Solve the following pairs by reducing them to a pair of linear equations:
{1/(3a+b)} +{1/(3a-b)} = 3/4; [1/{2(3a+b)}] - [1/{2(3a-b)}] = -1/8

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Solution

Hello ,
substitute 1/3a+b = x and 1/3a - b = y
then the eqs becomes , x + y = 3/4 and x- y = -1/4
Adding the 2 eqs we get x + y + x - y = 3/4 - 1/4
2x = 1/2 => x = 1/4
y = 1/2
ie.., 3a+b = 4 and 3a - b = 2
adding the 2 equations
6a +b - b = 1/4 + 1/2
6a = 3/4
a = 1 b = 1

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Question 1 (viii)
Solve the following pairs of equations by reducing them to a pair of linear equations:
$\frac{1}{3 x + y} + \frac{1}{3 x - y} = \frac{3}{4}$
$\frac{1}{2 (3 x + y)} - \frac{1}{2 (3 x - y)} = \frac{- 1}{8}$

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\frac{a}{4} - \frac{b}{3} = 0; \frac{3 a + 8}{5} = \frac{2 b - 1}{2}

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Solve the following pairs by reducing them to a pair of linear equations: {1/(3a+b)} +{1/(3a-b)} = 3/4; [1/{2(3a+b)}] - [1/{2(3a-b)}] = -1/8

Hello , substitute 1/3a+b = x and 1/3a - b = y then the eqs becomes , x + y = 3/4 and x- y = -1/4 Adding the 2 eqs we get x + y + x - y = 3/4 - 1/4 2x = 1/2 => x = 1/4 y = 1/2 ie.., 3a+b = 4 and 3a - b = 2 adding the 2 equations 6a +b - b = 1/4 + 1/2 6a = 3/4 a = 1 b = 1

Solve the following pairs by reducing them to a pair of linear equations:
{1/(3a+b)} +{1/(3a-b)} = 3/4; [1/{2(3a+b)}] - [1/{2(3a-b)}] = -1/8

Hello ,
substitute 1/3a+b = x and 1/3a - b = y
then the eqs becomes , x + y = 3/4 and x- y = -1/4
Adding the 2 eqs we get x + y + x - y = 3/4 - 1/4
2x = 1/2 => x = 1/4
y = 1/2
ie.., 3a+b = 4 and 3a - b = 2
adding the 2 equations
6a +b - b = 1/4 + 1/2
6a = 3/4
a = 1 b = 1