# How do you solve the following system: 3x + 2y = 1, 2x + 3y = 12?

Dec 30, 2015

The solution set is: $S = \left\{- 4.2 , 6.8\right\}$

#### Explanation:

To make it easier, we take first the second equation and isolate x:
$2 x + 3 y = 12 \implies 2 x = 12 - 3 y \implies x = \frac{12 - 3 y}{2}$.
Then, we change x in the first equation for its new "value" and solve it:
$3 x + 2 y = 1 \implies$
$3 \left(\frac{12 - 3 y}{2}\right) + 2 y = 1 \implies$
$18 - \frac{9 y}{2} + 2 y = 1 \implies$
$- \frac{9 y}{2} + \frac{4 y}{2} = - 17 \implies$
$- 5 \frac{y}{2} = - 17 \implies$
$- 5 y = - 34$
$y = \frac{34}{5} = 6.8$
We now have $y$, and to find $x$, we go back to the first equation:
$x = \frac{12 - 3 y}{2}$
$x = 6 - \frac{20.4}{2} = 6 - 10.2 = - 4.2$.
So, the solution set is: $S = \left\{- 4.2 , 6.8\right\}$

Dec 30, 2015

$x = - \frac{21}{5}$, $y = \frac{34}{5}$

#### Explanation:

I'm going to show you two methods to do this, namely Substitution and Elimination.

1. Substitution

As the name suggests, we'll substitute one of the variables in terms of another.

We'll have to write $x$ in terms of $y$ from the first equation, and then substitute that value in the second equation.

From the first equation we have:

$3 x + 2 y = 1$
$3 x = 1 - 2 y$

$x = \frac{1 - 2 y}{3}$

Now that we have a value for $x$, let's substitute that in the second equation. We get:

$2 \left(\frac{1 - 2 y}{3}\right) + 3 y = 12$

$\frac{2}{3} - \frac{4}{3} y + 3 y = 12$

$\frac{5}{3} y = \frac{34}{3}$

$y = \frac{34}{\cancel{3}} \cdot \frac{\cancel{3}}{5}$

$y = \frac{34}{5}$

We can now substitute this value in any of the two equations in the beginning to get $x$:

$3 x + 2 \left(\frac{34}{5}\right) = 1$

$3 x = - \frac{63}{5}$

$x = - \frac{21}{5}$

2. Elimination

In this method, we'll first eliminate either the $x$ or $y$ terms first. We can do this by making the coefficients of either variables equal.

I'm going to eliminate $x$.
In the first equation, the coefficient of $x$ is $3$, while in the second one it's $2$. What's the LCM of $3$ and $2$? $6$ of course!

So let's make the coefficients of $x$ equal to $6$ in both equations. We can do this by multiplying the first equation with $2$ and the second one with $3$.

$\left(3 x + 2 y = 1\right) \cdot 2 \Rightarrow 6 x + 4 y = 2$

$\left(2 x + 3 y = 12\right) \cdot 3 \Rightarrow 6 x + 9 y = 36$

Now, to eliminate the terms with $x$ we'll subtract the first equation from the second one.

$\left(6 x + 9 y\right) - \left(6 x + 4 y\right) = \left(36 - 2\right)$

$5 y = 34$

$y = \frac{34}{5}$

Now, like we did in the first step we can substitute this value of $y$ in any of the equations, to get the value of $x$,

$x = - \frac{21}{5}$