How do you solve the following system: $6 x + y = 2, 3 x - 4 y = - 10$ $6x + y = 2, 3x - 4y = -10$ ?

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Answer 1 · 2016-03-15T11:06:35

$x = - \frac{2}{27}$ $x = -2/27$ and $y = \frac{22}{9}$ $y = 22/9$

$6 x + y = 2$ $6x+y = 2$ ---------Eqn $1$ $1$
$3 x - 4 y = - 10$ $3x-4y=-10$ ---------Eqn $2$ $2$

Multiplying Eqn $2$ $2$ by $2$ $2$

$6 x - 8 y = - 20$ $6x-8y=-20$ ---------Eqn $3$ $3$

Subtracting Eqn $1$ $1$ from Eqn $3$ $3$
Note: Subtracting changes the bottom equation's sign

$cancel(6x)+y = 2$
$cancel(-6x)+8y=20$

$9y = 22 implies y = 22/9$

put $y = 22/9$ in Eqn $2$

$3x-4(22/9)=-10$

$3x-88/9=-10$

$27x-88=-90$

$27x=-90+88$

$27x = -2 implies x = -2/27$

How do you solve the following system: 6x + y = 2, 3x - 4y = -106x+y=2,3x−4y=−106x + y = 2, 3x - 4y = -10?