# How do you solve -4x – 6y = -28 and 2x + 3y = 14 ?

Jul 22, 2016

Here's what I got.

#### Explanation:

Your starting system of two equations with two unknowns looks like this

$\left\{\begin{matrix}- 4 x - 6 y = - 28 \\ \textcolor{w h i t e}{. .} 2 x + 3 y = \textcolor{w h i t e}{-} 14\end{matrix}\right.$

Notice that if you multiply the second equation by $\textcolor{b l u e}{- 2}$, you will get the first equation!

$2 x + 3 y = 14 \text{ } | \times \textcolor{b l u e}{\left(- 2\right)}$

$\textcolor{b l u e}{\left(- 2\right)} \cdot 2 x + \textcolor{b l u e}{\left(- 2\right)} \cdot 3 y = \textcolor{b l u e}{\left(- 2\right)} \cdot 14$

$- 4 x - 6 y = - 28$

This means that your system of equations has an infinite number of solutions. This is the case because if you were to subtract this new form of the second equation from the first equation, you'd get

$\left\{\begin{matrix}- 4 x - 6 y = - 28 \text{ } | - \\ - 4 x - 6 y = - 28\end{matrix}\right.$
$\frac{\textcolor{w h i t e}{a a a a a a a}}{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a}}$

$- 4 x - \left(- 4 x\right) - 6 y - \left(- 6 y\right) = - 28 - \left(- 28\right)$

$- \textcolor{red}{\cancel{\textcolor{b l a c k}{4 x}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{4 x}}} - \textcolor{red}{\cancel{\textcolor{b l a c k}{6 y}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{6 y}}} = - \textcolor{red}{\cancel{\textcolor{b l a c k}{28}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{28}}}$

which of course gives

$0 = 0$

So, when does $0$ equal to $0$? Well, pretty much always, which is why your system of equations is said to have and infinite number of solutions.

In other words, you can plug in any value you want for $x$ and for $y$ because $0 = 0$ will always be true.

Alternatively, you can think about the two equations given to you as describing the same line. To check that this is the case, rearrange both equations in slope-intercept form

$- 4 x - 6 y = - 28$

$- 6 y = 4 x - 28 \implies y = - \frac{2}{3} x + \frac{14}{3}$

Similarly, you have

$2 x + 3 y = 14$

$3 y = - 2 x + 14 \implies y = - \frac{2}{3} x + \frac{14}{3}$

This once again leads to the conclusion that the system has an infinite number of solutions. 