рд╣рд┐рдВрджреА
рдорд╣рд╛рд░рд╛рд╖реНрдЯреНрд░ рд╕реНрдЯреЗрдЯ рдмреЛрд░реНрдбрдПрд╕рдПрд╕рд╕реА (рдЕрдВрдЧреНрд░реЗрдЬреА рдорд╛рдзреНрдпрдо) реп рд╡реАрдВ рдХрдХреНрд╖рд╛

Solve the following equation. ЁЭСе2+12тБвЁЭСетИТ20/3тБвЁЭСетИТ5 =ЁЭСе2+8тБвЁЭСе+12/2тБвЁЭСе+3 - Algebra

Advertisements
Advertisements

рдкреНрд░рд╢реНрди

Solve the following equation.

`(x^2 + 12x - 20)/(3x - 5) = (x^2 + 8x + 12)/(2x + 3)`

рдпреЛрдЧ
Advertisements

рдЙрддреНрддрд░

`(x^2 + 12x - 20)/(3x - 5) = (x^2 + 8x + 12)/( 2x + 3 )`

Multiplying both sides by `1/4`, we get

`(x^2 + 12x - 20)/(12x - 20) = (x^2 + 8x + 12)/(8x + 12)`

Using dividendo, we get

`[(x^2 + 12x - 20) - (12x - 20)]/(12x - 20) = [(x^2 + 8x + 12) - (8x + 12)]/ (8x + 12)`

⇒ `[x^2 + 12x - 20 - 12x + 20 ]/(12x - 20) = [x^2 + 8x + 12 -  8x - 12]/(8x + 12)`

⇒ `x^2/(12x - 20) = x^2/(8x + 12)`

This equation is true for x = 0.

Therefore, x = 0 is a solution of the given equation.

If x ≠ 0, then x2 ≠ 0.

Dividing both sides by x2, we get

`1/(12x - 20)= 1/(8x + 12)`

⇒ 12x − 20 = 8x + 12

⇒ 12x − 8x = 20 + 12

⇒ 4x = 32

⇒ x = 8

Thus, the solutions of the given equation are x = 0 and x = 8.

shaalaa.com
Application of Properties of Equal Ratios
  рдХреНрдпрд╛ рдЗрд╕ рдкреНрд░рд╢реНрди рдпрд╛ рдЙрддреНрддрд░ рдореЗрдВ рдХреЛрдИ рддреНрд░реБрдЯрд┐ рд╣реИ?
рдЕрдзреНрдпрд╛рдп 4: Ratio and Proportion - Practice Set 4.3 [рдкреГрд╖реНрда ренреж]

APPEARS IN

рдмрд╛рд▓рднрд╛рд░рддреА Mathematics 1 [English] Standard 9 Maharashtra State Board
рдЕрдзреНрдпрд╛рдп 4 Ratio and Proportion
Practice Set 4.3 | Q (4) (i) | рдкреГрд╖реНрда ренреж
Share
Notifications

Englishрд╣рд┐рдВрджреАрдорд░рд╛рдареА


      Forgot password?
Use app×