Question

How do you solve `-17b+23=-4-8b`?

Answer 1

`color(green)("Very detailed explanation how to manipulate this type of algebraic")`
`color(green)("equation")`.

Taken one step at a time

Explanation:

Given `color(brown)(-17b+23=-4-8b)`
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`color(blue)("Stage 1")`

`color(green)("A quick note")`
`color(purple)("Changing the equation so that we have all the 'b' terms on the left")`
`color(purple)("of the equals sign")`
`color(purple)("If we can change the right hand side (RHS) so that "-8b" becomes 0. Then it has no effect on that side's -4")`

To change -8b onto 0 we add 8b so -8b becomes -8b+8b.
`color(green)("What we do to one side we do to the other:")`

`color(red)("Add "color(blue)(+8b)color(white)(.) underline("to both sides"))`

`color(brown)(-17b+23color(blue)(+8b)=-4-8b color(blue)(+8b)`

Giving:

`color(brown)(-17bcolor(blue)(+8b)+23=-4+0`

`-9b+23=-4`
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`color(blue)("Stage 2")`

Multiply both sides by (-1). This changes -9b into +9b

`color(brown)(+9b-23=+4)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Stage 3")`
Remove -23 from LHS

`color(red)("Add "color(blue)(+23)color(white)(.)underline("to both sides"))`

`color(brown)(+9b-23color(blue)(+23)=+4color(blue)(+23))`

`color(brown)(9b=27)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Stage 3")`
Divide both sides by 9 ( the same thing as `color(blue)(xx 1/9)` )

`color(brown)(9color(blue)(xx 1/9) xxb=27color(blue)(xx1/9))`

`color(brown)(9/(color(blue)(9))xxb = 27/(color(blue)(9))`

But `9/9=1 " and "27/9=3`

`color(green)(1xxb=3`

`b=3`
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