How do you solve $- 6 y - 2 x = 0$ $-6y - 2x = 0$ and $11 y + 3 x = 4$ $11y + 3x = 4$ using substitution?

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Answer 1 · 2017-01-05T21:46:04

$x = - 6 and y = 2$ $x=-6 and y=2$

$- 6 y - 2 x = 0 \to - 6 y = 2 x$ $-6y -2x =0 to -6y = 2x$
Therefore $- 3 y = x$ $-3y = x$

We can substitute $- 3 y = x$ $color(blue)(-3y = x)$ into the second equation;

$11 y + 3 (x) = 4$ $11y + 3 color(red)((x)) = 4$
$11 y + 3 (- 3 y) = 4$ $11y + 3 color(red)((-3y)) = 4$

This is simplified to;

$11 y - 9 y = 4$ $11y -9y =4$
$2 y = 4$ $2y=4$
$y = 2$ $y=2$

As we know that $- 3 y = x$ $color(blue)(-3y = x)$ , we can substitute the value of $y$ $y$ into that to get $x$ $x$

$- 3 y = x \to - 3 (2) = x$ $-3y = x to -3(2) =x$

$- 6 = x$ $-6=x$

How do you solve −6y−2x=0-6y - 2x = 0 and 11y+3x=411y + 3x = 4 using substitution?