How do you determine whether a linear system has one solution, many solutions, or no solution when given x+ 5y=1 and -3x +4y= 16?

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How do you determine whether a linear system has one solution, many solutions, or no solution when given x+ 5y=1 and -3x +4y= 16?

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Answer 1 · 2015-10-25T08:37:30

Rearrange in slope intercept form and notice that the slopes are different, so these equations represent two non-parallel lines that intersect at one point - that is one solution.

Explanation:

Given $x + 5 y = 1$ $x+5y=1$ , subtract $x$ $x$ from both sides, then divide both sides by $5$ $5$ to get:

$y = - \frac{1}{5} x + \frac{1}{5}$ $y = -1/5 x + 1/5$

Given $- 3 x + 4 y = 16$ $-3x+4y=16$ , add $3 x$ $3x$ to both sides, then divide both sides by $4$ $4$ to get:

$y = \frac{3}{4} x + 4$ $y = 3/4 x + 4$

So these equations represent lines with slope $- \frac{1}{5}$ $-1/5$ and $\frac{3}{4}$ $3/4$ .

Since the slopes are different, the lines will intersect at exactly one point.

graph{(x+5y-1)(-3x+4y-16) = 0 [-10, 10, -4.5, 5.5]}

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How do you determine whether a linear system has one solution, many solutions, or no solution when given x+ 5y=1 and -3x +4y= 16?

1 Answer

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