Problem 28
Question
Solve the equation and check your solution. $$-q+1=0$$
Step-by-Step Solution
Verified Answer
The solution to the equation is \(q=1\).
1Step 1: Rearrange the Equation
The equation can be rearranged to get \(q\) alone on one side. Add \(q\) to both sides of the equation to obtain: \(1=q\).
2Step 2: Check Solution
By substituting the calculated \(q\) into the original equation we have: \(-1+1=0\), which holds true.
Key Concepts
Solving Linear EquationsChecking SolutionBasic Algebraic Manipulation
Solving Linear Equations
Linear equations are mathematical statements that relate a constant and a variable using basic arithmetic operations. Solving a linear equation involves finding the value of the variable that makes the equation true.
In the given exercise, the equation is \(-q + 1 = 0\). To find the solution, you must isolate the variable (in this case, \(q\)) on one side of the equation. Here's how you can do this:
In the given exercise, the equation is \(-q + 1 = 0\). To find the solution, you must isolate the variable (in this case, \(q\)) on one side of the equation. Here's how you can do this:
- Add \(q\) to both sides. This gives: \(1=q\). This step removes the negative sign from the variable and transfers it to the constant side.
Checking Solution
Once we find a potential solution to a linear equation, it's important to verify that it indeed satisfies the original equation. This builds confidence in our answer and ensures we've solved the problem correctly.
For the equation \(-q + 1 = 0\), we found that \(q = 1\). Checking the solution involves substituting this value back into the original equation:
For the equation \(-q + 1 = 0\), we found that \(q = 1\). Checking the solution involves substituting this value back into the original equation:
- Replace \(q\) with 1: \(-1 + 1 = 0\). This simplifies to \(0 = 0\), which actually holds true.
Basic Algebraic Manipulation
Basic algebraic manipulation is a fundamental skill in mathematics that involves the rearrangement and simplification of equations. These operations make solving for unknown variables possible.
In our example, handling \(-q + 1 = 0\), required simple yet essential manipulation:
In our example, handling \(-q + 1 = 0\), required simple yet essential manipulation:
- Effectively moving the term containing \(q\) to one side of the equation by adding \(q\) to both sides.
- Ultimately, this rearrangement and simplification yielded the expression \(q = 1\).
Other exercises in this chapter
Problem 28
Solve the percent equation. What number is \(325 \%\) of 450 ?
View solution Problem 28
Solve the equation and check your solution. (Some of the equations have no solution.) $$3[(5 x+1)-4]=4(2 x-3)$$
View solution Problem 29
Solve and graph the inequality. $$\frac{2}{3} x \leq 12$$
View solution Problem 29
Solve for the specified variable. Solve for \(h: A=\frac{1}{2} b h\)
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