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How do you determine whether a linear system has one solution, many solutions, or no solution when given 5x+3y=19 and -5x+2y=21?

1 Answer

Oct 14, 2015

$x = - 1$ $x=-1$
$y = 8$ $y=8$

Explanation:

$5 x + 3 y = 19$ $5x+3y=19$
$- 5 x + 2 y = 21$ $-5x+2y=21$

If the two lines are parallel, then they have no solution.

If one equation is a multiple of another equation, then they have many solutions.

$5 x + 3 y = 19$ $5x+3y=19$ ---------------(1) Add the two equations.
$- 5 x + 2 y = 21$ $-5x+2y=21$ ---------------(2)

$5 y = 40$ $5y=40$
$y = \frac{40}{5} = 8$ $y=40/5=8$

Substitute $y = 8$ $y=8$ in equation (1)

$5 x + 3 (8) = 19$ $5x+3(8)=19$
$5 x + 24 = 19$ $5x+24=19$
$5 x = 19 - 24 = - 5$ $5x=19-24=-5$
$x = \frac{- 5}{5} = - 1$ $x=(-5)/5=-1$

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