How do you solve the following system: $2 x - 4 y = 6, 3 x - 3 y = 2$ $2x-4y=6 , 3x - 3y = 2$ ?

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Answer 1 · 2018-07-18T09:11:57

$x = - \frac{5}{3}$ $x=-5/3$ and $y = - \frac{7}{3}$ $y=-7/3$

Dividing by $2$ $2$ , $2 x - 4 y = 6$ $2x-4y=6$ gives us $x - 2 y = 3$ $x-2y=3$

or $x - y - y = 3$ $x-y-y=3$

Multiplying by $3$ $3$ , we get

$3 x - 3 y - 3 y = 9$ $3x-3y-3y=9$ , but $3 x - 3 y = 2$ $3x-3y=2$ ,

therefore $2 - 3 y = 9$ $2-3y=9$

or $2 - 9 = 3 y$ $2-9=3y$ i.e. $3 y = - 7$ $3y=-7$ and $y = - \frac{7}{3}$ $y=-7/3$

Hence $x - 2 y = 3$ $x-2y=3$ becomes $x - (- \frac{14}{3}) = 3$ $x-(-14/3)=3$

or $x = 3 - \frac{14}{3} = - \frac{5}{3}$ $x=3-14/3=-5/3$

Hence $x = - \frac{5}{3}$ $x=-5/3$ and $y = - \frac{7}{3}$ $y=-7/3$

Answer 2 · 2018-07-18T09:19:20

The solution is $S={(x=-5/3),(y=-7/3):}$

Solve the simutaneous equations by substitution

${(2x-4y=6),(3x-3y=2):}$

$<=>$ , ${(2x=4y+6),(3x-3y=2):}$

$<=>$ , ${(x=2y+3),(3(2y+3)-3y=2):}$

$<=>$ , ${(x=2y+3),(6y+9-3y=2):}$

$<=>$ , ${(x=2y+3),(3y=-9+2):}$

$<=>$ , ${(x=2*(-7/3)+3),(y=-7/3):}$

$<=>$ , ${(x=-5/3),(y=-7/3):}$

How do you solve the following system: 2x-4y=6 , 3x - 3y = 2 2x−4y=6,3x−3y=22x-4y=6 , 3x - 3y = 2 ?