Question #98353

1 Answer
Mar 27, 2017

Answer:

See explanation

Explanation:

Note that #4xx6=24#

#4xxEquation(1) ->24x-20y=-14#
#6xxEquation(2)->ul(24x+12y=color(white)(..)168) larr" subtract"#
#" "0x-32y= -182#

#-32y=-182#

Multiply both sides by (-1)

#32y=182#

Divide both sides by 32

#32/32 y=182/32#

but #32/32=1#

#y=91/16#

Where you can, always use fraction. They are more precise than decimals. Using decimals introduces rounding errors.

Just substitute for #y# in #Equation(1) or 2# to determine #x#