How do you solve and check your solutions to #-4q=52#?

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Answer 1 · 2016-10-07T15:19:19

#q=-13#

Given:#" "color(brown)(-4q=52)#

Divide both sides by #color(blue)( 4)#

#color(brown)(-4/(color(blue)(4)) xx q=52/(color(blue)(4)))#

But #-4/4=-1#

#-1xxq= 13#

#-q=13#

Multiply both sides by #(-1)# giving

#+q=-13#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check: Substitute #-13# for #q#

#=>color(brown)(-4q=52)" "->" "color(blue)(-4(-13)=52#

Consider the left hand side (LHS) of the equation.

Minus times minus gives plus (positive) and #4# is the same as #2xx2 #

so #4xx13" "# is the same as

#" "2xx2xx13 -> 2xx26->52 -> LHS #

So LHS=RHS=52