Question 622347

let's numbers be:

first: {{{x}}}

second: {{{y}}}}

if first number is eight more than a second number, we have

{{{x=y+8}}}............1

if three times the first is eight less than five times the second, we have

{{{3x+8=5y}}}............2

now, solve the system:

{{{x=y+8}}}............1...substitute in 2

{{{3x+8=5y}}}............2
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{{{3(y+8)+8=5y}}}............2...solve for {{{y}}}

{{{3y+24+8=5y}}}

{{{24+8=5y-3y}}}

{{{32=2y}}}


{{{highlight(16=y)}}}....now find {{{x}}}

{{{x=y+8}}}............1

{{{x=16+8}}}

{{{highlight(x=24)}}}