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

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Answer 1 · 2017-08-22T06:57:11

$x = - 4$ $x = -4$

$y = - 1$ $y = -1$

$y = x + 3 (1)$ $y = x+3" "(1)$

$x - 4 y = 0 (2)$ $x-4y = 0" "(2)$

Substitute the value of $y$ $y$ in equation $(2)$ $(2)$ . We get

$x - 4 (x + 3) = 0$ $x-4(x+3) = 0$

$x - 4 x - 12 = 0$ $x-4x-12 = 0$

$- 3 x = 12$ $-3x = 12$

$x = - 4$ $x = -4$

Substitute the value in equation $(1)$ $(1)$ . We get

$y = - 4 + 3$ $y = -4+3$

$y = - 1$ $y = -1$

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