Question

Find the sum of $$-59$$, $$-41$$, 73, $$-92$$, 81, $$-\left(-41\right)$$, $$-3$$.

Answer 1

Hint: Here, we need to find the sum of the given numbers. We will rearrange the numbers in the sun, and rewrite the expression. Then, we will convert the subtraction to addition using parentheses, and simplify to get the sum of the given numbers.

Complete step-by-step answer:
First, we will rewrite the negative integers in the given numbers.
The number $$-x$$ can be written as the product of the negative integer $$-1$$, and the positive integer $$x$$.
Therefore, rewriting the numbers $$-59$$, $$-41$$, $$-92$$, $$-\left(-41\right)$$, $$-3$$, we get
$$-59=\left(-1\times 59\right)$$
$$-41=\left(-1\times 41\right)$$
$$-92=\left(-1\times 92\right)$$
$$-3=\left(-1\times 3\right)$$
$$-\left(-41\right)=\left[-1\times \left(-41\right)\right]=-1\times -1\times 41$$
We know that $${\left(-1\right)}^n$$ is equal to 1 if $$n$$ is an even number, and is equal to $$-1$$ if $$n$$ is an odd number.
Therefore, we get
$$\Rightarrow -\left(-41\right)={\left(-1\right)}^2\times 41=1\times 41=41$$
Now, we will find the sum of the given numbers.
Writing the sum of the numbers as an expression, we get
$$\left(-59\right)+\left(-41\right)+73+\left(-92\right)+81+\left[-\left(-41\right)\right]+\left(-3\right)$$
Substituting $$-\left(-41\right)=41$$ in the expression, we get
$$\Rightarrow \left(-59\right)+\left(-41\right)+73+\left(-92\right)+81+\left[-\left(-41\right)\right]+\left(-3\right)=\left(-59\right)+\left(-41\right)+73+\left(-92\right)+81+41+\left(-3\right)$$
Rearranging the terms of the expression, we get
$$\Rightarrow \left(-59\right)+\left(-41\right)+73+\left(-92\right)+81+\left[-\left(-41\right)\right]+\left(-3\right)=73+81+41+\left(-59\right)+\left(-41\right)+\left(-92\right)+\left(-3\right)$$
$$\Rightarrow \left(-59\right)+\left(-41\right)+73+\left(-92\right)+81+\left[-\left(-41\right)\right]+\left(-3\right)=195+\left(-59\right)+\left(-41\right)+\left(-92\right)+\left(-3\right)$$
Substituting $$-59=\left(-1\times 59\right)$$, $$-41=\left(-1\times 41\right)$$, $$-92=\left(-1\times 92\right)$$, and $$-3=\left(-1\times 3\right)$$ in the expression, we get
$$\Rightarrow \left(-59\right)+\left(-41\right)+73+\left(-92\right)+81+\left[-\left(-41\right)\right]+\left(-3\right)=195+\left(-1\times 59\right)+\left(-1\times 41\right)+\left(-1\times 92\right)+\left(-1\times 3\right)$$
Converting subtraction to addition by factoring out $$-1$$, we get
$$\begin{array}{l}\Rightarrow \left(-59\right)+\left(-41\right)+73+\left(-92\right)+81+\left[-\left(-41\right)\right]+\left(-3\right)=195+\left(-1\right)\left(59+41+92+3\right)\\ \Rightarrow \left(-59\right)+\left(-41\right)+73+\left(-92\right)+81+\left[-\left(-41\right)\right]+\left(-3\right)=195-\left(59+41+92+3\right)\end{array}$$
Adding the terms of the expression, we get
$$\Rightarrow \left(-59\right)+\left(-41\right)+73+\left(-92\right)+81+\left[-\left(-41\right)\right]+\left(-3\right)=195-195$$
Subtracting 195 from 195, we get
$$\therefore (-59)+(-41)+73+(-92)+81+[-(-41)]+(-3)=0$$
Therefore, we get the sum of the numbers $$-59$$, $$-41$$, 73, $$-92$$, 81, $$-\left(-41\right)$$, and $$-3$$ as 0.

Note: We used the term ‘negative integer’ in the solution. An integer is a rational number that is not a fraction. For example: 1, $$-1$$, 3, $$-7$$, are integers. Integers can be positive or negative. Negative integers are the numbers $$-1$$, $$-5$$, $$-92$$, $$-41$$, etc.
We can also solve this question by making different pairs of numbers. We can make the pairs such that it is easy to add and subtract the terms, and thus, simplify the sum. The answer will be the same.