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Match each equation to the correct inverse operation needed to solve the equation.

1. [tex]3 = x - 1[/tex] [tex]\(\square\)[/tex] [tex]\(+\)[/tex]

2. [tex]\frac{w}{9} = 72[/tex] [tex]\(\square\)[/tex] [tex]\(\times\)[/tex]

3. [tex]d + \frac{1}{3} = \frac{2}{7}[/tex] [tex]\(\square\)[/tex] [tex]\(-\)[/tex]

4. [tex]5.1a = 15.3[/tex] [tex]\(\square\)[/tex] [tex]\(\div\)[/tex]


Sagot :

Sure, let's go through each equation step-by-step to determine the correct inverse operation needed to isolate the variable.

1. For the equation [tex]\(3 = x - 1\)[/tex]:
- In order to solve for [tex]\(x\)[/tex], we need to isolate [tex]\(x\)[/tex].
- Since [tex]\(x - 1\)[/tex] means [tex]\(x\)[/tex] decreased by 1, we need to add 1 to both sides of the equation to undo this subtraction.
- This gives us [tex]\(x = 3 + 1\)[/tex].

So, the inverse operation needed for this equation is addition: [tex]\(+\)[/tex]

2. For the equation [tex]\(\frac{w}{9} = 72\)[/tex]:
- In order to isolate [tex]\(w\)[/tex], we need to get rid of the division by 9.
- The way to undo division is to multiply by the same number.
- Thus, we multiply both sides by 9: [tex]\(w = 72 \times 9\)[/tex].

So, the inverse operation needed for this equation is multiplication: [tex]\(\)[/tex]

3. For the equation [tex]\(d + \frac{1}{3} = \frac{2}{7}\)[/tex]:
- To isolate [tex]\(d\)[/tex], we need to remove [tex]\(\frac{1}{3}\)[/tex] from the left side of the equation.
- This means we need to subtract [tex]\(\frac{1}{3}\)[/tex] from both sides.
- Thus, [tex]\(d = \frac{2}{7} - \frac{1}{3}\)[/tex].

So, the inverse operation needed for this equation is subtraction: [tex]\(-\)[/tex]

4. For the equation [tex]\(5.1a = 15.3\)[/tex]:
- To solve for [tex]\(a\)[/tex], we need to isolate it by removing the coefficient [tex]\(5.1\)[/tex].
- This means we need to divide both sides by [tex]\(5.1\)[/tex].
- Thus, [tex]\(a = \frac{15.3}{5.1}\)[/tex].

So, the inverse operation needed for this equation is division: [tex]\(/ \)[/tex]

Therefore, if we match each equation to the correct inverse operation, we have:

[tex]\[ 3 = x - 1\ \ \quad \ \ \quad + \][/tex]
[tex]\[ \frac{w}{9} = 72\ \ \quad
\][/tex]
[tex]\[ d + \frac{1}{3} = \frac{2}{7}\ \ \quad - \][/tex]
[tex]\[ 5.1a = 15.3\ \ \quad / \][/tex]
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