# How do solve the following linear system?:  2x+5y=1 , 4x+3y=1 ?

Jul 17, 2016

$x = \frac{1}{7} \mathmr{and} y = \frac{1}{7}$

#### Explanation:

To solve a system of two unknowns there are more than a way.
Get rid of $x$ and then we solve for $y$ by applying the following method:

Method:

1) Multiply one of the equations by an integer in such a way to obtain opposite coefficients of $x$ in both equations.

2)Add both equations to obtain an equation with one unknown $y$.

3) solve for $y$

4) Substitute $y$ in one of the original equations (equation before multiplying the integer).

5) Solve for $x$.

Let's apply this method:
$2 x + 5 y = 1 \left(e q .1\right)$

$4 x + 3 y = 1 \left(e q 2\right)$

Multiply eq.1 by $- 2$ so we have :
$- 4 x - 10 y = - 2 \left(e q 1\right)$
$4 x + 3 y = 1 \left(e q 2\right)$

$- 7 y = - 1$
then $y = \frac{1}{7}$

Substitute $y$ in eq.2 :
$4 x + 3 \left(\frac{1}{7}\right) = 1$
$\Rightarrow 4 x + \frac{3}{7} = 1$
$\Rightarrow 4 x = 1 - \frac{3}{7}$
$\Rightarrow 4 x = \frac{4}{7}$
$\Rightarrow x = \frac{1}{7}$

We can check if the values of $x$ and $y$ are right by substituting them in one of the equations
Substitute the values in eq1 :
 2(1/7) + 5(1/7) =? 1
 2/7 + 5/7 =? 1
 7/7 =?1  true

Therefore , $x = \frac{1}{7} \mathmr{and} y = \frac{1}{7}$