How do you solve the system of equations algebraically 2x+5y=4, 3x+6y=5?

Jul 11, 2018

Solution: $x = \frac{1}{3} , y = \frac{2}{3}$

Explanation:

2 x+ 5 y = 4 ; (1) , 3 x+ 6 y = 5 ;(2) . Multiplying equation (1) by

$3$ and equation (2) by $2$ we get ;

6 x+ 15 y = 12 ; (3) , 6 x+ 12 y = 10 ;(4) . Subtracting equation

(4) from equation (3) we get,

(6 x+ 15 y)-(6 x+12 y)= 12-10;  or

$\cancel{6 x} + 15 y - \cancel{6 x} - 12 y = 2 \mathmr{and} 3 y = 2 \therefore y = \frac{2}{3}$ Putting

$y = \frac{2}{3}$ in equation (2) we get, $3 x + 6 \cdot \frac{2}{3} = 5 \mathmr{and} 3 x + 4 = 5$ or

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

Solution: $x = \frac{1}{3} , y = \frac{2}{3}$ [Ans]