Question

solve the following system of equations
1/6x-1/2y=2;1/3x-4/5y=-2

Answer 1

It is easy when we follow the method of substitution and probably the quickest way to obtain the answer. So, let us start without any wait by using LaTeX.

$\begin{bmatrix}\dfrac{1}{6}x & - & \dfrac{1}{2}y & = & 2 \\ \\ \dfrac{1}{3}x & - & \dfrac{4}{5}y & = & - 2 \end{bmatrix}$

Isolate the variable of x to proceed further for substitution.

$\bf{\dfrac{1}{6}x - \dfrac{1}{2}y + \dfrac{1}{2}y = 2 + \dfrac{1}{2}y}$

$\bf{6 \times \dfrac{1}{6}x = 6 \times 2 + 6 \times \dfrac{1}{2}y}$

$\bf{\dfrac{6 \times 1}{6}x = 12 + \dfrac{6}{2}y}$

$\bf{x = 12 + 3y}$

Let us substitute back into our equation marked as 2nd equation in this current system, that is:

$\bf{\dfrac{1}{3} \Big(12 + 3y \Big) - \dfrac{4}{5}y = - 2}$

$\bf{\dfrac{12}{3} + \dfrac{3y}{3} - \dfrac{4}{5}y = - 2}$

$\bf{y + 4 - \dfrac{4}{5}y = - 2}$

$\bf{y - \dfrac{4}{5}y + 4 - 4 = - 2 - 4}$

$\bf{y - \dfrac{4}{5}y = - 6}$

$\bf{y \times 5 - \dfrac{4}{5}y \times 5 = - 6 \times 5}$

$\bf{5y - 4y = - 30}$

$\boxed{\mathbf{y = - 30}}$

$\bf{\since \quad y = - 30; \: \: x = 12 + 3y}$

$\mathbf{\therefore \quad x = 12 + 3 (- 30)}$

$\mathbf{x = 12 - 3 \times 30}$

$\mathbf{x = 12 - 90}$

$\boxed{\mathbf{x = - 78}}$

Hope it helps.