How do you solve the system of equations $8 x + 4 y = 13$ $8x+4y=13$ and $- 2 x = y + 4$ $-2x=y+4$ ?

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Answer 1 · 2017-05-03T13:46:18

See the entire solution process below:

First, sole the second equation for $y$ $y$ :

$- 2 x = y + 4$ $-2x = y + 4$

$- 2 x - 4 = y + 4 - 4$ $-2x - color(red)(4) = y + 4 - color(red)(4)$

$- 2 x - 4 = y + 0$ $-2x - 4 = y + 0$

$- 2 x - 4 = y$ $-2x - 4 = y$

$y = - 2 x - 4$ $y = -2x - 4$

Step 2) Substitute $- 2 x - 4$ $-2x - 4$ for $y$ $y$ in the first equation and solve for $x$ $x$ :

$8 x + 4 y = 13$ $8x + 4y = 13$ becomes:

$8 x + 4 (- 2 x - 4) = 13$ $8x + 4(-2x - 4) = 13$

$8 x + (4 \cdot - 2 x) - (4 \cdot 4) = 13$ $8x + (4 * -2x) - (4 * 4) = 13$

$8 x - 8 x - 16 = 13$ $8x -8x - 16 = 13$

$0 - 16 = 13$ $0 - 16 = 13$

$- 16 \neq 13$ $-16 != 13$

Therefore, there are no solutions to this problem. Or, the solution is the empty or null set: ${\emptyset}$ ${O/}$ .

This means the two lines defined by these equations are parallel and are not the same line.

How do you solve the system of equations 8x+4y=138x+4y=13 and −2x=y+4-2x=y+4?