# How do you solve 2x+y=0 and x-y=1 using substitution?

Jun 16, 2016

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

#### Explanation:

Write one of the equations with a single variable as the subject.
This could be :
$y = - 2 x \text{ or " x = 1 + y " or } y = x - 1$

I would use the fact that both equations can be written with y as the subject.

As $y = y$ it follows that

$x - 1 = - 2 x$
$\text{ } 3 x = 1$

$x = \frac{1}{3}$

Now that we know the value of $x$, substitute into one of the equations to find the value for $y$

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

Jun 16, 2016

$2 x + y = 0. \ldots \ldots \left(i\right)$
$x - y = 1. \ldots \ldots . . \left(i i\right)$

From $\left(i i\right)$ put $y = x - 1$ in $\left(i\right)$.

$2 x + \left(x - 1\right) = 0$
$\implies 3 x = 1$
$\implies x = \frac{1}{3}$

Now,put $x = \frac{1}{3}$ in either $\left(i\right)$ or $\left(i i\right)$ to find the value of $y$. I am going to put $x = \frac{1}{3}$ in $\left(i\right)$ but you can put in $\left(i i\right)$ for practice.

$2 \left(\frac{1}{3}\right) + y = 0$
$\implies y = - \frac{2}{3}$

$S o l u t i o n$ $S e t = \left\{\left(\frac{1}{3} , - \frac{2}{3}\right)\right\}$