Question

Solve the following pair of linear equations by the elimination method and the substitution method x + y is equal to 5 and 2 x minus 3 y is equal to 4

Answer 1

Answer :

The values are :

x = 6/5

y = 19/5

Given :

The equations are :

• x + y = 5
• 2x - 3y = 4

Task :

To solve the pairs of equation by both :

• The elimination method and
• The substitution method

Solution :

By elimination method :-

$\sf x + y = 5\\\\ \sf Multiplying \: \: by \: \: 2 \: \: on \: \: both \: \: side \\\\ \sf \implies 2x + 2y = 10 \: ........... (1)$

and

$\sf \implies 2x - 3y = 4 \: ..........(2)$

Subtracting (2) from (1) to eliminate the term 2x :

$\sf \implies 2x + 2y -(2x - 3y) = 10 - 4 \\\\ \sf \implies 2x + 2y - 2x + 3y = 6 \\\\ \sf \implies 5y = 6 \\\\ \sf \implies y = \dfrac{6}{5}$

Using the value of y in (1)

$\sf \implies x + \dfrac{6}{5}= 5 \\\\ \sf \implies x = 5 - \dfrac{6}{5}\\\\ \sf \implies x = \dfrac{25-6}{5}\\\\ \sf \implies x = \dfrac{19}{5}$

Substitution method :

$\sf x + y = 5 \\\\ \implies \sf x = 5 - y \: .......... (3)$

and

$\sf \implies 2x - 3y = 4 \: .........(4)$

Substituting the value of x from (3) in (4) :

$\sf \implies 2 (5 -y) - 3y = 4 \\\\ \sf \implies 10 - 2y - 3y = 4 \\\\ \sf \implies -5y = -6 \\\\ \sf \implies y = \dfrac{6}{5}$

Putting the value of y in (3) we have:

$\sf \implies x = 5 - \dfrac{6}{5} \\\\ \sf \implies x = \dfrac{19}{5}$

Answer 2

$\large\bold{\red{Value\:of-}}$

→$\sf x = \dfrac{19}{5}$

→ $\sf y = \dfrac{6}{5}$

Step-by-step explanation:

Given -

$\sf x + y = 5$ --------- ( a )

$\sf 2x - 3y = 4$ -------- ( b )

To Find -

$\sf Value\: of \:x\: and\: y$

Solution -

$\large{\red{\underline{\underline{\textsf{Elimination\:Method}}}}}$

→ Multiply equation ( a ) by 2

Now , 2x + 2y = 10 --------- ( c )

Now solving equation ( a ) and ( c )

2x + 2y = 10

2x - 3y = 04

_______________

(-) (+) = (-)

_______________

0 + 5y = 6

_______________

→ y = $\sf\dfrac{6}{5}$

Now , substituting value of y in eqn ( a )

→ x + y = 5

→ x + $\sf \dfrac{6}{5}$ = 5

→x = 5 - $\sf \dfrac{6}{5}$

→x = $\sf \dfrac{25 - 6}{5}$

→x = $\sf \dfrac{19}{5}$

$\rule{200}2$

$\large{\red{\underline{\underline{\textsf{Substitution\:Method}}}}}$

From eqn ( a )

$\sf x = 5 - y$

Substituting in eqn ( b )

$\sf 2x - 3y = 4$

$\sf 2 ( 5 - y ) - 3y = 4$

$\sf 10 - 2y - 3y = 4$

$\sf 10 - 4 = 5y$

$\sf y = \dfrac{6}{5}$

Now , substituting value of y in eqn ( a )

→ x + y = 5

→ x + $\sf \dfrac{6}{5}$ = 5

→x = 5 - $\sf \dfrac{6}{5}$

→x = $\sf \dfrac{25 - 6}{5}$

→x = $\sf \dfrac{19}{5}$

$\rule{200}2$