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2 years ago

How to solve this linear equation step by step please

Mathematics

1 answer:

mote1985 [20]2 years ago

7 0

$\bf \cfrac{3}{c^2-4c}-\cfrac{9}{2c^2+3c}=\cfrac{2}{2c^2-5c-12}\\\\\\ 
\begin{cases}
c^2-4c=&c(c-4)\\
2c^2+3c=&c(2c+3)\\\\
2c^2-5c-12=
&(2c+3)(c-4)
\end{cases}\\\\
-------------------------------\\\\
\textit{therefore, our LCD for the \underline{left-side} will have to be }\underline{c(c-4)(2c+3)}\\\\
-------------------------------\\\\$

$\bf \cfrac{3}{c(c-4)}-\cfrac{9}{c(2c+3)}=\cfrac{2}{(2c+3)(c-4)}
\\\\\\
\cfrac{3(2c+3)-9(c-4)}{c(c-4)(2c+3)}=\cfrac{2}{(2c+3)(c-4)}
\\\\\\
\cfrac{6c+9-(9c-36)}{c(c-4)(2c+3)}=\cfrac{2}{(2c+3)(c-4)}
\\\\\\
\cfrac{6c+9-9c+36}{1}=\cfrac{2[c(c-4)(2c+3)]}{(2c+3)(c-4)}
\\\\\\
-3c+45=2c\implies 45=5c\implies \cfrac{45}{5}=c\implies 9=c$

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<h3><u>Solution:</u></h3>

Given that, One of the same-side exterior angles formed by two lines and a transversal is equal to 1/6 of the right angle and is 11 times smaller than the other angle.

We have to prove that the lines are parallel.

If they are parallel, sum of the described angles should be equal to 180 as they are same side exterior angles.

Now, the 1st angle will be 1/6 of right angle is given as:

$\begin{array}{l}{\rightarrow 1^{\text {st }} \text { angle }=\frac{1}{6} \times 90} \\\\ {\rightarrow 1^{\text {st }} \text { angle }=15 \text { degrees }}\end{array}$

And now, 15 degrees is 11 times smaller than the other

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$\text {Other angle }=11 \times 15=165 \text { degrees }$

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