Question

solve the equation 3x-2y=5 5x-y=3
by substitute method​

Answer 1

Answer:

Question : (3x-2y=5 ; 5x-y=3) Solve by substitution method

▶ Explanation :-

=> 3x - 2y = 5 -------- ( 1 )

=> 5x - y = 3 --------- ( 2 )

=> 5x - y = 3

=> 5x = 3 + y

=> x = 3 + y / 5 -------- ( 3 )

Substituting eq ( 3 ) in eq ( 1 )

=> 3 ( 3 + y / 5 ) - 2y = 5

=> 9 + 3y / 5 - 2y = 5

=> 9 + 3y - 10y / 5 = 5

=> 9 - 7y = 5 × 5

=> 9 - 7y = 25

=> - 7y = 16

=> y = - 16 / 7

Substituting value of y in eq ( 3 )

=> x = 3 + ( - 16 / 7 ) / 5

=> x = 3 - 16 / 7 /5

=> x = 21 - 16 / 35

=> x = 5 / 35

=> x = 1 / 7

Answer 2

S O L U T I O N :

$\underline{\bf{Given\::}}$

• 3x - 2y = 5...............(1)
• 5x - y = 3.................(2)

$\underline{\bf{Explanation\::}}$

From equation (1),we get;

$\mapsto\tt{3x - 2y = 5}$

$\mapsto\tt{3x = 5+2y}$

$\mapsto\tt{x = \dfrac{5+2y}{3} ................(3)}$

∴ Putting the value of x in equation (2),we get;

$\mapsto\tt{5\bigg(\dfrac{5+2y}{3} \bigg)-y = 3}$

$\mapsto\tt{\dfrac{25+10y}{3} -y = 3}$

$\mapsto\tt{25 + 10y - 3y = 9}$

$\mapsto\tt{25 + 7y= 9}$

$\mapsto\tt{ 7y= 9-25}$

$\mapsto\tt{ 7y= -16}$

$\mapsto\bf{y = -16/7}$

∴ Putting the value of y in equation (3),we get;

$\mapsto\tt{x = \dfrac{5+2\bigg(-\dfrac{16}{7} \bigg)}{3} }$

$\mapsto\tt{x = \dfrac{5+\bigg(-\dfrac{32}{7} \bigg)}{3} }$

$\mapsto\tt{x = \dfrac{5-\dfrac{32}{7} }{3} }$

$\mapsto\tt{x = \dfrac{\dfrac{35-32}{7} }{3} }$

$\mapsto\tt{x = \dfrac{\dfrac{3}{7} }{3} }$

$\mapsto\tt{x = \dfrac{\cancel{3}}{7} \times \dfrac{1}{\cancel{3} }}$

$\mapsto\bf{x = 1/7}$

Thus,

The value of x & y will be 1/7 and -16/7 .