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To solve the equations of the form :
ax + by = c ...(i)
and bx + ay = d ...(ii)
where a ¹ b, we follow the following steps :
StepI : Add (i) and (ii) and obtain (a + b)x + (b + a) y = c + d, i.e., x + y = c d a b +
+ ...(iii)
StepII : Subtract (ii) from (i) and obtain (a – b)x – (a – b) y = c – d, i.e., x – y = c – d a – b ...(iv)
StepIII : Solve (iii) and (iv) to get x and y.
Ex.12 Solve for x and y : 47x + 31y = 63, 31x + 47y = 15.
Sol. We have, 47x + 31y = 63 ...(i) and 31x + 47y = 15 ...(ii) Adding (i) and (ii), we get : 78x + 78y = 78 Þ x + y = 1 ...(iii) Subtracting (ii) from (i), we get : 16x – 16y = 48 Þ x – y = 3 ...(iv) Now, adding (iii) and (iv), we get : 2x = 4 Þ x = 2 Putting x = 2 in (iii), we get : 2 + y = 1 Þ y = – 1 Hence, the solution is x = 2 and y = –1
Equations which contain the variables, only in the denominators, are called reciprocal equations. These equations can be of the following types and can be solved by the under mentioned method :
TypeI : c' ∈ a,b,c,a',b',c'∈ R
Put 1
x
u = and 1
v
y = and find the value of x and y by any method described earlier.
Then u = 1
x and v = 1
y TypeII : au + bv = cuv and a'u + b'v = c'uv " a,b,c,a',b',c'ÎR Divide both equations by uv and equations can be converted in the form explained in (I).
TypeIII : a b k lx m y cx dy + =
+ + ,
a ' b ' k ' lx m y cx dy + =
+ +
" a,b,k,a',b',k'ÎR
Put 1
u lx m y =
+
and
1
v cx dy =
+
Then equations are au + bv = k and a'u + b'v = k'
Find the values of u and v and put in Ix + my = 1
u and cx + dy = 1
v Again solve for x and y, by any method explained earlier.
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