Question

solve the system of equations.\\(10x - 3y - 81 = 0\\)\\(- 5x - 7y - 19 = 0\\)\\(x = \square\\)\\(y = \square\\)

Explanation:

Step1: Rearrange first equation

$10x - 3y = 81$

Step2: Rearrange second equation

$-5x - 7y = 19$

Step3: Eliminate $x$: multiply eq2 by 2

$2(-5x - 7y) = 2(19) \implies -10x - 14y = 38$

Step4: Add eq1 and new eq2

$(10x - 3y) + (-10x - 14y) = 81 + 38$
$\implies -17y = 119$

Step5: Solve for $y$

$y = \frac{119}{-17} = -7$

Step6: Substitute $y=-7$ into eq1

$10x - 3(-7) = 81 \implies 10x + 21 = 81$

Step7: Solve for $x$

$10x = 81 - 21 = 60 \implies x = \frac{60}{10} = 6$

Answer:

$x = 6$
$y = -7$