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eimsori [14]
2 years ago
7

What is the solution to the equation 2(×+2. 4)=6. 4

Mathematics
2 answers:
Trava [24]2 years ago
7 0

Answer:

0.8

Step-by-step explanation:

2x + 4.8 = 6.4

2x = 6.4 - 4.8

2x = 1.6

x = 0.8

Leno4ka [110]2 years ago
7 0

Answer:

  • x = <u>0</u><u>.</u><u>8</u>

Step-by-step explanation:

In the question we're given with an equation that is<u> </u><u>2</u><u> </u><u>(</u><u> </u><u>x </u><u>+</u><u> </u><u>2</u><u>.</u><u>4</u><u> </u><u>)</u><u> </u><u>=</u><u> </u><u>6</u><u>.</u><u>4</u><u> </u>. And we're asked to find the solution of the equation that means we have to find the value of <u>x </u>.

<u>Solution</u><u> </u><u>:</u><u> </u><u>-</u>

\longmapsto \: 2(x + 2.4) = 6.4

<u>Step </u><u>1</u><u> </u><u>:</u> Removing parenthesis by multiplying 2 with x and 2.4 :

\longmapsto \: 2x + 4.8 = 6.4

<u>Step </u><u>2</u><u> </u><u>:</u> Subtracting 4.8 on both sides :

\longmapsto \: 2x +  \cancel{4.8} -  \cancel{4.8 }= 6.4 - 4.8

On further calculations :

\longmapsto \:2x = 1.6

<u>Step </u><u>3 </u><u>:</u> Dividing by 2 on both sides :

\longmapsto \: \frac{ \cancel{2}x}{ \cancel{2 }}  =  \frac{1.6}{2}

Calculating further :

Here , we have removed decimal from 1.6 . So it'll become 16 and 10 it'll divided by 10

\longmapsto \:x =   \cancel { \frac{16}{2}}  \times  \frac{1}{10}

<u>Step </u><u>4</u><u> </u><u>:</u> Cancelling 16 by 2 :

\longmapsto \:x =  \frac{8}{10}

\longmapsto \:   \green{\underline{ \green  {\boxed{x = 0.8}}}}

  • <u>Therefore</u><u> </u><u>value </u><u>of </u><u>x </u><u>is </u><u>0</u><u>.</u><u>8</u><u> </u><u>.</u>

<u>Verifying</u><u> </u><u>:</u><u> </u><u>-</u>

We are verifying our answer by substituting value of x in given equation. So ,

  • 2 ( x + 2.4 ) = 6.4

  • 2 ( 0.8 + 2.4 ) = 6.4

  • 2 ( 3.2 ) = 6.4

  • 6.4 = 6.4

  • L.H.S = R.H.S

  • Hence, Verified .

<u>Therefore</u><u>,</u><u> our</u><u> </u><u>value</u><u> for</u><u> x</u><u> is</u><u> correct</u><u> </u><u>.</u>

<h2><u>#</u><u>K</u><u>e</u><u>e</u><u>p</u><u> </u><u>Learning</u></h2>
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