Question

please help me in solving this out

Answer 1

Answer:

Option(B)

Step-by-step explanation:

(34)

Given: (2x - 3y + 1)/2 = (x + 9y + 8)/3 = (4x + 7y + 12)/5

(i)

(2x - 3y + 1)/2 = (x + 9y + 8)/3

⇒ 3(2x - 3y + 1) = 2(x + 9y + 8)

⇒ 6x - 9y + 3 = 2x + 18y + 16

⇒ 4x - 27y = 13

(ii)

(x + 9y + 8)/3 = (4x + 7y + 12)/5

⇒ 5(x + 9y + 8) = 3(4x + 7y + 12)

⇒ 5x + 45y + 40 = 12x + 21y + 36

⇒ -7x + 24y = -4

⇒ 7x - 24y = 4

On solving (i) * 7 & (ii) * 4, we get

28x- 189y = 91

28x - 96y = 16

----------------------------

         -93y = 75

               y = -25/31

Substitute y = -25/31 in (i), we get

4x - 27y = 13

⇒ 4x - 27(-25/31) = 13

⇒ 4x = -272/31

⇒ x = -68/31

Thus,

x + y = (-68/31) - (25/31)

        = -93/31

      = -3

Hope it helps!

Answer 2

$\frac{2x-3y+1}{2}=\frac{x+9y+8}{3}=\frac{4x+7y+12}{5}\\consider\:\frac{2x-3y+1}{2}=\frac{x+9y+8}{3}\\3(2x-3y+1)=2(x+9y+8)\\6x-9y+3=2x+18y+16\\4x-27y-13=0\:==>\:(EQ\:01)\\x=\frac{27y+13}{4}\\consider\:\frac{x+9y+8}{3}=\frac{4x+7y+12}{5}\\5(x+9y+8)=3(4x+7y+12)\\5x+45y+40=12x+21y+36\\7x-24y-4=0\:==>(EQ\:02)\\replace\:the\:x\:value\:from\:in\:EQ\:02\\7\left(\frac{27y+13}{4}\right)-24y-4=0\\we\:get\:y=-\frac{25}{31}\\By\:putting\:this\:value\:of\:y\:in\:either\:EQ 01 or 02\:we\:get\:x=-\frac{68}{31}\\x+y=-\frac{68}{31}+\left(-\frac{25}{31}\right)=-\frac{93}{31}=\:-3$

Hope it hepls :))