How do you solve the system of equations 2x+8y=6 and -5x-20y=-15?

Nov 19, 2016

This system of equation doesn't have any solutions.

Explanation:

$2 x + 8 y = 6$
$- 5 x - 20 y = - 15$

Let's simplify these equations: divide the first one by $2$ andd the second by $- 5$:

$x + 4 y = 3$
$x + 4 y = 15$

Now, using any of the equations, we must express one variable by the other, let's say $x$ by $y$ from the second equation:

$x = 15 - 4 y$

Using this we solve the other equation:

$x + 4 y = 3$
$15 - 4 y + 4 y = 3$
$15 = 3$

We got $15 = 3$ which obviously isn't true so this equation has no solution. It follows that this system of equation doesn't have any solutions as well.

Nov 19, 2016

$x$ can have any value and the equations will work.
The y- values will depend on the x chosen.

Explanation:

$2 x + 8 y = 6 \text{ and } - 5 x - 20 y = - 15$

There are different options available to solve the equations.
$\rightarrow \text{ }$ elimination
$\rightarrow \text{ }$ substitution $\rightarrow$ single variable
$\rightarrow \text{ }$ equating
$\rightarrow \text{ }$matrices
$\rightarrow \text{ }$ graphically

To be able to use substitution, a single variable in one of the equations is a useful indicator

$2 x + 8 y = 6 \text{ "div2 rarr " } x + 4 y = 3$

Make x the subject of the equation:

$\textcolor{red}{x = 3 - 4 y}$

Replace $\textcolor{red}{x}$ in the second equation by $\textcolor{red}{\left(3 - 4 y\right)}$

$\text{ } - 5 \textcolor{red}{x} - 20 y = - 15$
$\textcolor{w h i t e}{\times \times} \downarrow$
$- 5 \textcolor{red}{\left(3 - 4 y\right)} - 20 y = - 15$

$- 15 + 20 y - 20 y = - 15$

$- 15 = - 15$

$x$ can have any value