# How do you solve the simultaneous equations 3x-4y=11 and 5x+6y=12?

Aug 1, 2015

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

#### Explanation:

You can solve this system of equations by using the multiplication method.

The first thing that you need to do is pick a variable to eliminate first, then figure out the least common multiple (LCM) of its coefficients.

Let's say that you want to eliminate $x$ and solve for $y$ first. The two coefficients of $x$ are $3$ and $5$, which means that they're LCM will be equal to $15$.

So, multiply the first equation by $5$ and the second equation by $- 3$ to get

$5 \cdot \left(3 x - 4 y\right) = 5 \cdot 11$

$15 x - 20 y = 55$

and

$\left(- 3\right) \cdot \left(5 x + 6 y\right) = - 3 \cdot 12$

$- 15 x - 18 y = - 36$

Add the left side and the right side of these two equations separately to get

$\textcolor{red}{\cancel{\textcolor{b l a c k}{15 x}}} - 20 y - \textcolor{red}{\cancel{\textcolor{b l a c k}{15 x}}} - 18 y = 55 - 36$

$- 38 y = 19 \implies y = \frac{19}{- 38} = \textcolor{g r e e n}{- \frac{1}{2}}$

Now use this value of $y$ in one of the two equations to determine the value of $x$.

$3 x - 4 \left(- \frac{1}{2}\right) = 11$

$3 x + 2 = 11 \implies x = \frac{11 - 2}{3} = \textcolor{g r e e n}{3}$

The solutions to this system of equations are

$\left\{\begin{matrix}x = 3 \\ y = - \frac{1}{2}\end{matrix}\right.$