Before we begin adding and subtracting expressions, recall the rule for adding and subtracting integers.
Rules of Addition
To add a positive number, the operation is addition,
\(2 +(+4) = 2+4\)
To add a negative number, the operation is subtraction,
\(5+(-1)=5-1\)
Rules of Subtraction
To subtract a positive number, the operation is subtraction,
\(9-(+7) = 9-7\)
To subtract a negative number, the operation is addition,
\(8-(-2) = 8+2\)
Example 1
Simplify \((4x+5y) + (6x-2y)\).
Solution
We will need to collect like terms but first we need to remove the brackets.
The plus sign between the brackets means that we need to add everything inside of the second set of brackets, so we have
\((4x+5y) + (6x-2y) = 4x + 5y +(6x) + (-2y)\)
Don't forget how to add and subtract integers. When adding a negative value, the operation is subtraction. Now we have an expression with like terms.
\(\begin{align*} 4x + 5y +(6x) + (-2y) &= 4x +5y +6x -2y \end{align*}\)
We can identify like terms and add or subtract their coefficients to get,
\(\begin{align*} \class{hl1}{4x} \class{hl2}{+5y} \class{hl1}{+6x} \class{hl2}{-2y} &= \class{hl1}{4x+6x} \class{hl2}{+5y -2y} \\ &= \class{hl1}{10x} \class{hl2}{-3y} \end{align*}\)
The final expression, \(\class{hl1}{10x} \class{hl2}{-3y}\), is the simplest expression we can get to. We cannot combine the two terms further as they are not like terms.
In summary,
\(\begin{align*} (4x+5y) + (6x-2y) &= 4x + 5y +(6x) + (-2y) \\ &= \class{hl1}{4x} \class{hl2}{+5y} \class{hl1}{+6x} \class{hl2}{-2y} \\ &= \class{hl1}{4x+6x} \class{hl2}{+5y -2y} \\ &= \class{hl1}{10x} \class{hl2}{-3y} \end{align*}\)
Example 2
Simplify \((4g+5h) - (2g -3h)\).
Solution
In this example, the first set of brackets can be removed. The second set of brackets, however, has a minus sign in front. We need to subtract everything in the second set of brackets, so we have
\((4g+5h) - (2g -3h) = 4g+5h - (2g) -(-3h)\)
When subtracting a positive value, the operation is subtraction. When subtracting a negative value, the operation is addition.
\(4g+5h - (2g) -(-3h) = 4g+5h -2g + 3h\)
We can identify like terms and add or subtract their coefficients to get,
\(\begin{align*} \class{hl1}{4g} \class{hl2}{+5h} \class{hl1}{-2g} \class{hl2}{+ 3h} &= \class{hl1}{4g-2g} \class{hl2}{+5h + 3h} \\ &= \class{hl1}{2g} \class{hl2}{+8h} \end{align*}\)
In summary,
\(\begin{align*} (4g+5h) - (2g -3h) &= 4g+5h - (2g) -(-3h) \\ &= \class{hl1}{4g} \class{hl2}{+5h} \class{hl1}{-2g} \class{hl2}{+ 3h} \\ &= \class{hl1}{4g-2g} \class{hl2}{+5h + 3h} \\ &= \class{hl1}{2g} \class{hl2}{+8h} \end{align*}\)