What value of x is in the solution set of â€""5x â€"" 15 > 10 20x? â€""2 â€""1 0 1.

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Ответ:

Hfruit

14.05.2021

Mathematics

Given:

The expression is:

$\left(\dfrac{11}{19}n-8\right)+\left(\dfrac{1}{19}n+10\right)=\left(\_\_\_+\dfrac{1}{19}n\right)+\left(-8+\_\_\_\right)$

$=\_\_\_n+\_\_\_$

To find:

The missing values.

Solution:

We have,

$\left(\dfrac{11}{19}n-8\right)+\left(\dfrac{1}{19}n+10\right)$

It can be written as

$\left(\dfrac{11}{19}n-8\right)+\left(\dfrac{1}{19}n+10\right)=\left(\dfrac{11}{19}n+\dfrac{1}{19}n\right)+\left(-8+10\right)$

So, the values for the first two boxes are $\dfrac{11}{19}n$ and 10 respectively.

On simplification, we get

$\left(\dfrac{11}{19}n-8\right)+\left(\dfrac{1}{19}n+10\right)=\left(\dfrac{11+1}{19}n\right)+\left(2\right)$

$\left(\dfrac{11}{19}n-8\right)+\left(\dfrac{1}{19}n+10\right)=\dfrac{12}{19}n+2$

So, the values for third and fourth boxes are $\dfrac{12}{19}$ and 2 respectively.

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