2x + 3y -7 = 0
[a - b]x + [a + b]y - 3a + b - 2 = 0
Here `a_1 = 2, b_1 = 3,c_1 = -7`
`a_2 = [a - b], b_2 = [a + b], c_2 = -[3a + b - 2]`
`a_1/a_2 = 2/[a - b], b_1/b_2 = 3/[a + b], c_1/c_2 = [-7]/[-[3a + b - 2]] = [-7]/[3a + b - 2]`
For the equation to have infinitely many solutions, we have:
`a_1/a_2 = b_1/b_2 = c_1/c_2`
`2/[a - b] = 7/[3a + b -2]`
6a + 2b - 4 = 7a - 7b
a- 9b = -4 ......[1]
`2/[a -b] = 3/[a + b]`
a - 5b = 0 .....[2]
Subtracting [1] from [2], we obtain
4b = 4
b = 1
Substituting the value of b in equation [2], we obtain
a - 5 x 1 = 0
a = 5
Thus, the values of a and b are 5 and 1 respectively.
Solution
Step 1: Calculate the ratios of values of coefficients of the linear equations.
Given equations are:
2x+3y=7a-bx+ a+by-3a+b-2=0
On solving both the equations,
a1a2=2a-b b1b2=3a+bc1c2=-7-3a+b -2
Step 2: Applying the conditions for infinitely many solutions.
The condition for infinitely many solutions is a1a2=b1b2=c1c2
So, 2a-b=3a+b
On simplifying the above equation,
⇒2a+b=3 a-b⇒2a+2b=3a-3b⇒ a=5b
and 2a-b=-7-3a+b-2
On simplifying the above equation,
⇒-23a+b-2=-7a-b⇒-6a-2b+ 4=-7a+7b⇒ a=9b-4
Step 3: Substituting a=5b in the above equation
⇒5b=9b-4⇒4b=4⇒b=1
and
⇒a=51=5
Hence, values of a and bare 5&1.
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Updated On: 27-06-2022
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Text Solution
Answer : a = 5, b = 1
Answer
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