Question

just want the steps of these problem (x+3)/(x-2)=(x+4)/(x-3),thanks

Answer 1

Given:

$\sf{ \frac{x + 3}{x - 2} = \frac{x + 4}{x - 3} }$

On cross multiplication,we get:

→x²-9=x²+4x-2x-8

Regrouping and cancelling like terms,

→4x-2x=8-9

→2x= -1

→x= -1/2

Answer 2

Answer :-

x = - 1/2

Solution :-

$\sf \dfrac{x + 3}{x - 2} = \dfrac{x + 4}{x - 3}$

By cross multiplication :-

⇒ (x + 3)(x - 3) = (x + 4)(x - 2)

For Left Hand Side of above equation

• Remember an identity i.e, (x + y)(x - y) = x² - y². Here y = 3. So substitue the value of y in the Right Hand Side of identity[ (x + y)(x - y) = x² - y² ]

⇒ x² - 3² = (x + 4)(x - 2)

⇒ x² - 9 = (x + 4)(x - 2)

⇒ x² - 9 = (x + 4){x + (-2)}

For Right Hand Side of above equation

• Remember an identity i.e, (x + a)(x + b) = x² + (a + b)x + ab. Here a = 4 and b = - 2. So substitue the value of y and x in the Right Hand Side of identity[ (x + a)(x + b) = x² + (a + b)x + ab ]

⇒ x² - 9 = x² + {4 + (-2)}x + 4(-2)

• Now first remove the brackects

⇒ x² - 9 = x² + (4 - 2)x - 8

• Add the numbers which are brackets

⇒ x² - 9 = x² + (2)x - 8

⇒ x² - 9 = x² + 2x - 8

• Cancell x² on both sides

⇒ - 9 = 2x - 8

• Transpose - 9 to RHS and 2x to LHS

⇒ - 2x = - 8 + 9

• Add numerical terms in RHS

⇒ - 2x = 1

• Transpose - 2 to RHS

⇒ x = 1/-2

⇒ x = - 1/2