How do you solve this equation for #b#: #A = 1/2h(b + b_1)#?

Answer 1

See a solution process below:

First multiply each side of the equation by #color(red)(2)# to eliminate the fraction while keeping the equation balanced:

#color(red)(2) xx A = color(red)(2) xx 1/2h(b + b_1)#

#2A = cancel(color(red)(2)) xx 1/color(red)(cancel(color(black)(2)))h(b + b_1)#

#2A = 1h(b + b_1)#

#2A = h(b + b_1)#

Next, divide each side of the equation by #color(red)(h)# to eliminate the parenthesis while keeping the equation balanced:

#(2A)/color(red)(h) = (h(b + b_1))/color(red)(h)#

#(2A)/h = (color(red)(cancel(color(black)(h)))(b + b_1))/cancel(color(red)(h))#

#(2A)/h = b + b_1#

Now, subtract #color(red)(b_1)# from each side of the equation to solve for #b# while keeping the equation balanced:

#(2A)/h - color(red)(b_1) = b + b_1 - color(red)(b_1)#

#(2A)/h - b_1 = b + 0#

#(2A)/h - b_1 = b#

#b = (2A)/h - b_1#

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Answer 2

James Atkins

To solve the equation ( A = \frac{1}{2}h(b + b_1) ) for ( b ), you can follow these steps:

Multiply both sides of the equation by 2 to eliminate the fraction: ( 2A = h(b + b_1) ).
Divide both sides of the equation by ( h ): ( \frac{2A}{h} = b + b_1 ).
Subtract ( b_1 ) from both sides of the equation: ( \frac{2A}{h} - b_1 = b ).

So, the solution for ( b ) is given by ( b = \frac{2A}{h} - b_1 ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.