How do you solve #-4r - 11 = 4r + 21#?

1 Answer
Oct 17, 2016

Answer:

#r" "=" "-4#

Explanation:

The objective is to have just have one #r# and for it to be on one side of the = and everything else on the other.

I am going to use the first principles from which the short cuts are derived.

#color(brown)("For add or subtract change the value into 0 and it ends up on the other side of =")#
#color(brown)("For divide or multiply change the value to 1 and it ends up on the other side of =")#

You will see what I mean.
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#"Given: "-4r-11=4r+21#

As the right hand side (RHS) #4r# is positive we change the left hand side (LHS) one to 0

#color(green)("Add "color(blue)(4r)" to both sides")#

#color(brown)(-4rcolor(blue)(+4r)-11" "=" "4rcolor(blue)(+4r)+21#

#" "0-11" "=" "4rcolor(blue)(+4r)+21#

#" "-11" "=" "8r+21#
............................................................................................................

#color(green)("Subtract "color(blue)(21)" from both sides")#

#color(brown)(" "-11color(blue)(-21)" "=" "8r+21color(blue)(-21))#

#" "-11-21" "=" "8r+0#

#" "8r" "=" "-32#
.........................................................................................................

#color(green)("Divide both sides by "color(blue)(8))#

#color(brown)(" "8/(color(blue)(8))r" "=" "-32/(color(blue)(8))#

But #8/8=1# giving

#" "r" "=" "-4#

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Shortcut methods:

For add or subtract move the other side of = and reverse its sign

For multiply or divide move the other side of = and reverse its sine (action). So add becomes subtract and multiply becomes divide and so on

Examples:
#" "color(brown)(2x=3" "->" " x=3/2)" "color(blue)(x/2=3" "->" "x=2xx3)#

#color(brown)(2+x=3" "->" " x=3-2)color(blue)(" "x-2=3" "->" "x=3+2)#