Find the sum of all positive integral values of a for which every solution to the system of equation x+ay=3 ax+4y=6 satisfy the inequality x greater than 1 y greater than 0
Answers
Given : system of equation x+ay=3 ax+4y=6 satisfy the inequality x > 1 y > 0
To Find : sum of all positive integral values of a
Solution:
x+ay=3 => 4x + 4ay = 12
ax + 4y = 6 => a²x + 4ay = 6a
4x - a²x = 12 - 6a
=> x = (12 - 6a)/(4 - a²)
(12 - 6a)/(4 - a²) > 1
=> 6(2 - a) /( 2 + a)(2 - a) > 1
=> 6 /(2 + a) > 1
=> 6 > 2 + a
=> 4 > a
=> a < 4
x+ay=3 => ax + a²y = 3a
ax + 4y = 6
=> y (a² - 4) = 3a - 6
=> y = (3a - 6)/(a² - 4)
=> y = 3(a - 2)/(a - 2)(a + 2)
=> y = 3/(a + 2) a ≠ 2
y > 0
=> 3/(a + 2) > 0
if a + 2 > 0
=> a > - 2
a ∈ ( - 2 , 4 ) - { 2}
positive integral values of a
= 1 & 3
Sum = 1 + 3 = 4
sum of all positive integral values of a = 4
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