How do you write #y=3/8x-3/4# in standard form?

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Answer 1 · 2015-06-28T06:50:55

#y=3/8x-3/4# in standard form is #3x-8y=6#.

Standard form for a linear equation is #Ax+By=C#.

#y=3/8x-3/4#

Simplify #3/8x# to #(3x)/8#.

Subtract #(3x)/8# from both sides of the equation.

#(-3x)/8+y=-3/4#

Multiply both sides by #-1#.

#(-3x)/8(-1)+y(-1)=-3/4(-1)# =

#(3x)/8-y=3/4#

Multiply both sides by #8#.

#((3x)*8)/8-y(8)=((3)*8)/4# =

#((3x)*cancel8^1)/(cancel 8^1)-y(8)=((3)*cancel8^2)/(cancel4^1)# =

#3x-8y=6#