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hichkok12 [17]
3 years ago
13

What is the x value for the solution -x+2y=6 6y =x+18

Mathematics
1 answer:
Alika [10]3 years ago
5 0

For the given set of equations, x = 0 and y = 3.

Step-by-step explanation:

Step 1:

From the question,

-x + 2y = 6, take this as equation 1,

6y =x+18, x - 6y = -18, take this as equation 2.

Step 2:

When we add equation 1 and equation 2, the x variable is canceled out and y can be calculated.

-4y = -12, y = \frac{-12}{-4}, y = 3.

Step 3:

Substituting this value of y in any of the previous equations we will get x's value.

Here this value of y i.e y = 3 is substituted in equation 1.

-x + 6 = 6, -x = 6 -6, -x=0, x=0.

So we have x = 0 and y = 3.

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Read 2 more answers
Simplify: [5×(25)^n+1 - 25 × (5)^2n]/[5×(5)^2n+3 - (25)^n+1​]
blagie [28]

\green{\large\underline{\sf{Solution-}}}

<u>Given expression is </u>

\rm :\longmapsto\:\dfrac{5 \times  {25}^{n + 1}  - 25 \times  {5}^{2n} }{5 \times  {5}^{2n + 3}  -  {25}^{n + 1} }

can be rewritten as

\rm \:  =  \: \dfrac{5 \times  { {(5}^{2} )}^{n + 1}  -  {5}^{2}  \times  {5}^{2n} }{5 \times  {5}^{2n + 3}  -  {( {5}^{2} )}^{n + 1} }

We know,

\purple{\rm :\longmapsto\:\boxed{\tt{  {( {x}^{m} )}^{n}  \: = \:   {x}^{mn}}}} \\

And

\purple{\rm :\longmapsto\:\boxed{\tt{ \:  \:   {x}^{m} \times  {x}^{n} =  {x}^{m + n} \: }}} \\

So, using this identity, we

\rm \:  =  \: \dfrac{5 \times  {5}^{2n + 2}  - {5}^{2n + 2} }{{5}^{2n + 3 + 1}  -  {5}^{2n + 2} }

\rm \:  =  \: \dfrac{{5}^{2n + 2 + 1}  - {5}^{2n + 2} }{{5}^{2n + 4}  -  {5}^{2n + 2} }

can be further rewritten as

\rm \:  =  \: \dfrac{{5}^{2n + 2 + 1}  - {5}^{2n + 2} }{{5}^{2n + 2 + 2}  -  {5}^{2n + 2} }

\rm \:  =  \: \dfrac{ {5}^{2n + 2} (5 - 1)}{ {5}^{2n + 2} ( {5}^{2}  - 1)}

\rm \:  =  \: \dfrac{4}{25 - 1}

\rm \:  =  \: \dfrac{4}{24}

\rm \:  =  \: \dfrac{1}{6}

<u>Hence, </u>

\rm :\longmapsto\:\boxed{\tt{ \dfrac{5 \times  {25}^{n + 1}  - 25 \times  {5}^{2n} }{5 \times  {5}^{2n + 3}  -  {25}^{n + 1} }  =  \frac{1}{6} }}

3 0
2 years ago
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