Explanation Given information: A system of equation { a x + b y + c z = a + b + c a 2 x + b 2 y + c 2 z = a c + a b + b c a b x + b c y = b c + a c , assuming a ≠ 0 , b ≠ 0 , c ≠ 0 Explanation: Consider a x + b y + c z = a + b + c …(1) a 2 x + b 2 y + c 2 z = a c + a b + b c … (2) a b x + b c y = b c + a c … (3) Consider equation (2), ⇒ a 2 x + b 2 y + c 2 z = ( a c + b c ) + a b From equation (2), the above becomes ⇒ a 2 x + b 2 y + c 2 z = a b x + b c y + a b ⇒ ( a 2 − a b ) x + ( b 2 − b c ) y + c 2 z = a b … (4) Multiply equation (1) by c ⇒ a c x + b c y + c 2 z = a c + b c + c 2 … (5) Subtract equation (5) from (4) ( a 2 − a b ) x + ( b 2 − b c ) y + c 2 z = a b − a c x + b c y + c 2 z = a c + b c + c 2 _ ( a 2 − a b − a c ) x + ( b 2 − b c − b c ) y = a b − a c − b c − c 2 ⇒ ( a 2 − a b − a c ) x + ( b 2 − 2 b c ) y = a b − a c − b c − c 2 … (6) Consider equation (3), a b x + b c y = b c + a c ⇒ x = b c + a c − b c y a b Plug the value of x in equation (6), ⇒ ( a 2 − a b − a c ) ( b c + a c − b c y a b ) + ( b 2 − 2 b c ) y = a b − a c − b c − c 2 ⇒ ( a − b − c ) ( b c + a c − b c y b ) + ( b 2 − 2 b c ) y = a b − a c − b c − c 2 ⇒ ( a − b − c ) ( c + a c b − c y ) + ( b 2 − 2 b c ) y = a b − a c − b c − c 2 ⇒ a c + a 2 c b − a c y − b c − a c + b c y − c 2 − a c 2 b + c 2 y + b 2 y − 2 b c y = a b − a c − b c − c 2 ⇒

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ISBN: 9780135228982

Author: Sullivan

Publisher: PEARSON

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Chapter 11.1, Problem 87AYU

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The value of x,y, and z, where {ax+by+cz=a+b+ca2x+b2y+c2z=ac+ab+bcabx+bcy=bc+ac assuming a≠0,b≠0,c≠0.

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Chapter 11.1, Problem 86AYU

Chapter 11.1, Problem 88AYU

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Chapter 11 Solutions

EBK PRECALCULUS

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The value of x , y , and z , where { a x + b y + c z = a + b + c a 2 x + b 2 y + c 2 z = a c + a b + b c a b x + b c y = b c + a c assuming a ≠ 0 , b ≠ 0 , c ≠ 0 .

Solution Summary: The author explains the value of x, y, and z.

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The value of x,y, and z, where {ax+by+cz=a+b+ca2x+b2y+c2z=ac+ab+bcabx+bcy=bc+ac assuming a≠0,b≠0,c≠0.

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Chapter 11 Solutions