If (x1,y1)(x_1, y_1)(x1,y1) is the solution of the pair of equations: x10+y5−1=$$0\frac{x}{10}$$ + $$\frac{y}{5}$$ - 1 = 010x+5y−1=0 andx8+y6=15 \quad $$\frac{x}{8}$$ + $$\frac{y}{6}$$ = 15 \\8x+6y=15 and y1=λx1+5, y_1 = \lambda x_1 + 5,y1=λx1+5, then the value of λ \lambdaλ is:

Given:x10+y5−1=0(1)x8+y6=15(2)y1=λx1+$$5\\frac{x}{10}$$ + $$\\frac{y}{5}$$ - 1 = 0 \\quad $$\\text{(1)}$$ \\\\$$\\frac{x}{8}$$ + $$\\frac{y}{6}$$ = 15 \\quad $$\\text{(2)}$$ \\\\y_1 = \\lambda x_1 + 5 \\\\10x+5y−1=0(1)8x+6y=15(2)y1=λx1+5Solution:From (1):x10+y5=1 \\quad $$\\frac{x}{10}$$ + $$\\frac{y}{5}$$ = 1 \\\\10x+5y=1=&gtx10+2y10=1=&gtx+2y10=1=&gtx+2y=10.....A\\Rightarrow $$\\frac{x}{10}$$ + $$\\frac{2y}{10}$$ = 1 \\\\\\Rightarrow $$\\frac{x + 2y}{10}$$ = 1 \\\\\\Rightarrow x + 2y = 10 {.....A} \\\\=&gt10x+102y=1=&gt10x+2y=1=&gtx+2y=10.....AFrom (2):x8+y6=15=&gt3x24+4y24=15=&gt3x+4y24=15=&gt3x+4y=360........B$$\\frac{x}{8}$$ + $$\\frac{y}{6}$$ = 15 \\\\\\Rightarrow $$\\frac{3x}{24}$$ + $$\\frac{4y}{24}$$ = 15 \\\\\\Rightarrow $$\\frac{3x + 4y}{24}$$ = 15 \\\\\\Rightarrow 3x + 4y = 360 {........B} \\\\8x+6y=15=&gt243x+244y=15=&gt243x+4y=15=&gt3x+4y=360........BFrom (A): x=10−2y \\quad x = 10 - 2y \\\\x=10−2ySubstitute in (B):3(10−2y)+4y=36030−6y+4y=360−2y=330=&gty=−165x=10−2(−165)=10+330=3403(10 - 2y) + 4y = 360 \\\\30 - 6y + 4y = 360 \\\\-2y = 330 \\Rightarrow y = -165 \\\\x = 10 - 2(-165) = 10 + 330 = 340 \\\\3(10−2y)+4y=36030−6y+4y=360−2y=330=&gty=−165x=10−2(−165)=10+330=340Now use: y1=λx1+5\\quad y_1 = \\lambda x_1 + 5 \\\\y1=λx1+5−165=λ(340)+5λ(340)=−170λ=−170340=−12-165 = \\lambda(340) + 5 \\\\\\lambda(340) = -170 \\\\\\lambda = $$\\frac{-170}{340}$$ = -$$\\frac{1}{2}$$ \\\\−165=λ(340)+5λ(340)=−170λ=340−170=−21Correct answer is (c)