A fraction when added to 7/3 gives 4. What is the fraction?
The problem asks us to find a fraction that, when added to \( \frac{7}{3} \), results in the number 4. We can represent the unknown fraction with a variable, let's say \( x \). The problem can then be translated into a simple algebraic equation:
\( x + \frac{7}{3} = 4 \)
Our goal is to isolate the variable \( x \) on one side of the equation to find its value. To do this, we need to eliminate the term \( \frac{7}{3} \) from the left side. We can achieve this by subtracting \( \frac{7}{3} \) from both sides of the equation. Remember, whatever operation you perform on one side of an equation, you must perform the same operation on the other side to maintain equality.
\( x = 4 - \frac{7}{3} \)
Now, we need to perform the subtraction of a whole number and a fraction. To subtract a fraction from a whole number, we first need to express the whole number as a fraction with the same denominator as the fraction being subtracted. The denominator of \( \frac{7}{3} \) is 3. So, we need to express 4 as a fraction with a denominator of 3.
We know that any whole number can be written as a fraction with a denominator of 1 (e.g., \( 4 = \frac{4}{1} \)). To change the denominator from 1 to 3, we multiply both the numerator and the denominator by 3:
\( 4 = \frac{4 \times 3}{1 \times 3} = \frac{12}{3} \)
Now substitute this back into our equation:
\( x = \frac{12}{3} - \frac{7}{3} \)
Subtracting fractions with the same denominator is straightforward: subtract the numerators and keep the denominator the same.
\( x = \frac{12 - 7}{3} \)
\( x = \frac{5}{3} \)
So, the unknown fraction is \( \frac{5}{3} \). Now let's look at the options provided and see which one matches \( \frac{5}{3} \).
We will convert the options to improper fractions or compare them with \( \frac{5}{3} \).
To convert a mixed number \( a\frac{b}{c} \) to an improper fraction, the formula is \( \frac{(a \times c) + b}{c} \).
For \( 1\frac{2}{3} \), \( a=1, b=2, c=3 \).
Improper fraction = \( \frac{(1 \times 3) + 2}{3} = \frac{3 + 2}{3} = \frac{5}{3} \).
This matches our calculated value for \( x \).
Therefore, the fraction is \( 1\frac{2}{3} \).
| Operation | Equation |
|---|---|
| Starting equation | \( x + \frac{7}{3} = 4 \) |
| Subtract \( \frac{7}{3} \) from both sides | \( x = 4 - \frac{7}{3} \) |
| Express 4 as a fraction with denominator 3 | \( 4 = \frac{12}{3} \) |
| Substitute and subtract | \( x = \frac{12}{3} - \frac{7}{3} = \frac{5}{3} \) |
| Convert to mixed number | \( \frac{5}{3} = 1\frac{2}{3} \) |
| Concept | Description | Example |
|---|---|---|
| Fraction | Represents a part of a whole. Written as \( \frac{\text{numerator}}{\text{denominator}} \). | \( \frac{3}{4} \) (3 parts out of 4) |
| Improper Fraction | A fraction where the numerator is greater than or equal to the denominator. | \( \frac{5}{3} \), \( \frac{7}{7} \) |
| Mixed Number | A combination of a whole number and a proper fraction. | \( 1\frac{2}{3} \) (1 whole and 2/3) |
| Converting Mixed to Improper Fraction | Multiply whole number by denominator, add numerator, put over original denominator: \( a\frac{b}{c} = \frac{(a \times c) + b}{c} \) | \( 2\frac{1}{4} = \frac{(2 \times 4) + 1}{4} = \frac{9}{4} \) |
| Solving Linear Equations | Use inverse operations to isolate the variable. | If \( x + a = b \), then \( x = b - a \). |
When solving equations involving fractions, a common strategy is to find a common denominator for all terms. In our case, the denominators were 3 and 1 (for the whole number 4). The least common multiple (LCM) of 3 and 1 is 3. We could have also solved the equation by multiplying every term by the LCM, 3, to clear the denominators:
Original equation: \( x + \frac{7}{3} = 4 \)
Multiply every term by 3:
\( 3 \times (x + \frac{7}{3}) = 3 \times 4 \)
Distribute the 3 on the left side:
\( (3 \times x) + (3 \times \frac{7}{3}) = 12 \)
\( 3x + 7 = 12 \)
Now, this is a two-step linear equation without fractions. Subtract 7 from both sides:
\( 3x = 12 - 7 \)
\( 3x = 5 \)
Divide both sides by 3:
\( x = \frac{5}{3} \)
This method also gives the same result, \( \frac{5}{3} \), which confirms our previous calculation. This demonstrates that there can be multiple valid approaches to solving the same mathematical problem.
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