How do you determine whether a linear system has one solution, many solutions, or no solution when given 6x +y= -6 and 4x+3y= 17?

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Answer 1 · 2015-10-13T17:17:14

In this case we can reformulate the equations as two slope intercept equations describing lines of different slope and therefore one solution.

Slope intercept form of the equation of a line is:

$y = m x + c$ $y = mx + c$

where $m$ $m$ is the slope and $c$ $c$ the intercept.

Starting with $6 x + y = - 6$ $6x+y = -6$ , subtract $6 x$ $6x$ from both sides to get:

$y = - 6 x - 6$ $y = -6x-6$

This is a line with slope $- 6$ $-6$ and intercept $- 6$ $-6$ .

Starting with $4 x + 3 y = 17$ $4x+3y=17$ , first subtract $4 x$ $4x$ from both sides to get:

$3 y = - 4 x + 17$ $3y = -4x+17$

Then divide both sides by $3$ $3$ to get:

$y = - \frac{4}{3} x + \frac{17}{3}$ $y = -4/3x + 17/3$

This is a line with slope $- \frac{4}{3}$ $-4/3$ and intercept $\frac{17}{3}$ $17/3$

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

graph{(6x+y+6)(4x+3y-17) = 0 [-20.78, 19.22, -4.16, 15.84]}