Question

how to solve (a-b)x + (a+b)y = (a2-2ab-b2) , (a+b) (x+y)=a2+b2. by cross multiplication method

Answer 1

Step-by-step explanation:

Cross multiplication method of Solution of two linear equation

$a_1x + b_1y + c_1 = 0 \\ \\ a_2x + b_2y + c_2 = 0 \\ \\ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{a_2c_1 - a_1c_2} = \frac{1}{a_1b_2 - a_2b_1}$

$(a-b)x + (a+b)y - ( {a}^{2} -2ab- {b}^{2} ) = 0 \\ \\ (a+b)x+(a + b)y - {a}^{2} - {b}^{2} = 0 \\ \\ \frac{x}{(a + b)( - {a}^{2} - {b}^{2} ) - (a + b)( -( {a}^{2} -2ab- {b}^{2} ) } \\ \\ = \frac{y}{- ( {a}^{2} -2ab- {b}^{2} ) (a + b) - ( - {a}^{2} - {b}^{2} )(a + b) } = \frac{1}{(a - b)(a + b) - (a + b)(a + b)} \\ \\ \frac{x}{ - {a}^{3} - a {b}^{2} - {a}^{2}b - {b}^{3} + {a}^{3} - 2 {a}^{2} b - a {b}^{2} + {a}^{2}b - 2a {b}^{2} - {b}^{3} } = \frac{y}{ - {a}^{3} - {a}^{2}b + 2 {a}^{2}b + 2a {b}^{2} + a {b}^{2} + {b}^{3} + {a}^{3} + {a}^{2} b + a {b}^{2} + {b}^{3} } = \frac{1}{ {a}^{2} - {b}^{2} - {a}^{2} - {b}^{2} - 2ab} \\ \\ \frac{x}{ - 4a {b}^{2} - 3 {a}^{2} b - 2{b}^{3} } = \frac{y}{3 {a}^{2}b + 4a {b}^{2} + 2 {b}^{3} } = \frac{1}{ - 2 {b}^{2} - 2ab} \\ \\ \frac{x}{ - 4a {b}^{2} - 3 {a}^{2} b - 2{b}^{3} } = \frac{1}{ - 2 {b}^{2} - 2ab} \\ \\ x = \frac{(a+b)( - 2 {b}^{2} - 2ab) }{- 2 {b}^{2} - 2ab} \\ \\ x = a+b \\ \\ \frac{y}{3 {a}^{2}b + 4a {b}^{2} + 2 {b}^{3} } = \frac{1}{ - 2 {b}^{2} - 2ab} \\ \\ y = \frac{3 {a}^{2}b + 4a {b}^{2} + 2 {b}^{3} }{- 2 {b}^{2} - 2ab} \\ \\ y = \frac{-2ab }{ (b + a)}$

Hope it helps you.

Answer 2

Step-by-step explanation:

Cross multiplication method of Solution of two linear equation

\begin{gathered}a_1x + b_1y + c_1 = 0 \\ \\ a_2x + b_2y + c_2 = 0 \\ \\ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{a_2c_1 - a_1c_2} = \frac{1}{a_1b_2 - a_2b_1} \\ \\\end{gathered}

a

1

x+b

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y+c

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=0

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x+b

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y+c

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=0

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1

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−b

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c

1

x

=

a

2

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1

−a

1

c

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=

a

1

b

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−a

2

b

1

1