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How do you determine whether a linear system has one solution, many solutions, or no solution when given 3x-2y= 10 and -6x +4y= -20?

1 Answer

Oct 23, 2015

See the explanation.

Explanation:

$3 x - 2 y = 10$ $3x-2y=10$
$- 6 x + 4 y = - 20 ∣ : (- 2)$ $-6x+4y=-20 | :(-2)$

$3 x - 2 y = 10$ $3x-2y=10$
$3 x - 2 y = 10$ $3x-2y=10$

So, we get system of two equal equations and because of that, it has many solutions.

Let's pick one unknown as a parameter: $x = K$ $x=K$ , then:

$3 K - 2 y = 10 \Rightarrow 2 y = 3 K - 10 \Rightarrow y = \frac{3 K - 10}{2} = \frac{3}{2} K - 5$ $3K-2y=10 => 2y=3K-10 => y=(3K-10)/2=3/2K-5$

Every pair $(K, \frac{3}{2} K - 5)$ $(K,3/2K-5)$ where $K \in R$ $K in R$ is solution of given system.

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