Question

\frac{x+y-8}{2}=\frac{x+2y-14}{3}=\frac{3x-y}{4} 2 x+y−8 ​ = 3 x+2y−14 ​ = 4 3x−y
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Answer 1

       $\underline{\bold{Answer:}}$

       $(2,6)$

       $\underline{\bold{Step-by-step-explaination:}}$

       $\text{The given equation is:}$

       $\boxed{\frac{x+ y- 8}{2} = \frac{x+2y-14}{3} = \frac{3x-y}{4} }$

       $\text{We can solve this by taking two equations at a time}$

       $\text{Taking the first two equations}$

$\implies \boxed{\frac{x+ y- 8}{2} = \frac{x+2y-14}{3}}$

       $\text{Cross multiplying...}$

$\implies \boxed{(x+ y- 8)(3) = (x+2y-14)(2)}$

       $\text{Simplifying the equation}$

$\implies \boxed{3x+ 3y- 24 = 2x+4y-28}$

       $\text{Transposing the terms from R.H.S into L.H.S}$

$\implies \boxed{3x+ 3y- 24 - 2x - 4y + 28 = 0}$

       $\text{Simplifying the equation}$

$\implies \boxed{x- y+ 4 = 0}\text{ ------(i)}$

       $\text{Taking the first and the last equation}$

$\implies \boxed{\frac{x+ y- 8}{2} = \frac{3x-y}{4} }$

       $\text{Cross multiplying...}$

$\implies \boxed{(x+y-8)(4) = (3x-y)(2)}$

       $\text{Simplifying the equation}$

$\implies \boxed{(x+y-8)(2) = 3x-y}$

       $\text{Further simplifying the equation}$

$\implies \boxed{2x+2y-16 = 3x-y}$

       $\text{Transposing the terms from R.H.S into L.H.S}$

$\implies \boxed{2x+2y-16 -3x+y=0}$

       $\text{Simplifying the equation}$

$\implies \boxed{-x+3y-16=0} \text{ ------(ii)}$

       $\text{Now, taking the last two equations}$

$\implies\boxed{\frac{x+2y-14}{3} = \frac{3x-y}{4} }$

       $\text{Cross multiplying...}$

$\implies\boxed{(x+2y-14)(4) = (3x-y)(3)}$

       $\text{Simplifying the equation}$

$\implies\boxed{4x+8y-56 = 9x-3y}$

       $\text{Transposing the terms from R.H.S into L.H.S}$

$\implies\boxed{4x+8y-56 -9x+3y=0}$

       $\text{Simplifying the equation}$

$\implies\boxed{-5x+11y-56 =0} \text{ ------(iii)}$

       $\text{Now that we've got 3 linear equations i.e (i), (ii) and (iii) we can now}\\\text{normally solve them}$

       $\text{Adding (i) and (ii)}$

$\implies \boxed{x-y+4 -x + 3y - 16= 0}$

       $\text{Simplifying the equation}$

$\implies \boxed{ 2y - 12= 0}$

       $\text{Transposing 14 to R.H.S}$

$\implies \boxed{ 2y = 12}$

       $\text{Transposing 2 to R.H.S}$

$\implies \boxed{y = \frac{12}2}$

       $\text{Simplifying the equation}$

$\implies \boxed{y = 6}$

       $\text{Now substituting the value of }y \text{ in (i)}$

$\implies \boxed{x- 6+ 4 = 0}$

       $\text{Simplifying the equation}$

$\implies \boxed{x- 2 = 0}$

       $\text{Transposing 3 to R.H.S}$

$\implies \boxed{x= 2 }$

       $\text{Now we must substitute the values of } x\text{ and }y\text{, obtained by} \\\text{solving (i) and (ii), into (iii) in order to confirm the answer}$      

$\implies\boxed{-5(2)+11(6)-56 =0}$

       $\text{Simplifying the equation}$

$\implies\boxed{-10+66-56 =0}$

       $\text{Further simplifying the equation}$

$\implies\boxed{0 = 0}$

$\implies (2,6) \text{ is the solution}$