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algol [13]
3 years ago
13

Help Me With This Question Please! I'm Answering This Question For A Hard Time! :(

Mathematics
1 answer:
Rom4ik [11]3 years ago
3 0

Answer:

Step-by-step explanation:

The main goal in any equation is to get the unknown variable all by itself.  In this case that means moving the fraction of 7/8 over to the other side of the equals sign.  On the left, the fraction is being subtracted; that means that to move it to the other side, we add it since adding is the opposite of subtracting.

x-\frac{7}{8}+\frac{7}{8}=4+\frac{7}{8}

On the left side, if you have 7/8 and then take that 7/8 away, it's gone completely.  What we have now is

x=4+\frac{7}{8}

We need a common denominator so we can add 4 (which is 4/1) to 7/8.  8 is the common denominator.

x=\frac{4}{1}(\frac{8}{8})+\frac{7}{8} and

x=\frac{32}{8}+\frac{7}{8} giving us

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keeping in mind that parallel lines have exactly the same slope, let's check for the slope of the equation above

x-2y=8\implies -2y=-x+8\implies y=\cfrac{-x+8}{-2} \\\\\\ y=\cfrac{-x}{-2}+\cfrac{8}{-2}\implies y=\stackrel{\stackrel{m}{\downarrow }}{\cfrac{1}{2}}x-4 \qquad \impliedby \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}

so we're really looking for the equation of a line whose slope is 1/2 and passes through the point (2 , -6)

(\stackrel{x_1}{2}~,~\stackrel{y_1}{-6})\qquad\qquad \stackrel{slope}{m}\implies \cfrac{1}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-6)}=\stackrel{m}{\cfrac{1}{2}}(x-\stackrel{x_1}{2}) \\\\\\ y+6=\cfrac{1}{2}x-1\implies y=\cfrac{1}{2}x-7

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You can use point  (-1, -6) and (0, -8)

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Answer:

Step-by-step explanation:

Hello friend !!!

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