Problem 29
Question
contain linear equations with constants in denominators Solve each equation. $$ \frac{x+1}{3}-5-\frac{x+2}{7} $$
Step-by-Step Solution
Verified Answer
The solution to the equation is \(x = 26\).
1Step 1: Make the Denominators the Same
To simplify the equation, the denominators must be the same. Find the least common multiple of 3 and 7, which is 21. Multiply each term by 21 to clear the denominators: \[21*\frac{x+1}{3}-21*5-21*\frac{x+2}{7}\] which simplifies to: \[7(x + 1) - 5*21 - 3(x+2)\]
2Step 2: Simplify the equation
Now, simplify the equation and try to get all terms involving x on one side of the equation and constants on the other side. Then, apply the distributive property: \[7x + 7 - 105 - 3x - 6 = 0\] leading to: \[4x - 104 = 0\]
3Step 3: Isolate the variable x
Lastly, the equation can be solved for x by adding 104 to both sides: \[4x - 104 + 104 = 104\] This simplifies down to: \[4x = 104\] and then dividing both sides by 4 to isolate x: \[x = \frac{104}{4} = 26\]
Key Concepts
Least Common MultipleDistributive PropertyIsolate the Variable
Least Common Multiple
When dealing with equations that have different denominators, finding the least common multiple (LCM) is essential. The LCM helps to simplify the problem by making all denominators the same. In this process, you identify the smallest number that is a multiple of both denominators. For example, in the equation \( \frac{x+1}{3}-\frac{x+2}{7} \), we need to find the LCM of 3 and 7.
The LCM of 3 and 7 is 21, because 21 is the smallest number that both 3 and 7 can divide into without leaving a remainder. By multiplying every term in the equation by 21, we eliminate the fractions. This makes it easier to manipulate the equation, allowing us to focus on simplifying and solving for the variable.
The LCM of 3 and 7 is 21, because 21 is the smallest number that both 3 and 7 can divide into without leaving a remainder. By multiplying every term in the equation by 21, we eliminate the fractions. This makes it easier to manipulate the equation, allowing us to focus on simplifying and solving for the variable.
Distributive Property
The distributive property is a fundamental concept in algebra that allows us to simplify expressions and combine like terms. It states that for any numbers \(a\), \(b\), and \(c\), the equation \(a(b + c) \) is equivalent to \( ab + ac \).
In the given equation, after finding the LCM and multiplying through to eliminate fractions, we reach:
In the given equation, after finding the LCM and multiplying through to eliminate fractions, we reach:
- \( 7(x+1) \)
- \( -3(x+2) \)
- \( 7(x+1) \) becomes \( 7x + 7 \)
- \( -3(x+2) \) becomes \( -3x - 6 \)
Isolate the Variable
To solve linear equations, a key goal is to isolate the variable, allowing us to find its value. This process involves several steps that simplify the equation, ensuring that the variable stands alone on one side.
Once the equation is simplified using the least common multiple and distributive property, you work towards keeping the variable \(x\) on one side. In our example, we transform the equation into:
Once the equation is simplified using the least common multiple and distributive property, you work towards keeping the variable \(x\) on one side. In our example, we transform the equation into:
- \( 4x - 104 = 0 \)
- \( 4x = 104 \)
- \( x = \frac{104}{4} = 26 \)
Other exercises in this chapter
Problem 28
Graph each equation .Let $x=-3,-2,-1,0,1,2, and 3. $$ y=x^{3}-1 $$
View solution Problem 29
Use interval notation to express solution sets and graph each solution set on a number line. Solve linear inequality. \(3 x-7 \geq 13\)
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The bus fare in a city is \(\$ 1.25 .\) People who use the bus have the option of purchasing a monthly discount pass for \(\$ 15.00 .\) With the discount pass,
View solution Problem 29
Solve cach equation in Exercises \(15-34\) by the square root property. $$(3 x+2)^{2}=9$$
View solution