Q. negative plus a positive equals

Answer

Let \(x>0\) and \(y>0\). Consider \( -x + y \).

– If \(y > x\): \( -x + y = y – x > 0\) (positive).
– If \(y = x\): \( -x + y = 0\).
– If \(y < x\): \( -x + y = -(x - y) < 0\) (negative).

Detailed Explanation

Problem: “negative plus a positive” — step-by-step explanation

We will explain, in full detail, how to evaluate a sum in which one addend is negative and the other is positive. Write the negative number as \(-a\) with \(a>0\), and the positive number as \(b\) with \(b>0\).

  1. Rewrite the sum in a subtraction form.

    The expression \(-a + b\) can be rewritten by moving terms so that the positive quantity comes first:

    \[ -a + b = b – a. \]

    Explanation: adding a negative number is the same as subtracting its absolute value. So \(-a + b\) is the same operation as subtracting \(a\) from \(b\).

  2. Compare the absolute values.

    To determine the sign of the result, compare \(a\) and \(b\):

    • If \(b > a\), the positive number has the larger absolute value, so the result is positive.
    • If \(b < a\), the negative number (in absolute value \(a\)) is larger, so the result is negative.
    • If \(b = a\), the numbers cancel and the result is zero.
  3. Compute the numerical result by subtracting the smaller absolute value from the larger, and attach the sign of the larger.

    Using the comparison above, compute the difference and determine the sign:

    • Case \(b > a\): \[ -a + b = b – a \quad\text{(positive result)}. \]
    • Case \(b < a\): \[ -a + b = -(a - b) \quad\text{(negative result)}. \]
    • Case \(b = a\): \[ -a + b = 0. \]
  4. Worked examples (each shown step-by-step).

    1. Example 1: Evaluate \(-3 + 5\).

      Step 1: Rewrite as subtraction: \[ -3 + 5 = 5 – 3. \]

      Step 2: Compare absolute values: \(5 > 3\), so result is positive.

      Step 3: Subtract: \[ 5 – 3 = 2. \]

      Answer: \[ -3 + 5 = 2. \]

    2. Example 2: Evaluate \(-7 + 4\).

      Step 1: Rewrite: \[ -7 + 4 = 4 – 7. \]

      Step 2: Compare absolute values: \(7 > 4\), so result is negative.

      Step 3: Subtract the smaller from the larger and attach the negative sign: \[ 7 – 4 = 3 \Rightarrow – (7 – 4) = -3. \]

      Answer: \[ -7 + 4 = -3. \]

    3. Example 3: Evaluate \(-6 + 6\).

      Step 1: Rewrite: \[ -6 + 6 = 6 – 6. \]

      Step 2: Compare absolute values: \(6 = 6\), so they cancel.

      Step 3: Subtract: \[ 6 – 6 = 0. \]

      Answer: \[ -6 + 6 = 0. \]

  5. Summary (rule to apply every time).

    To add a negative and a positive:

    1. Rewrite as a subtraction: \(-a + b = b – a\).
    2. Compare absolute values \(a\) and \(b\).
    3. Subtract the smaller absolute value from the larger.
    4. The sign of the result is the sign of the number with the larger absolute value; if equal, the result is zero.
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FAQs

What is the basic rule for adding a negative and a positive number?

If signs differ, subtract the smaller absolute value from the larger; the result takes the sign of the number with the larger absolute value. Example: \( -5 + 3 = -(5-3) = -2 \); \( -3 + 5 = +(5-3) = 2 \).

How do I do this on a number line?

Start at the first number, move right for positives and left for negatives. Example: for \( -4 + 6 \), start at \( -4 \) and move 6 units right to land at \( 2 \).

What happens if the absolute values are equal?

They cancel and the sum is zero: \( -a + a = 0 \) for any \( a>0 \). Example: \( -7 + 7 = 0 \).

How is subtraction related to adding a negative?

Subtraction is adding the opposite: \( a - b = a + (-b) \). So \( -2 + 5 \) can be seen as \( 5 + (-2) \) or \( 5-2 = 3 \).

How do I add several mixed-sign numbers?

Sum all positives and sum all negatives separately, then add those two results (or subtract their absolute values and take the sign of the larger sum). Example: \( -3 + 7 - 4 + 2 = (7+2) + (-3-4) = 9 + (-7) = 2 \).

How do I handle decimals or fractions?

Same rule: compare absolute values, subtract, and give sign of larger. Example decimals: \( -2.5 + 1.2 = -(2.5-1.2) = -1.3 \). For fractions: \( -\tfrac{3}{4} + \tfrac{1}{2} = -\tfrac{1}{4} \).

How can I use absolute value to think about it?

Write as \( \text{sign}(\text{larger})\cdot(|\text{larger}|-|\text{smaller}|) \). Example: \( -8+3 = -(|8|-|3|) = -(8-3) = -5 \).

How does this apply to real-life (money/temperature)?

Interpret negatives as owing or below zero and positives as receiving or above zero. Example: owing \$6 and getting \$4 leaves owing \$2: \( -6 + 4 = -2 \).
Find sums of negatives and positives.
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