Problem 47 - Introductory and Intermediate Algebra for College Students 4th

Question

Solve each equation. Using the addition property of equality. Be sure to check your proposed solutions. $$7-5 x+8+2 x+4 x-3=2+3 \cdot 5$$

Step-by-Step Solution

Answer

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

Step 1: Simplify the Left Hand Side

Combine the like terms. In the equation, \(7-5x+8+2x+4x-3\), combine -5x, 2x, and 4x to get \(1x\). Similarly, combine 7, 8, and -3 to get \(12\). Our simplified left hand side is \(1x+12\).

Step 2: Simplify the Right Hand Side

In the equation, \(2+3 \cdot 5\), use the order of operations (BIDMAS/BODMAS) to do multiplication before addition. Multiply 3 by 5 to get 15 and then add 2. The simplified right hand side is \(17\).

Step 3: Solve for x

Having simplified both sides of the equation to \(1x+12=17\), we subtract 12 from both sides to isolate \(x\). This gives us \(x = 17 - 12\). Therefore, \(x = 5\).

Step 4: Check the Solution

Replace \(x\) in the original equation with 5 and assess whether both sides of the equation are equal. The left side of the equation becomes \(7-5*5+8+2*5+4*5-3\) which simplifies to 17. The right side of the equation is \(2+3*5\) which also simplifies to 17. Since both sides equal 17, our solution is correct.

Key Concepts

Addition Property of Equality

Understanding the addition property of equality is crucial when solving algebraic equations. This property states that you can add or subtract the same amount to both sides of an equation without changing its equality. For instance, if you have an equation like \( a = b \), and you add \( c \) to both sides, it becomes \( a + c = b + c \). In the exercise provided, during step 3, we subtracted 12 from both sides to isolate \( x \). Doing so applied the addition property of equality, showing us that \( x + 12 - 12 = 17 - 12 \) simplifies to \( x = 5 \). This step is essential as it helps maintain the balance of the equation while moving us closer to finding the value of \( x \).

Remember, it’s not just about adding; subtracting a number on both sides is also applying the addition property of equality because subtraction is the addition of a negative number. Thus, when solving equations, always consider this property to keep your equations balanced.

Order of Operations

Solving an algebraic equation correctly often depends on the order of operations, sometimes remembered by the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction). This guideline informs us of the sequence in which to solve different parts of an equation.

For example, in the exercise's second step, we looked at the right-hand side of the equation \( 2 + 3 \cdot 5 \). Here, it's vital to perform multiplication before addition, which is in alignment with the order of operations. Therefore, we first calculated \( 3 \cdot 5 = 15 \) and then added 2 to get 17, giving us the simplified right-hand side. Failure to follow the correct order could lead to an incorrect solution. Students should prioritize operations within parentheses, then calculate exponents, followed by multiplication and division from left to right, and finally addition and subtraction from left to right.

Combining Like Terms

One of the first steps in solving an algebraic equation is often to combine like terms. Like terms are terms whose variables (and their exponents) are the same. They can be combined by adding or subtracting their coefficients—the numerical parts.

In the provided exercise, step 1 required combining like terms on the left-hand side of the equation. Terms with \( x \) as the variable, specifically \( -5x, 2x, \) and \( 4x \), were combined to form \( 1x \) or simply \( x \). Similarly, constant terms without variables, such as 7, 8, and -3, were combined to give us 12. This process of combining like terms simplifies the equation and makes it more manageable to solve, as seen by the transformation from a complicated expression to a simple \( x + 12 \). Always look for and combine like terms to simplify your equations as much as possible before proceeding with solving for the variable.