How do you solve for #x# in the equation #a=bx+c#?

1 Answer
Aug 9, 2016

#x=(a-c)/b#

Explanation:

#color(green)("Using first principles method")#

Subtract #color(blue)(c)# from both sides

#color(brown)(acolor(blue)(-c)=bx+c color(blue)(-c)#

#a-c=bx+0#

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

#color(brown)((a-c)/(color(blue)(b))=b/(color(blue)(b))xxx)#

But #b/b=1# giving

#(a-c)/b=x#

#x=(a-c)/b#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(green)("Using shortcut method")#

Move #c# to the other side of = and change its sign from + to - (opposite action)

#a-c=bx#

As #b# is multiply move it to the other side of = and change it to divide (opposite action).

#(a-c)/b=x#