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

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Answer 1 · 2016-03-20T16:27:27

The solutions are:

$x = \frac{7}{3}$ $color(green)(x = 7/3$
$y = \frac{1}{3}$ $color(green)( y = 1/ 3$

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

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

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

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

$1 + 9 y = 4$ $1 + 9y = 4$

$9 y = 4 - 1$ $9y = 4 -1$

$9 y = 3$ $9y = 3$

$y = \frac{3}{9} = \frac{1}{3}$ $y = 3 / 9 = color(green)(1/ 3$

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

$x = 1 + 4 \cdot (\frac{1}{3})$ $x = 1 + 4 * (1/3)$

$x = 1 + (\frac{4}{3})$ $x = 1 + ( 4 /3)$

$x = \frac{3}{3} + (\frac{4}{3})$ $x = 3/3 + ( 4 /3)$

$x = \frac{7}{3}$ $color(green)(x = 7/3$

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