How do you solve $x - 4y - 1 = 0$ and $x + 5y - 4 = 0$ using substitution?

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Answer 1 · 2016-03-20T16:27:27

The solutions are:

$color(green)(x = 7/3$
$color(green)( y = 1/ 3$

$x- 4y - 1 = 0$
$x- 4y = 1$
$color(green)(x= 1 + 4y$ ..........equation $(1)$

$x + 5y - 4 = 0$
$x + 5y = 4$ .............equation $(2)$

Substituting equation $(1)$ in $(2):$

$color(green)(1 + 4y ) + 5y = 4$

$1 + 9y = 4$

$9y = 4 -1$

$9y = 3$

$y = 3 / 9 = color(green)(1/ 3$

Finding the value of $x$ from equation $(1)$ :
$color(green)(x= 1 + 4y$

$x = 1 + 4 * (1/3)$

$x = 1 + ( 4 /3)$

$x = 3/3 + ( 4 /3)$

$color(green)(x = 7/3$

How do you solve x - 4y - 1 = 0x - 4y - 1 = 0 and x + 5y - 4 = 0 x + 5y - 4 = 0 using substitution?