I am a paid user of meritnation Q=Solve each of the following systems of equations by the method of - Maths - Pair of Linear Equations in Two Variables

Question

I am a paid user of meritnation
Q=Solve each of the following systems of equations
by the method of cross-multiplication
(a)= x(a - b + ab/a-b) =y(a+b
-ab/a+b)
x+y=2a2
(b) = 57/x+y + 6/x-y=5
38/x+y + 21/x -y =9
(c)=a2 /x - b2 /y
=0
a2 b/x +b2 a/y
=a+b
(d)=mx - ny=m2
+n2
x+y=2m
(e)=bx/a + ay/b =a2
+b2
x + y=2ab

Priyanka Kedia · Accepted Answer

$(a) T h e g i v e n e q u a t i o n b e a s, x (a - b + \frac{a b}{a - b}) = y (a + b - \frac{a b}{a + b}) \Rightarrow x (a - b + \frac{a b}{a - b}) - y (a + b - \frac{a b}{a + b}) = 0 ....... (1) x + y = 2 a^{2} ...... (2) N o w, b y c r o s s m u l t i p l i c a t i o n m e t h o d, w e h a v e, \frac{x}{- (a + b - \frac{a b}{a + b}) \times (- 2 a^{2}) - 1 \times 0} = \frac{y}{0 \times 1 + 2 a^{2} \times (a - b + \frac{a b}{a - b})} = \frac{1}{(a - b + \frac{a b}{a - b}) \times 1 + (a + b - \frac{a b}{a + b}) \times 1} \Rightarrow \frac{x}{(a + b - \frac{a b}{a + b}) \times 2 a^{2}} = \frac{y}{2 a^{2} \times (a - b + \frac{a b}{a - b})} = \frac{1}{a - b + \frac{a b}{a - b} + a + b - \frac{a b}{a + b}} \Rightarrow \frac{x}{(\frac{{(a + b)}^{2} - a b}{a + b}) \times 2 a^{2}} = \frac{y}{2 a^{2} \times (\frac{{(a - b)}^{2} + a b}{a - b})} = \frac{1}{2 a + \frac{a^{2} b + a b^{2} - a^{2} b + a b^{2}}{a^{2} - b^{2}}} \Rightarrow \frac{x}{(\frac{a^{2} + b^{2} + 2 a b - a b}{a + b}) \times 2 a^{2}} = \frac{y}{2 a^{2} \times (\frac{a^{2} + b^{2} - 2 a b + a b}{a - b})} = \frac{1}{\frac{2 a (a^{2} - b^{2}) + 2 a b^{2}}{a^{2} - b^{2}}} \Rightarrow \frac{x}{(\frac{a^{2} + b^{2} + a b}{a + b}) \times 2 a^{2}} = \frac{y}{2 a^{2} \times (\frac{a^{2} + b^{2} - a b}{a - b})} = \frac{1}{\frac{2 a^{3}}{a^{2} - b^{2}}} \Rightarrow \frac{x}{(\frac{a^{2} + b^{2} + a b}{a + b}) \times 2 a^{2}} = \frac{1}{\frac{2 a^{3}}{a^{2} - b^{2}}} a n d \frac{y}{2 a^{2} \times (\frac{a^{2} + b^{2} - a b}{a - b})} = \frac{1}{\frac{2 a^{3}}{a^{2} - b^{2}}} \Rightarrow \frac{x}{(a^{2} + b^{2} + a b)} = \frac{1}{\frac{a}{(a - b)}} a n d \frac{y}{(a^{2} + b^{2} - a b)} = \frac{1}{\frac{a}{(a + b)}} \Rightarrow x = \frac{(a - b) (a^{2} + b^{2} + a b)}{a} a n d y = \frac{(a + b) (a^{2} + b^{2} - a b)}{a} \Rightarrow x = \frac{(a^{3} - b^{3})}{a} a n d y = \frac{(a^{3} + b^{3})}{a} \Rightarrow x = a^{2} - \frac{b^{3}}{a} a n d y = a^{2} + \frac{b^{3}}{a}$

$(e) . T h e g i v e n e q u a t i o n b e a s, \frac{b}{a} x + \frac{a}{b} y = a^{2} + b^{2} ...... (1) x + y = 2 a b ...... (2) N o w, b y c r o s s - m u l t i p l i c a t i o n m e t h o d, w e h a v e, \frac{x}{\frac{a}{b} (- 2 a b) + (a^{2} + b^{2})} = \frac{y}{- (a^{2} + b^{2}) + \frac{b}{a} (2 a b)} = \frac{1}{\frac{b}{a} - \frac{a}{b}} \Rightarrow \frac{x}{- 2 a^{2} + (a^{2} + b^{2})} = \frac{y}{- (a^{2} + b^{2}) + 2 b^{2}} = \frac{a b}{b^{2} - a^{2}} \Rightarrow \frac{x}{b^{2} - a^{2}} = \frac{y}{b^{2} - a^{2}} = \frac{a b}{b^{2} - a^{2}} \Rightarrow \frac{x}{1} = \frac{y}{1} = \frac{a b}{1} T h u s, \frac{x}{1} = \frac{a b}{1} \Rightarrow x = a b a n d \frac{y}{1} = \frac{a b}{1} \Rightarrow y = a b H e n c e, x = a b a n d y = a b$

$(d) . T h e g i v e n e q u a t i o n b e a s, m x - n y = m^{2} + n^{2} ...... (1) x + y = 2 m ...... (2) N o w, b y c r o s s m u l t i p l i c a t i o n m e t h o d, w e h a v e, \frac{x}{- n (- 2 m) + (m^{2} + n^{2})} = \frac{y}{- (m^{2} + n^{2}) + 2 m^{2}} = \frac{1}{m + n} \Rightarrow \frac{x}{2 m n + (m^{2} + n^{2})} = \frac{y}{m^{2} - n^{2}} = \frac{1}{m + n} \Rightarrow \frac{x}{{(m + n)}^{2}} = \frac{y}{(m - n) (m + n)} = \frac{1}{m + n} \Rightarrow \frac{x}{(m + n)} = \frac{y}{(m - n)} = 1 \Rightarrow \frac{x}{(m + n)} = 1 \Rightarrow x = (m + n) A n d, \frac{y}{(m - n)} = 1 \Rightarrow y = (m - n)$