Problem 98
Question
Solve: \(4(x-5)=22+2(6 x+3)\) (Section \(2.3,\) Example 3 )
Step-by-Step Solution
Verified Answer
After analyzing and carefully working through each step of the algebraic equation, the solution is \(x=-6\).
1Step 1: Expand Terms
Expand the terms in the bracket on both side of the equation: \(4 * x - 4 * 5 = 22 + 2 * 6x + 2 * 3\). Simplify each side of the equations to get: \(4x - 20 = 22 + 12x + 6\)
2Step 2: Simplify the Expression
Simplify the right side of the equation: \(4x - 20 = 28 + 12x\). After simplification, rearrange to get like terms on each side: \(4x - 12x = 28 + 20\). This, when simplified further gives: \(-8x = 48\).
3Step 3: Solve for x
To find the value of x, divide each side of the equation by -8: \(-8x/-8=48/-8\). This simplifies to: \(x=-6\)
Key Concepts
Algebraic ExpressionsEquation SimplificationIsolating Variables
Algebraic Expressions
Algebraic expressions are the cornerstone of algebra. They consist of variables, numbers, and arithmetic operations such as addition, subtraction, multiplication, and division. For example, in the expression \(4(x-5)\), the variable \(x\) represents an unknown value, and it is being multiplied by 4 and subtracted by 20 (which is \(4\times5\)). Understanding how to manipulate these expressions is essential for solving equations.
When you come across an equation such as \(4(x-5)=22+2(6x+3)\), your first task is to expand the algebraic expressions. This means applying the distributive property, which involves multiplying the number outside the parenthesis with each term inside it. Remember, each operation within the expression affects how it can be simplified or expanded. By mastering algebraic expressions, you lay the groundwork for solving more complex problems in algebra.
When you come across an equation such as \(4(x-5)=22+2(6x+3)\), your first task is to expand the algebraic expressions. This means applying the distributive property, which involves multiplying the number outside the parenthesis with each term inside it. Remember, each operation within the expression affects how it can be simplified or expanded. By mastering algebraic expressions, you lay the groundwork for solving more complex problems in algebra.
Equation Simplification
The simplification of an equation is a process that makes an equation easier to solve. Taking the initial equation \(4(x-5)=22+2(6x+3)\), expanding the terms is just the first step to simplification. After expansion, you combine like terms, which are terms that have the same variable raised to the same power. For instance, \(4x\) and \(-20\) on the left side, as well as \(12x\) and \(28\) on the right side, are combined by addition or subtraction.
Why Simplify?
- To reduce the equation to a form that is easier to understand and work with.
- It helps to identify like terms for easier rearrangement.
- Simplification allows for the straightforward application of further algebraic techniques, such as factoring or isolating variables.
Isolating Variables
Isolating the variable is the final step in solving linear equations. It involves manipulating the equation to get the unknown variable, in this case \(x\), by itself on one side of the equals sign. The goal is to have the variable equal to a numerical value or a simpler expression.
The equation we have after simplification is \(-8x=48\). To isolate \(x\), you need to perform the same operation on both sides of the equation to keep it balanced. Since \(x\) is being multiplied by -8, you do the opposite - you divide by -8. Dividing both sides by -8 gives you \(x=-6\). This step is the cornerstone of algebra, and it embodies the principles of balance and equivalence. It is important to practice this technique with various types of equations to become proficient in algebra.
The equation we have after simplification is \(-8x=48\). To isolate \(x\), you need to perform the same operation on both sides of the equation to keep it balanced. Since \(x\) is being multiplied by -8, you do the opposite - you divide by -8. Dividing both sides by -8 gives you \(x=-6\). This step is the cornerstone of algebra, and it embodies the principles of balance and equivalence. It is important to practice this technique with various types of equations to become proficient in algebra.
Other exercises in this chapter
Problem 96
Factor completely: \(12 x^{2}+14 x-6\) (Section 6.5, Example 2)
View solution Problem 97
$$\text { Divide: } \frac{x^{2}-x-6}{3 x-3} \div \frac{x^{2}-4}{x-1}$$
View solution Problem 99
Will help you prepare for the material covered in the next section. $$\text { Factor: } x^{2}+8 x+16$$
View solution Problem 100
Will help you prepare for the material covered in the next section. $$\text { Factor: } x^{2}-14 x+49$$
View solution