2\xd7 x – 1 = x + 2 , urgent

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2× x – 1 = x + 2
urgent​

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The value of your 1st Degree equation will be 3. This equation is pretty simple, and we can solve it in two ways, which are pretty simple too! The first is like this…

  • We move the variable to the left side, and add the opposites. Then we move the constant to the right side, and add the opposites. So we put similar terms in evidence, and we add.

Now we have the second mode, which is like this, we multiply 2 by x, (unknown value), which in this case is 3, and we subtract by 1. So we add, x to 2, which makes 5. So the first value must also return 5, to be “5 = 5”. Lets get down to calculations! Calculating with the first mode:

2x – 1 = x + 2

2x – 1 – X = 2

2x – X = 2 + 1

x = 2 + 1

  \: \boxed{x = 3}

So the unknown value is going to be 3, well lets do the other calculation to see if the result is the same! Second way to calculate:

2•x-1 = x + 2

2•3 – 1 = x + 2

6 – 1 = x + 2

5 = x + 2

5 = 3 + 2

5 = 5

  \: \boxed{x=3}

Here, the unknown value was also 3, that is, both ways are correct, and the value of your equation will be equal to 3.

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