Question

pls answer step by step
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Answer 1

Answer:

value if b is 1/-2

Step-by-step explanation:

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Answer 2

$\longrightarrow$ $\frac{ 1 }{ (x-5)(x-3) } = \frac{ \frac{ 1 }{ 2 } }{ (x-5) } + \frac{ B }{ (x-3) }$

Multiply both sides of the equation by $\left(x-5\right)\left(x-3\right)$.

$\longrightarrow$ $\small \frac{ 1 }{ \cancel{ (x-5)(x-3) }} × \cancel{ \left(x-5\right)\left(x-3\right)} = \frac{ \frac{ 1 }{ 2 } }{ (\cancel{x-5}) } × \left(\cancel{x-5}\right)\left(x-3\right) + \frac{ B }{ (\cancel{x-3}) } × \left(x-5\right)\left(\cancel{x-3}\right)$

The least common multiple of $\left(x-5\right)\left(x-3\right),x-5,x-3$ .

And we get,

$\longrightarrow$ $1=\left(x-3\right)\times \left(\frac{1}{2}\right)+\left(x-5\right)B$

Use the distributive property to multiply $x-3$ by $\frac{1}{2}$

$\longrightarrow$ $1=\frac{1}{2}x-\frac{3}{2}+\left(x-5\right)B$

Use the distributive property to multiply x - 5 by B.

$\longrightarrow$ $1=\frac{1}{2}x-\frac{3}{2}+xB-5B$

Swap sides so that all variable terms are on the left hand side.

$\longrightarrow$ $\frac{1}{2}x-\frac{3}{2}+xB-5B=1$

Subtract $\frac{1}{2}x$ from both sides.

$\longrightarrow$ $\frac{1}{2} x - \frac{3}{2} + xB-5B \underline{- \frac{1}{2} x }= 1 \underline{ - \frac{1}{2} x}$

And we get,

$\longrightarrow$ $-\frac{3}{2}+xB-5B=1-\frac{1}{2}x$

Now, Add $\frac{3}{2}$ to both sides.

$\longrightarrow$ $-\frac{3}{2}+xB-5B \underline{ + \frac{3}{2} }=1-\frac{1}{2}x \underline{ + \frac{3}{2} }$

Then we get,

$\longrightarrow$ $xB-5B=1-\frac{1}{2}x+\frac{3}{2}$

$\longrightarrow$ $xB-5B=\frac{5}{2}-\frac{1}{2}x$

Combine all terms containing B.

$\longrightarrow$ $\left(x-5\right)B=\frac{5}{2}-\frac{1}{2}x$

The equation is in standard form.

$\longrightarrow$ $\left(x-5\right)B=\frac{5-x}{2}$

$\longrightarrow$ $B=\frac{5-x}{2\left(x-5\right)}$

$\longrightarrow$ $B=\frac{5-x}{ - 2\left(5 - x\right)}$

$\longrightarrow$ $B = - \frac{ \cancel{5 - x}}{ - 2( \cancel{5 - x})}$

$\longrightarrow$ $\underline{ \boxed{ \bf B = - \frac{1}{2} }}$

I hope it helps you...