Question 760154

if one number (let it be {{{x}}}) is {{{1}}} less than a second number (let it be {{{y}}}), then

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

if twice the second number is {{{20}}} less than {{{4}}} times the first, then

{{{2y+20=4x}}}....eq.2

solve the system:

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

{{{2y+20=4x}}}....eq.2
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 {{{x+1=y}}}....eq.1...substitute in eq.2

{{{2(x+1)+20=4x}}}....eq.2

{{{2x+2+20=4x}}}

{{{22=4x-2x}}}

{{{22=2x}}}

{{{22/2=x}}}

{{{highlight(11=x)}}}

now find {{{x+1=y}}}....eq.1

{{{11+1=y}}}

{{{highlight(12=y)}}}