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maria [59]
3 years ago
9

Create a system of linear equations that when solved algebraically, prove to have no solution.

Mathematics
2 answers:
sweet [91]3 years ago
8 0

Basically, you just need to make them have the same slope, but a different y - intercept. That means they'll never intersect, so will have no solution.

Say one equation is

y = 3x + 5

That means the other can be something like

y = 3x + 6

Right? Because if

3x + 5 = 3x + 6

and you subtract 3x from both sides

5 = 6

which can't be true.

Natalija [7]3 years ago
7 0
The answer is y=3x+7 and y=3x+4
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Write the standard form of the equation that is parallel to y = -6x + 5 and goes through point (4, 4).
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6x + y = 28.  The standard form of the equation that is parallel to y = -6x + 5 and goes through point (4, 4) is 6x + y = 28.

The equation is written in the slope-intercept form y = mx +b. So:

y = -6x + 5

The slope m = -6

Since the slopes of parallel lines are the same, we are looking for a slope line m = -6 and goes through point (4, 4).

With the slope-intercept form:

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