Problem 40 - Introductory and Intermediate Algebra for College Students 4th

Question

Solve each equation using both the addition and multiplication properties of equality. Check proposed solutions. $$-x-5=5$$

Step-by-Step Solution

Answer

The solution to the equation \(-x-5=5\) is \(x=-10\).

Step 1: Isolate the constant term on the left side of the equation

Use the addition property of equality to get rid of the -5 on the left side of the equation by adding 5 to both sides. This turns the equation into \(-x = 5+5 = 10\).

Step 2: Isolate the x-term

Use the multiplication property of equality to solve for \(x\). Since the coefficient for \(x\) is -1, divide both sides of the equation by -1 to get \(x = -10/1 = -10\).

Step 3: Check the proposed solution

Substitute the proposed solution, \(x = -10\), back into the original equation. The original equation \(-(-10)-5 =5 \) simplifies to \(10-5=5\), which is a true statement, confirming that the proposed solution is correct.

Key Concepts

Addition Property of Equality

Understanding the addition property of equality is fundamental in solving algebraic equations. It states that you can add the same number to both sides of an equation without changing the equality. This property is used when you need to move numbers from one side of the equation to the other to isolate the variable.
For example, in the equation \( -x-5=5 \), we can use the addition property to eliminate the -5 by adding 5 to both sides, leading to \( -x=10 \). The equation balances because whatever we do to one side, we also do to the other. This simple yet powerful tool is key to streamlining equations and paves the way for further simplification.

Multiplication Property of Equality

Similarly, the multiplication property of equality plays a crucial role when we deal with equations. It indicates that we can multiply or divide both sides of an equation by the same non-zero number, and the equality will still hold. This property is particularly useful for dealing with coefficients attached to variables.
Take the next step in our equation, where we now have \( -x=10 \). To isolate the variable, we divide both sides by -1 (the coefficient of x), effectively using the multiplication property (remember, dividing is the same as multiplying by the reciprocal). Therefore, \( x = -10 \) becomes the isolated solution. This process turns coefficients into a neutral '1', setting the variable free, so to speak.

Isolating Variables

Isolating the variable is the most critical step in solving equations. It involves using algebraic properties, like the addition and multiplication properties of equality, to get the variable on one side of the equation and the numerical values on the other. This step helps us to determine the value of the unknown variable.
In our case, once we've applied the addition property to move the constant term and the multiplication property to deal with the coefficient, we have isolated the variable \( x \). This methodical approach of one step at a time - addition/subtraction first, followed by multiplication/division - is the most effective and clearest path to finding the solution.

Checking Solutions in Algebra

Once we believe we’ve solved the equation, verifying our proposed solution ensures accuracy and boosts understanding. We do this by substituting the solution back into the original equation to check if it satisfies the equality.
For instance, if we insert \( x = -10 \) into the original equation \( -x-5=5 \), we should confirm that both sides equal the same number. After substitution, we get \( 10-5=5 \), a true statement. This step is important to validate our work and is an excellent habit for catching mistakes early on. It's the algebraic equivalent of double-checking your answer before turning in a test.