How do you rewrite #w=3(2a + b) - 4# to solve for #a#?

Answer 1

Ethan Adkins

#color(blue)(a=(w+4-3b)/6)#

If #w=3(2a+b)-4# then #color(white)("XXX")w+4=3(2a+b)#

#color(white)("XXX")(w+4)/3=2a+b#

#color(white)("XXX")(w+4)/3-b=2acolor(white)("xx")rarrcolor(white)("xx")(w+4-3b)/3=2a#

#color(white)("XXX")(w+4-3b)/6=acolor(white)("xxx")"or"color(white)("xxx")a=(w+4-3b)/6#

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Answer 2

Eva Abramson

To rewrite ( w = 3(2a + b) - 4 ) to solve for ( a ), follow these steps:

Expand the expression inside the parentheses: [ 2a + b ]
Multiply (3) by the result of the expansion: [ 3(2a + b) ]
Subtract (4) from the previous result: [ 3(2a + b) - 4 ]
Distribute (3) across (2a) and (b): [ 6a + 3b - 4 ]
Move terms not containing (a) to the right side of the equation: [ 6a = w - 3b + 4 ]
Finally, divide both sides by (6) to isolate (a): [ a = \frac{w - 3b + 4}{6} ]

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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.