How do you solve the following system?:  2x + 3y = 1 , -3x + 7y = 1

May 19, 2018

Solution: $x = \frac{4}{23} , y = \frac{5}{23}$

Explanation:

 2 x + 3 y =1 ; (1) , -3 x +7 y =1 ; (2) Multiplying equation

(1) by $3$ and equation (2) by $2$ we get,

 6 x + 9 y =3 ; (3) , -6 x +14 y =2 ; (4) Adding equation

(3) and equation (4) we get, $23 y = 5 \therefore y = \frac{5}{23}$. Putting

$y = \frac{5}{23}$ in equation (1) we get, $2 x + 3 \cdot \frac{5}{23} = 1$ or

$2 x = 1 - \frac{15}{23} \mathmr{and} 2 x = \frac{23 - 15}{23} \mathmr{and} 2 x = \frac{8}{23} \mathmr{and} x = \frac{4}{23}$

Solution: $x = \frac{4}{23} , y = \frac{5}{23}$[Ans]

May 19, 2018

$y = \frac{5}{23}$

$x = \frac{4}{23}$

Explanation:

$2 x + 3 y = 1$
$- 3 x + 7 y = 1$

Probably the easiest way to solve this is by elimination; notice there is a GCD of $2 x$ and $- 3 x$ of $6 x$ so let's multiply both equations by 3 and 2 respectively:

$3 \left(2 x + 3 y = 1\right)$
$2 \left(- 3 x + 7 y = 1\right)$

$6 x + 9 y = 3$
$- 6 x + 14 y = 2$

now add the equations together, notice the x terms are eliminated:

$23 y = 5$

$y = \frac{5}{23}$

now use either original equations with the y value you found to solve for x:

$2 x + 3 \left(\frac{5}{23}\right) = 1$

$x = \frac{4}{23}$