How do you solve the following system?: $8 x - 5 y = 9, 5 x - 7 y = 2$ $8x -5y =9 , 5x -7y = 2$

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Answer 1 · 2018-03-23T12:36:10

Solution: $x = \frac{53}{31}, y = \frac{29}{31}$ $x=53/31, y=29/31$

$8 x - 5 y = 9; (1); 5 x - 7 y = 2; (2)$ $8x-5y=9 ; (1) ; 5x-7y=2 ; (2)$ . Multiplying equation(1)

by $5$ $5$ and equation(2) by $8$ $8$ we get,

$40 x - 25 y = 45; (3); 40 x - 56 y = 16; (4)$ $40x-25y=45 ; (3) ; 40x-56y=16 ; (4)$ Subtracting

equation(4) from equation(3) we get,

$31y=29 :. y= 29/31$ Putting $y= 29/31$ in equation (1)

we get $8x-5*29/31=9 or 8x= 9+145/31or 8x= 424/31$ or

$x= 424/(31*8) = 53/31:.$ Solution: $x=53/31, y=29/31$ [Ans]

How do you solve the following system?: 8x -5y =9 , 5x -7y = 28x−5y=9,5x−7y=28x -5y =9 , 5x -7y = 2