Classify each of the following equations. 2.5 x + 3.1 = 1.2 ( x − 2 ) 2.5x+3.1=1.2\left(x-2\right)

Welcome back. For this problem, we want to classify the equation 2.5 x plus 3.1 equals 1.2 times x minus two. To classify this equation, we're going to have to solve it to see how many solutions it has, so let's start there. This equation has decimals, so we'll have to multiply by a power of 10 to clear those decimals first. We'll check each term to see how many decimal places it has, take the highest number, and then multiply by that power of 10. Starting with 2.5 x, 2.5 has one decimal place. Next, we have 3.1, which has one decimal place, and then we have 1.2 times x minus two. The only decimal is 1.2, and it has one decimal place. We can see that the highest number is one, so we need to multiply both sides of the equation by 10 to the first power which is 10. So we get 10 times 2.5 x plus 3.1 equals 10 times 1.2, times x minus two. Let's distribute that 10 on both sides of the equation, starting with 10 times 2.5 x. When we multiply by 10, we move the decimal point one place to the right, So 10 times 2.5 is 25, and we get 25 x. Next, we'll do 10 times 3.1. Again, we move the decimal point one place to the right, so we get plus 31. And then on the right side of the equation, we have 10 times 1.2 times x minus two. For 10 times 1.2, again, we move the decimal point one place to the right, and we get 12 times x minus two. Now that we have no more decimals, we can move on to simplifying the equation. We can keep the left side as is. We have 25 x plus 31 equals and then on the right side, let's distribute that 12. So we have 12 times x, which is 12 x, and then 12 times negative two, which is minus 24. Now we wanna get all of our variable terms on one side of the equation and all of our constant terms on the other side. We currently have both variable and constant terms on both sides, so let's try to get variables to the left and constants on the right, and let's start with constants. We wanna get rid of this plus 31 on the left side of the equation, so we can subtract 31 on both sides of the equation. It will cancel the plus 31 on the left, leaving us with 25 x. And on the right side of the equation, we have 12 x minus 24 minus 31, which is minus 55. Next, we want to get variables to the left, so we want to get rid of this 12 x on the right, which we can do by subtracting 12 x on both sides of the equation. This is gonna cancel the 12 x on the right side of the equation, leaving us with negative 55. And on the left side of the equation, we have 25 x minus 12 x, which is 13 x. Next, we wanna isolate our variable, so we have to get rid of this coefficient of 13. So we can divide both sides of the equation by 13. When we do, it cancels this 13 on the left leaving us with just x, and on the right we get negative 55 over 13. We can't simplify this fraction at all, so our solution is just x equals negative 55 thirteenths. We can see that we have one solution, this negative 55 over 13. Because we only have one solution, we have a conditional equation because this equation is only true when x equals that one value of negative 55 over 13. Great work.

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Multiple Choice

Classify each of the following equations.
$2.5 x + 3.1 = 1.2 (x - 2)$

Conditional

Identity

Contradiction

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Start by writing down the given equation: \$2.5x + 3.1 = 1.2\left(x - 2\right)$.

Distribute the \$1.2$ on the right side to both terms inside the parentheses: $1.2 \times x$ and $1.2 \times (-2)$, resulting in $1.2x - 2.4$.

Rewrite the equation with the distributed terms: \$2.5x + 3.1 = 1.2x - 2.4$.

Bring all terms involving $x$ to one side and constants to the other side by subtracting \$1.2x$ from both sides and subtracting $3.1$ from both sides: $2.5x - 1.2x = -2.4 - 3.1$.

Simplify both sides to get a linear equation in the form $ax = b$. Then analyze the result: if $a \neq 0$, the equation is conditional (true for some $x$); if $a = 0$ and $b = 0$, it is an identity (true for all $x$); if $a = 0$ and $b \neq 0$, it is a contradiction (no solution).

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Classify each of the following equations.2.5x+3.1=1.2(x−2)2.5x+3.1=1.2\left(x-2\right)

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Classify each of the following equations.
$2.5 x + 3.1 = 1.2 (x - 2)$