# How do you solve the simultaneous equations 6x + 2y = 50 and 2x + 4y = 20?

Aug 9, 2015

$\left\{\begin{matrix}x = 8 \\ y = 1\end{matrix}\right.$

#### Explanation:

$6 x + 2 y = 50 \text{ " " " " } \left(1\right)$

Can be written as

$6 x + 2 y - 50 = 0$

$2 x + 4 y = 20 \text{ " " " } \left(2\right)$

Can be written as

$2 x + 4 y - 20 = 0$

To eliminate $x$, let us multiply equation $\left(2\right)$ by $3$; then it becomes -

$6 x + 12 y - 60 = 0$

Since both equations are equal to zero; we shall have them like this-

$6 x + 2 y - 50 = 6 x + 12 y - 60$

Take the unknowns to left and the constants to the right.

$\cancel{6 x} + 2 y - \cancel{6 x} - 12 y = - 60 + 50$

$- 10 y = - 10 \implies y = \frac{\left(- 10\right)}{\left(- 10\right)} = 1$

Substitute $y = 1$ in equation $\left(1\right)$

$6 x + 2 \cdot 1 = 50$

$6 x = 50 - 2$

$x = \frac{48}{6} = 8$