Determining Whether a Whole Number is a Solution to an Equation
Learning Outcomes
- Determine whether a whole number is a solution to an equation
Determine Whether a Number is a Solution of an Equation
Solving an equation is like discovering the answer to a puzzle. An algebraic equation states that two algebraic expressions are equal. To solve an equation is to determine the values of the variable that make the equation a true statement. Any number that makes the equation true is called a solution of the equation. It is the answer to the puzzle!Solution of an Equation
A solution to an equation is a value of a variable that makes a true statement when substituted into the equation. The process of finding the solution to an equation is called solving the equation.[latex]\begin{array}{}\\ \hfill x+2=7\hfill \\ \hfill 5+2\stackrel{?}{=}7\hfill \\ \\ \hfill 7=7\quad\checkmark \hfill \end{array}[/latex]
Since [latex]5+2=7[/latex] is a true statement, we know that [latex]5[/latex] is indeed a solution to the equation. The symbol [latex]\stackrel{?}{=}[/latex] asks whether the left side of the equation is equal to the right side. Once we know, we can change to an equal sign [latex]\text{(=)}[/latex] or not-equal sign [latex]\text{(\not = ).}[/latex]
Determine whether a number is a solution to an equation.
- Substitute the number for the variable in the equation.
- Simplify the expressions on both sides of the equation.
- Determine whether the resulting equation is true.
- If it is true, the number is a solution.
- If it is not true, the number is not a solution.
example
Determine whether [latex]x=5[/latex] is a solution of [latex]6x - 17=16[/latex]. Solution| [latex]6x--17=16[/latex] | |
| Substitute [latex]\color{red}{5}[/latex] for x. | [latex]6\cdot\color{red}{5}--17=16[/latex] |
| Multiply. | [latex]30--17=16[/latex] |
| Subtract. | [latex]13=16[/latex] |
try it
[ohm_question]146455[/ohm_question]example
Determine whether [latex]y=2[/latex] is a solution of [latex]6y - 4=5y - 2[/latex].Answer: Solution Here, the variable appears on both sides of the equation. We must substitute [latex]2[/latex] for each [latex]y[/latex].
| [latex]6y--4=5y--2[/latex] | |
| Substitute [latex]\color{red}{2}[/latex] for y. | [latex]6(\color{red}{2})--4=5(\color{red}{2})--2[/latex] |
| Multiply. | [latex]12--4=10--2[/latex] |
| Subtract. | [latex]8=8[/latex] |
try it
[ohm_question]146456[/ohm_question]Licenses & Attributions
CC licensed content, Original
- Question ID 146456, 146455. Authored by: Lumen Learning. License: CC BY: Attribution. License terms: IMathAS Community License.
CC licensed content, Shared previously
- Introduction to Algebraic Equations (L5.1). Authored by: James Sousa (Mathispower4u.com). License: Public Domain: No Known Copyright.
CC licensed content, Specific attribution
- Prealgebra. Provided by: OpenStax License: CC BY: Attribution. License terms: Download for free at http://cnx.org/contents/[email protected].
