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# Solve Each of the Following Systems of Equations by the Method of Cross-multiplication : 2(Ax – By) + a + 4b = 0 2(Bx + Ay) + B – 4a = 0 - CBSE Class 10 - Mathematics

ConceptAlgebraic Methods of Solving a Pair of Linear Equations Cross - Multiplication Method

#### Question

Solve each of the following systems of equations by the method of cross-multiplication :

2(ax – by) + a + 4b = 0

2(bx + ay) + b – 4a = 0

#### Solution

The given system of equation may be written as

2(ax – by) + a + 4b = 0

2(bx + ay) + b – 4a = 0

here

a_1 = 2a, b_1 = -2b, c_1 = a+ 4b

a_2 = 2b, b_2 = 2a, c_2 = b - 4a

By cross multiplication, we have

=> x/(((-2b)(b - 4a) - (2a)(a + 4b))) = (-y)/((2b)(b - 4a) - (2a) (a + 4a) ) = 1/(4a^2 + 4b^2)

=> x/(-2b^2 + 8ab - 2a^2 - 8ab) = (-y)/(2ab - 8a^2 - 2ab - 8b^2) = 1/(4a^2 + 4b^2)

=> x/(-2a^2 - 2b^2) = (-y)/(-8a^2 - 8b^2) = 1/(4a^2 + 4b^2)

Now

x/(-2a^2 - 2b^2) = 1/(4a^2 + 4b^2)

=> x = (-2a^2 - 2b^2)/(4a^2 + 4b^2)

= (-2(a^2 - b^2))/(4(a^2 + b^2))

= (-1)/2

And

(-y)/(-8a^2 - 8b^2) = 1/(4a^2 + 4b^2)

=> -y = (-8a^2 - 8b^2)/(4a^2 + 4b^2)

=> -y = (-8(a^2 - b^2))/(4(a^2 + b^2))

=> -y = (-8)/4

=> y = 2

Hence x = (-1)/2, y = 2 is the solution of the given system of the equations.

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Solution Solve Each of the Following Systems of Equations by the Method of Cross-multiplication : 2(Ax – By) + a + 4b = 0 2(Bx + Ay) + B – 4a = 0 Concept: Algebraic Methods of Solving a Pair of Linear Equations - Cross - Multiplication Method.
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