Question 424971


{{{y=2x-3}}}.....1...
{{{y=-(1/2)x+4}}}....2

first multiply second equation by {{{4}}} then add 1 and 2


{{{y=2x-3}}}.....1...
{{{4y=4(-1/2)x+4*4}}}....2


{{{y=2x-3}}}.....1...
{{{4y=cross(4)2(-1/cross(2))x+16}}}....2

{{{y=2x-3}}}.....1...
{{{4y=-2x+16}}}....2
------------------------ now add them

{{{y+4y=2x+(-2x)-3+16}}}

{{{5y=cross(2x)+cross((-2x))-3+16}}}


{{{5y=13}}}}

{{{y=13/5}}}

{{{y=2.6}}}....now find {{{x}}}


{{{y=2x-3}}}.....1...

{{{2.6+3=2x}}}

{{{5.6=2x}}}

{{{5.6/2= x}}}

{{{2.8= x}}}


check:

{{{4y=-2x+16}}}....2

{{{4*2.6=-2*2.8+16}}}

{{{10.4 = -5.6+16}}}

{{{10.4 = 10.4}}}


2.


using substitution method

*[invoke linear_substitution "x", "y", 1, -2, 3, 2, -4, 0]