Question

what should be added to (-7/20) to get (-2/5)​

Answer 1

Answer:

-1 SᕼOᑌᒪᗪ ᗷE ᗩᗪᗪEᗪ

Step-by-step explanation:

ᑭᒪS ᗰᗩᖇK ᗰE ᗩS ᗷᖇᗩIᑎᒪIEST

ᗩᑎᗪ ᖴOᒪᒪOᗯ ᖴOᖇ ᗰOᖇE

Answer 2

★ To obtain the rational number (−2/5), (−7/20) must be added with (−1/20).

Step-by-step explanation

Analysis:-

⠀⠀⠀In the question, it has been stated that a rational number should be added to (−7/20). When it's done, we will obtain the rational number (−2/5). We have been asked to find what the rational number is.

Solution:-

⠀⠀⠀Let's consider the rational number as "x". So, as per the analysis, the equation that forms in this case is:

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{ \boxed{ \tt \pmb{ \dfrac{ -7 }{20} + x = \dfrac{ - 2}{5} }}}$

Here the value of x is the required answer. So, let's start solving the equation.

1) Write the equation nearly in a paper.

$\twoheadrightarrow{ \sf{ \dfrac{ - 7}{20} + x = \dfrac{ - 2}{5} }}$

2) Now transpose the like terms. Here, (−7/20) and (−2/5) are like terms. So, on transposing (−7/20) to the R.H.S., we get,

$\twoheadrightarrow{ \sf{x = \dfrac{ - 2}{5} + \dfrac{7}{20} }}$

Note: While performing transposition in solving an equation, the mathematical operators will be changed like:

• (+) changes to (−) and vice-versa.
• (×) changes to (÷) and vice-versa.

Hence, (−2/5) has been written as (2/5) in the R.H.S.

3) Simplify the R.H.S. in the equation. Since the denominators aren't same, find the L.C.M. of the denominators.

$\begin{gathered} \: \: \: \: \: \: \begin{gathered}\begin{gathered} \begin{array}{c|c} \underline{\sf{5}}& \underline{\sf{ \: \: \: 5 \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: 20 \: \: \: \: }} \\ \underline{\sf{2}}&\underline{\sf{\: \: \: 1 \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \: 4 \: \: \: \:}} \\ \underline{\sf{2}}&\underline{\sf{\: \: \: 1 \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \:2 \: \: \: \:}} \\ \underline{}&\sf{\: \: \: 1 \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: 1 \: \: \: \:} \\ \end{array}\end{gathered}\end{gathered} \\ \\ \sf{L.C.M. = 5 \times 2 \times 2} \\\sf{ \: \: = {20}}\end{gathered}$

We've obtained the L.C.M. of the denominators, i.e., 20. Now, let's make their denominators same.

$\twoheadrightarrow{ \sf{x = \bigg( \dfrac{ - 2}{5} \times \dfrac{4}{4} \bigg) + \bigg( \dfrac{7}{20} \times \dfrac{1}{1} \bigg) }}$

$\twoheadrightarrow{ \sf{x = \bigg( \dfrac{ - 8}{20} + \dfrac{7}{20} \bigg) }}$

$\twoheadrightarrow{ \sf{ x = \bigg( \dfrac{ - 8 + 7}{20} \bigg)}}$

$\twoheadrightarrow \underline{ \boxed{ \frak{ \pmb{ \red{ x = \dfrac{ - 1}{20} }}}}}$

This is the required answer.

Verification:-

⠀⠀⠀Plug in the value of x in the equation formed.

$\twoheadrightarrow{ \sf{ \bigg(\dfrac{ - 7}{20} + x \bigg)= \dfrac{ - 2}{5} }}$

$\twoheadrightarrow{ \sf{ \bigg( \dfrac{ - 7}{20} + \dfrac{ - 1}{20} \bigg) = \dfrac{ - 2}{5} }}$

$\twoheadrightarrow{ \sf{ \bigg( \dfrac{ - 7 + ( - 1)}{20} \bigg) = \dfrac{ - 2}{5} }}$

$\twoheadrightarrow{ \sf{ \bigg( \dfrac{ - 7 - 1}{20} \bigg) = \dfrac{ - 2}{5} }}$

$\twoheadrightarrow{ \sf{ \bigg(\dfrac{ - 8}{20} \bigg)= \dfrac{ - 2}{5} }}$

$\twoheadrightarrow{ \sf{ \dfrac{ \cancel4( - 2)}{ \cancel4(5)} = \dfrac{ - 2}{5} }}$

$\twoheadrightarrow \underline{ \boxed{ \frak{ \dfrac{ - 2}{5} = \dfrac{ - 2}{5} }}}$

The values of L.H.S. ans R.H.S are same. Hence, the answer obtained is correct.

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