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sashaice [31]
2 years ago
10

Determine if the following system of equations has no solutions 4x+3y= -8 -8x-6y= 16

Mathematics
1 answer:
cupoosta [38]2 years ago
3 0

\begin{cases} 4x+3y=-8\\\\ -8x-6y=16 \end{cases}~\hspace{10em} \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill

4x+3y=-8\implies 3y=-4x-8\implies y=\cfrac{-4x-8}{3}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{4}{3}} x-\cfrac{8}{3} \\\\[-0.35em] ~\dotfill\\\\ -8x-6y=16\implies -6y=8x+16\implies y=\cfrac{8x+16}{-6} \\\\\\ y=\cfrac{8}{-6}x+\cfrac{16}{-6}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{4}{3}} x-\cfrac{8}{3}

one simple way to tell if both equations do ever meet or have a solution is by checking their slope, notice in this case the slopes are the same for both, meaning the lines are parallel lines, however, notice both equations are really the same, namely the 2nd equation is really the 1st one in disguise.

since both equations are equal, their graph will be of one line pancaked on top of the other, and the solutions is where they meet, hell, they meet everywhere since one is on top of the other, so infinitely many solutions.

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Hey there!

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KEEP: 8 – 6 because that helps solve for your r-value

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• FOURTH: ANSWER THE QUESTION

8 – 6 = 2

FIFTH: ANSWER

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The equation of the line is y = -2x + 16 which passes through the point (4, 8) and has a slope of -2.

<h3>What is a linear equation?</h3>

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.

It is given that:

The line that passes through the point (4, 8) and has a slope of -2.

On comparing the standard equation:

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Here m is the slope of the line and c is the y-intercept.

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Learn more about the linear equation here:

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