Question 1049211
{{{y}}}={{{x-13}}} (1)
{{{2x+3y}}}={{{1}}} (2)


first Method Substitution


{{{y}}}={{{x-13}}} subsitute eq(1) into eq(2)

{{{2x+3y}}}={{{1}}}
{{{2x+3(x-13)}}}={{{1}}} distribute the 3

{{{2x+3x-39)}}}={{{1}}}
{{{5x-39)}}}={{{1}}} add 39 to both sides

{{{5x)}}}={{{1+39}}}

{{{5x)}}}={{{40}}} divide both sides by 5

{{{5x)}}}={{{40}}}
{{{x)}}}={{{8}}}

if {{{x)}}}={{{8}}} then {{{y}}}={{{x-13}}} becomes
{{{y}}}={{{8-13}}}

{{{y}}}={{{-5}}}

second method elimination

{{{y}}}={{{x-13}}} (1) turn this equation into standard form



{{{y}}}-{{{x}}}={{{-13}}} (1)
{{{2x+3y}}}={{{1}}} (2)

now re-arrange
{{{-x}}}+{{{y}}}={{{-13}}} (1)
{{{2x+3y}}}={{{1}}} (2)


multiply eq(1) by 2 on both sides
{{{-2x}}}+{{{2y}}}={{{-26}}} (1)

add eq(1) and (2)

{{{-2x}}}+{{{2y}}}={{{-26}}} 
{{{2x+3y}}}={{{1}}} 
{{{5y}}}={{{-25}}} 
{{{y}}}={{{-5}}} 

then find the value of x

{{{-x}}}+{{{-5}}}={{{-13}}} 
{{{-x}}}={{{-13}}}+{{{5}}}
{{{x}}}={{{8}}}