If the equation $\frac{1}{2} (6 x - 4) - (3 - x) = a x + x + b$ has infinite number of solutions. Find the value of $a - b$ ,

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Solution

Given, $\frac{1}{2} (6 x - 4) - (3 - x) = a x + x + b$
$\Rightarrow 3 x - 2 - 3 + x = a x + x + b$
$\Rightarrow 3 x + x - 2 - 3 = a x + x + b$
$\Rightarrow 3 x + x - 5 = a x + x + b$
As seeing both these side in equation, we get
$a = 3$ and $b = - 5$
Then $a - b = 3 - (- 5) = 3 + 5 = 8$

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If the equation 12(6x−4)−(3−x)=ax+x+b has infinite number of solutions. Find the value of a−b ,

Given, 12(6x−4)−(3−x)=ax+x+b⇒3x−2−3+x=ax+x+b⇒3x+x−2−3=ax+x+b⇒3x+x−5=ax+x+bAs seeing both these side in equation, we geta=3 and b=−5Then a−b=3−(−5)=3+5=8

If the equation $\frac{1}{2} (6 x - 4) - (3 - x) = a x + x + b$ has infinite number of solutions. Find the value of $a - b$ ,

Given, $\frac{1}{2} (6 x - 4) - (3 - x) = a x + x + b$
$\Rightarrow 3 x - 2 - 3 + x = a x + x + b$
$\Rightarrow 3 x + x - 2 - 3 = a x + x + b$
$\Rightarrow 3 x + x - 5 = a x + x + b$
As seeing both these side in equation, we get
$a = 3$ and $b = - 5$
Then $a - b = 3 - (- 5) = 3 + 5 = 8$