Question 700301

two supplementary angles {{{alpha}}} and {{{beta}}} add up to {{{180}}}°

so, {{{alpha+beta=180}}}°

if  {{{alpha=6y + 3}}} and {{{beta=4y -13}}}, than we have


{{{6y + 3+4y -13=180}}}°...solve for {{{y}}}


{{{10y -10=180}}}°

{{{10y =180+10}}}°

{{{10y =190}}}°

{{{y =190/10}}}°

{{{y =19}}}°......now find {{{alpha}}} and {{{beta}}}


{{{alpha=6y + 3}}}° ...plug in {{{y}}} value

{{{alpha=6*19 + 3}}}°

{{{alpha=114 + 3}}}°

{{{highlight(alpha=117)}}}°



and {{{beta=4*19 -13}}}°

{{{beta=76 -13}}}°

{{{highlight(beta=63)}}}°