How do you determine whether a linear system has one solution, many solutions, or no solution when given 2x+3y= -12 and 2x-y = -4?

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How do you determine whether a linear system has one solution, many solutions, or no solution when given 2x+3y= -12 and 2x-y = -4?

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Answer 1 · 2015-11-06T20:44:14

If the gradients are different in a 2 equation linear system then 1 solution

If they are parallel then no solution. They will not cross.

Explanation:

In your case it is given that:
$2 x + 3 y = - 12$ $2x+3y=-12$ ..................................(1)
$2 x - y = - 4$ $2x-y=-4$ .......................................(2)

Write as:

$y = - \frac{2}{3} x - 4$ $y=-2/3 x -4$ ...........................( $1_{a}$ $1_a$ )
$y = 2 x + 4$ $y= 2x+4$ ....................................( $2_{a}$ $2_a$ )

The gradient of ( $1_{a}$ $1_a$ ) is negative so its plot has a downward slope when progressing from left to right.

The gradient of ( $2_{a}$ $2_a$ ) is positive so its plot has an upward slope as you progress from left to right.

As there is no restraint on the magnitude of $x$ $x$ we have $\infty \leq x \leq \infty$ $oo<=x<=oo$ so at some point they will cross and then carry on. This will produce just 1 solution.

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How do you determine whether a linear system has one solution, many solutions, or no solution when given 2x+3y= -12 and 2x-y = -4?

1 Answer

Explanation:

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