Solve the following simultaneous equations.
(i) 3x + 4y = 18; 4x + 3y = 17

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Solution

$3 x + 4 y = 18 . . . (i) 4 x + 3 y = 17 . . . (ii) Here we observe that the coefficients of x and y in the first equation are interchanged in the second equation . Adding these two equations, we get : 3 x + 4 y = 18 4 x + 3 y = 17 7 x + 7 y = 35 ∴ 7 (x + y) = 35 ∴ x + y = 5 . . . (iii) Subtracting eq (ii) from (i), we get : 3 x + 4 y = 18 4 x + 3 y = 17 - - - - x + y = 1 ∴ x - y = - 1 . . . (iv) From (iii) and (iv) we get : x + y = 5 and x - y = - 1 Adding (iii) and (iv), we get : x + y = 5 x - y = - 1 2 x = 4 ∴ x = 2 Substituting x = 2 in eq (iii), we get : 2 + y = 5 ∴ y = 3 Hence, x = 2 and y = 3 is the solution of the given equations .$

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\frac{x}{3} + \frac{y}{4} = 4; \frac{x}{2} - \frac{y}{4} = 1

\frac{x}{3} + 5 y = 13; 2 x + \frac{y}{2} = 19

\frac{2}{x} + \frac{3}{y} = 13; \frac{5}{x} - \frac{4}{y} = - 2

4 x + 5 y = 10; 5 x + 4 y = 17

4 x - 3 y = 8

3 x - 4 y = - 1

3 x - 4 y = 10 : 4 x + 3 y = 5

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