Finding Integers in Algebraic Equations
By Kathleen Cantor, 18 Jul 2020
Integers include any whole number found on a number line. They also include any negative number that falls on a number line. Whole numbers are strictly positive. Integers do not include anything that is a fraction or decimal.
- -4 is an integer.
- 0 is an integer.
- 5.3 is not an integer.
It is worth noting that some people learn differently when it comes to whole numbers. Some do not believe that 0 is a whole number and that whole numbers automatically include negatives. If this criterion applies to you, simply adjust the definition of an integer provided here.
It’s that easy! It stays that easy when finding them in algebraic equations, too. This is because integers are simply elements added to mathematical equations. Because they are so identifiable, there’s no question when you come across one -- except maybe when it’s hidden in a variable.
Parts of an Algebraic Equation
An algebraic equation incorporates more advanced elements than your basic arithmetic equation. Also referred to as algebraic expressions, these make use of variables, integers and other numbers, and at least one algebraic operation.
2x + 3y(15-3) = 30
The variables in the equation are x and y. Variables stand in place of numbers for which we must solve. We do not yet know if the variable is an integer or not. Remember, integers do not include fractions and decimals.
The coefficient -- the number by which we are multiplying our variable -- for term 2x is 2. For 3y it is 3.
Our algebraic operations include addition (+), subtraction (-), and multiplication (3y(15-3)...).
However, we can identify integers in our equation right away. These are 2, 3, 15, and 30.
Revealing the Variable
There’s only one way to determine if a variable is an integer or not, and that’s to solve the equation. Not every problem you’ll solve will provide you with variable values. In fact, you’ll most likely be creating your own algebraic equations where you need to solve for a variable in real life. This is especially true if calculating profit. You can use any variable to represent your value.
In this case, a=$12,000 (total revenue), b= $10,000.45 (total expense), and c=profit.
a - b = c
Insert your values into the profit equation above.
$12,000 - $10,000.45 = c
Solve for c.
$1,999.55 = c
In this case, c is not representing an integer. If your values are not whole numbers and include parts -- if there are cents with your dollars -- unless the parts negate each other, your profit will not be an integer.
Solving for a Variable’s Value
Because we have to solve to find a variable that could be hiding an integer, let’s learn a simple way to solve for the variable. This equation will use like terms with one variable.
14x + 13(7*6) - 6x = 610
1. Combine Like Terms
First, let’s combine the like terms in this equation: 14x and -6x. Only when a variable is the same can you combine them.
8x + 13(7*6) = 610
2. Follow Order of Operations
Following Order of Operations, we solve what is in parentheses first, followed by multiplication.
8x + 13(42) = 610
8x + 546 = 610
3. Isolate the Variable
Next, we need to start isolating the variable or removing the remaining numbers and the coefficient from the left side of the equation. The rule goes that anything done to one side of the equation must also be done to the other. In this case, we will subtract 546 from both sides.
8x = 64
Finally, to fully isolate the variable, divide 8x by 8 and 64 by 8.
4. Prove Your Answer
According to this answer, our variable is an integer. Let’s prove our equation by placing the variable in the equation and solving for the result, 610.
14(8) + 13(7*6) - 6(8) = 610
112 + 13(42) - 48
112 + 546 - 48
658 - 48 = 610
Our integer works!
Finding integers in algebraic equations is all about knowing what is considered an integer or not. It also relies on your ability to solve for variables. Again, never assume variables are integers. There are plenty of cases when they aren’t.
See the 1 Comment below.