Question

Solve the following system of equaions by substitution method: x+2y=5 and 2x+y=4.​

Answer 1

Given :

• x + 2y = 5
• 2x + y = 4

To Find :

• The value of 'x' and 'y' by substitution method.

According to the question,

⇒ x + 2y = 5 - - - - - (i)

⇒ 2x + y = 4 - - - - - (ii)

From equation (i),

⇒ x = 5 - 2y

Putting the value of 'x' in equation (ii),

⇒ 2x + y = 4

⇒ 2(5 - 2y) + y = 4

⇒ 10 - 4y + y = 4

⇒ 10 - 3y = 4

⇒ - 3y = 4 - 10

⇒ - 3y = - 6

⇒ y = 2

Now,

⇒ x + 2y = 5

⇒ x + 2(2) = 5

⇒ x = 5 - 4

⇒ x = 1

• So,the value of x is 1 and y is 2.

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Answer 2

$\underline{\underline{\mathscr{Given\::-}}}$

• Eqⁿ (1) = x + 2y = 5

• Eqⁿ (2) = 2x + y = 4

$\underline{\underline{\mathscr{To\:Find\::-}}}$

• We have to solve these equations with substitution method.

$\underline{\underline{\mathscr{Solution\::-}}}$

★ From Eqⁿ (1), we get :-

$\qquad\leadsto\quad{\bf{\pink{x = 5 - 2y}}}$

★ Putting value of 'x' in Eqⁿ (2) :-

$\qquad\leadsto\quad\sf 2(5 - 2y) + y = 4$

$\qquad\leadsto\quad\sf (2\:\times\:5) - (2\:\times\:2y) + y = 4$

$\qquad\leadsto\quad\sf 10 - 4y + y = 4$

$\qquad\leadsto\quad\sf 10 - 3y = 4$

$\qquad\leadsto\quad\sf -3y = 4 - 10$

$\qquad\leadsto\quad\sf -3y = -6$

$\qquad\leadsto\quad\sf y = \dfrac{-6}{-3}$

$\qquad\leadsto\quad\sf y = {\cancel{\dfrac{-6}{-3}}}\:(Cancelling)$

$\qquad\leadsto\quad{\bf{\purple{y = 2}}}$

Now,

★ Putting value of 'y' in Eqⁿ (2) :-

$\qquad\leadsto\quad\sf 2x + 2 = 4$

$\qquad\leadsto\quad\sf 2x = 4 - 2$

$\qquad\leadsto\quad\sf 2x = 2$

$\qquad\leadsto\quad\sf x = \dfrac{2}{2}$

$\qquad\leadsto\quad\sf x = {\cancel{\dfrac{2}{2}}}\:(Cancelling)$

$\qquad\leadsto\quad{\bf{\red{x = 1}}}$

$\therefore\:{\underline{\mathscr{Hence,\:value\:of\:x\:and\:y\:=\:\bf{1}\:\mathscr{and}\:\bf{2}}}}$

★ More to know :-

• Substitution method is used to solve linear equations. In this method value of one variable from one eqⁿ us substituted in the other equation.

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