Problem 27
Question
SOLVING EQUATIONS Use division to solve the equation. $$ 10 x=110 $$
Step-by-Step Solution
Verified Answer
The solution to the equation is \(x = 11\).
1Step 1: Identify the equation
The equation given in the problem is \(10x = 110\). This equation tells us that 10 times some unknown number \(x\) is equal to 110.
2Step 2: Isolate the variable
The goal here is to isolate \(x\) on one side of the equation. This can be done by dividing both sides of the equation by 10. When you divide both sides of the equation by 10, you get \(x = 110/10\).
3Step 3: Solve for the variable
To solve for \(x\), divide 110 by 10. The solution is \(x = 11\).
Key Concepts
Algebraic EquationsIsolate the VariableDivision in Algebra
Algebraic Equations
In mathematics, algebraic equations are like puzzles where you need to find the missing piece, usually represented by a variable such as x or y. An algebraic equation consists of two expressions set equal to each other and involves numbers and variables combined using operations such as addition, subtraction, multiplication, and division.
For instance, in the equation 10x = 110, we have an expression 10x that equals 110. The '10' is known as a coefficient, which multiplies the variable x, and '110' is a constant. The primary goal is to determine the value of x that makes the equation true. Solving algebraic equations requires systematic methods to simplify them and find the values of the variables they contain.
For instance, in the equation 10x = 110, we have an expression 10x that equals 110. The '10' is known as a coefficient, which multiplies the variable x, and '110' is a constant. The primary goal is to determine the value of x that makes the equation true. Solving algebraic equations requires systematic methods to simplify them and find the values of the variables they contain.
- Recognize the structure of the equation.
- Identify variables and constants.
- Apply arithmetic operations to simplify.
Isolate the Variable
Isolating the variable is a fundamental technique used to solve algebraic equations. This means rearranging the equation so the variable you’re solving for is by itself on one side of the equals sign, showing what it must equal to satisfy the equation.
Let's consider our example 10x = 110. To isolate x, we need to undo the multiplication by 10. The operation opposite to multiplication is division, so we divide both sides of the equation by 10, respecting the balance principle. This balance principle is crucial: whatever you do to one side of the equation, you must do to the other to maintain equality.
Let's consider our example 10x = 110. To isolate x, we need to undo the multiplication by 10. The operation opposite to multiplication is division, so we divide both sides of the equation by 10, respecting the balance principle. This balance principle is crucial: whatever you do to one side of the equation, you must do to the other to maintain equality.
Step-by-step isolation:
- Identify the operation applied to the variable.
- Perform the inverse operation on both sides.
- Simplify to get the variable alone.
Division in Algebra
Division in algebra works similarly to basic arithmetic division but involves variables. In an equation, when a term with a variable has a numerical coefficient, division is used to cancel out this coefficient and solve for the variable.
In the example 10x = 110, the variable x is multiplied by 10. To isolate x, we divide the entire equation by the coefficient 10. This is shown as x = 110/10, simplifying to x = 11 after the division. It’s important to divide both sides of the equation by the same number to keep the equation balanced.
In the example 10x = 110, the variable x is multiplied by 10. To isolate x, we divide the entire equation by the coefficient 10. This is shown as x = 110/10, simplifying to x = 11 after the division. It’s important to divide both sides of the equation by the same number to keep the equation balanced.
- Use division to 'undo' multiplication.
- Keep the equation balanced by dividing both sides equally.
- Reduce coefficients to 1, effectively isolating the variable.
Other exercises in this chapter
Problem 27
Solve the equation. Round the result to the nearest hundredth. Check the rounded solution. $$ -35 m+75=48 $$
View solution Problem 27
Solve the equation. $$ 5 x-4 x=-6 x+3 $$
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Solve the equation. $$ x-8=-13 $$
View solution Problem 28
Check to see if the given value of the variable is or is not a solution of the inequality. $$ 4+k \geq 32 ; k=30 $$
View solution