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Finding Solution for Consistent Pair of Linear Equations

148 x+231 y=9...

Question

148x+231y=910

231x+ 148y=610

find x and y

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Solution

148x + 231y = 610 ---- (1)
231x + 148y = 527 ---- (2)

On adding eq. (1) and (2), we get

379x + 379y = 1137
x + y = 3 ---- (3)

On subtracting eq. (1) from (2), we get
83x - 83y = -83
x - y = -1 ---- (4)

On adding eq. (3) and (4), we get

2x = 2
x = 1

On putting value of "x" in eq (4), we get

1 - y = -1
y = 1 + 1
y = 2 A

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Similar questions

Q. Solve the following simultaneous equations.
(1)

\frac{2}{x} + \frac{2}{3 y} = \frac{1}{6}; \frac{3}{x} + \frac{2}{y} = 0

\frac{7}{2 x + 1} + \frac{13}{y + 2} = 27; \frac{13}{2 x + 1} + \frac{7}{y + 2} = 33

\frac{148}{x} + \frac{231}{y} = \frac{527}{x y}; \frac{231}{x} + \frac{148}{y} = \frac{610}{x y}

\frac{7 x - 2 y}{x y} = 5; \frac{8 x + 7 y}{x y} = 15

\frac{1}{2 (3 x + 4 y)} + \frac{1}{5 (2 x - 3 y)} = \frac{1}{4}; \frac{5}{(3 x + 4 y)} - \frac{2}{(2 x - 3 y)} = - \frac{3}{2}

If x ≠ y, (x,y) ≠ (y,x) and (x,y) = (y,x), if x = y.

x

y :

\frac{15}{x - y} + \frac{22}{x + y} = 5, \frac{40}{x - y} + \frac{55}{x + y} = 13, x \neq y

x \neq - y .

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\frac{44}{x + y} + \frac{30}{x - y} = 10

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x + y \neq 0

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If $x \neq y, (x, y) \neq (y, x) and (x, y) = (y, x), if x = y$ .

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