The solution of the system of equations given by \(\cfrac{x+y-8}2\)

1.	The solution of the system of equations given by \(\cfrac{x+y-8}2\) = \(\cfrac{x+2y-14}3\) = \(\cfrac{3x+y-12}{11}\) is ………………A) x = 3, y = 8 B) x = 4, y = 4 C) x = 2, y = 6 D) x = 6, y = 8
Answer» Correct option is (C) x = 2, y = 6 Given system of equations is \(\frac{x+y-8}{2}=\frac{x+2y-14}{3}=\frac{3x+y-12}{11}\) _________(1) \(\Rightarrow\) \(\frac{x+y-8}{2}=\frac{x+2y-14}{3}\) \(\Rightarrow\) \(3(x+y-8)=2(x+2y-14)\) \(\Rightarrow\) 3x + 3y - 24 = 2x + 4y - 28 \(\Rightarrow\) x - y = -4 _________(2) From (1), we also have \(\frac{x+y-8}{2}=\frac{3x+y-12}{11}\) \(\Rightarrow\) 11x + 11y - 88 = 6x + 2y - 24 \(\Rightarrow\) 5x + 9y = 64 _________(3) \(\Rightarrow\) 5 (y - 4) + 9y = 64 (From (2)) \(\Rightarrow\) 5y - 20 + 9y = 64 \(\Rightarrow\) 14y = 64+20 = 84 \(\Rightarrow\) y = \(\frac{84}{14}\) = 6 Then from (2), we have x = y - 4 = 6 - 4 = 2 Hence, the solution of given system of equations is x = 2 & y = 6. Correct option is C) x = 2, y = 6

The solution of the system of equations given by \(\cfrac{x+y-8}2\) = \(\cfrac{x+2y-14}3\) = \(\cfrac{3x+y-12}{11}\) is ………………A) x = 3, y = 8 B) x = 4, y = 4 C) x = 2, y = 6 D) x = 6, y = 8

Answer»

Correct option is (C) x = 2, y = 6

Given system of equations is

\(\frac{x+y-8}{2}=\frac{x+2y-14}{3}=\frac{3x+y-12}{11}\) _________(1)

\(\Rightarrow\) \(\frac{x+y-8}{2}=\frac{x+2y-14}{3}\)

\(\Rightarrow\) \(3(x+y-8)=2(x+2y-14)\)

\(\Rightarrow\) 3x + 3y - 24 = 2x + 4y - 28

\(\Rightarrow\) x - y = -4 _________(2)

From (1), we also have

\(\frac{x+y-8}{2}=\frac{3x+y-12}{11}\)

\(\Rightarrow\) 11x + 11y - 88 = 6x + 2y - 24

\(\Rightarrow\) 5x + 9y = 64 _________(3)

\(\Rightarrow\) 5 (y - 4) + 9y = 64 (From (2))

\(\Rightarrow\) 5y - 20 + 9y = 64

\(\Rightarrow\) 14y = 64+20 = 84

\(\Rightarrow\) y = \(\frac{84}{14}\) = 6

Then from (2), we have

x = y - 4

= 6 - 4 = 2

Hence, the solution of given system of equations is x = 2 & y = 6.

Correct option is C) x = 2, y = 6