$

putting in $(1)$ , we get

$\Rightarrow x + 3 (5 - 2 x) = 10$

$\Rightarrow x + 15 - 6 x = 10$

$\Rightarrow - 5 x = - 5$

$\Rightarrow x = 1$

$1 + 3 y = 10$
$y = 3$

Suggest Corrections

Similar questions

\frac{3}{x} + \frac{1}{y} = 7; \frac{5}{x} - \frac{4}{y} = 6. (x \neq 0, y \neq 0)

\frac{1}{2 x} + \frac{1}{3 y} = 2 (x \neq 0, y \neq 0) \frac{1}{3 x} + \frac{1}{y} = \frac{11}{3}

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

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

x - y = 1

2 x - 3 y = 2

Join BYJU'S Learning Program