In custody deven depicted negatively and grostesuely
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Answer:
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Explanation:
GivenQuestion−
Find the values of a and b for which the simultaneous linear equations x + 2y = 1 and (a - b)x + (a + b)y = a + b - 2 have infinitely many solutions.
\large\underline{\sf{Solution-}}Solution−
Given pair of linear equation is
\rm :\longmapsto\:x + 2y = 1 - - - (1):⟼x+2y=1−−−(1)
and
\rm :\longmapsto\:(a - b)x + (a + b)y = a + b - 2 - - - (2):⟼(a−b)x+(a+b)y=a+b−2−−−(2)
Now,
Comparing the given two equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we get
a₁ = 1
b₁ = 2
c₁ = 1
a₂ = a - b
b₂ = a + b
c₂ = a + b - 2
Now, we know that,
System of two equations a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 have infinitely many solutions iff
\boxed{ \bf{ \: \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}}}a2a1=b2b1=c2c1
So, on substituting the values, we get
\rm :\longmapsto\:\dfrac{1}{a - b} = \dfrac{2}{a + b} = \dfrac{1}{a + b - 2}:⟼a−b1=a+b2=a+b−21
On taking first and second member, we get
\rm :\longmapsto\:\dfrac{1}{a - b} = \dfrac{2}{a + b}:⟼a−b1=a+b2
\rm :\longmapsto\:a + b = 2(a - b):⟼a+b=2(a−b)
\rm :\longmapsto\:a + b = 2a - 2b:⟼a+b=2a−2b
\rm :\longmapsto\:a - 2a = - b- 2b:⟼a−2a=−b−2b
\rm :\longmapsto\: - a = - 3b:⟼−a=−3b
\bf :\longmapsto\: a = 3b - - - - - (1):⟼a=3b−−−−−(1)
Now, On taking first and third member, we get
\rm :\longmapsto\:\dfrac{1}{a - b} = \dfrac{1}{a + b - 2}:⟼a−b1=a+b−21
\rm :\longmapsto\:a + b - 2 = a - b:⟼a+b−2=a−b
\rm :\longmapsto \: b - 2 = - b:⟼b−2=−b
\rm :\longmapsto \: b + b = 2:⟼b+b=2
\rm :\longmapsto \: 2b= 2:⟼2b=2
\bf :\longmapsto \: b= 1:⟼b=1
On substituting the value of b in equation (1), we get
\bf :\longmapsto\:a = 3 \times 1 = 3:⟼a=3×1=3
\begin{gathered} \red{\begin{gathered}\begin{gathered}\rm :\longmapsto\:\bf\: Hence-\begin{cases} &\sf{a \: = \: 3} \\ \\ &\sf{b \: = \: 1} \end{cases}\end{gathered}\end{gathered}}\end{gathered}:⟼Hence−⎩⎪⎪⎨⎪⎪⎧a=3b=1
refer the attachment hope it will help
