Linear Equations One Variable Practice Problems & Solutions

Master solving linear equations with one variable through step-by-step practice problems. Learn isolation techniques, verification methods, and common solution strategies.

πŸ“šMaster Linear Equations with One Variable Through Targeted Practice
  • Solve equations using addition and subtraction to isolate variables
  • Apply multiplication and division operations to clear coefficients and fractions
  • Verify solutions by substituting back into original equations
  • Handle equations with negative coefficients and zero on one side
  • Combine multiple operations to solve complex linear equations step-by-step
  • Identify and correct common mistakes in equation solving processes

Understanding Linear Equations (One Variable)

Complete explanation with examples

The solution for an equation is a numerical value that, when inserted in the place of the unknown (or the variable), will render both members of the equation equal, that is, we will obtain a "true statement". In first degree equations with one unknown, there can only be one solution.

Example:

Xβˆ’1=5X - 1 = 5

Solution of an equation x-1=5

This is an equation with one unknown or variable indicated by the letter XX. The equation is composed of two members separated by the use of the equal sign = = . The left member is everything to the left of the sign = = , and the right member is everything to the right of the sign.

Our goal is to isolate the variable (or clear the variable) X X in order that it only remains in one of the members of the equation. In doing so we should be able to determine its value. In this article we will learn how to use the four mathematical operations(addition, subtraction, multiplication and division) to isolate the variable X X .

Detailed explanation

Practice Linear Equations (One Variable)

Test your knowledge with 42 quizzes

Solve for X:

\( -5+x=-3 \)

Examples with solutions for Linear Equations (One Variable)

Step-by-step solutions included
Exercise #1

Solve the equation

5xβˆ’15=30 5x-15=30

Step-by-Step Solution

We start by moving the sections:

5X-15 = 30
5X = 30+15

5X = 45

 

Now we divide by 5

X = 9

Answer:

x=9 x=9

Video Solution
Exercise #2

4x:30=2 4x:30=2

Step-by-Step Solution

To solve the given equation 4x:30=2 4x:30 = 2 , we will follow these steps:

  • Step 1: Recognize that 4x:304x:30 implies 4x30=2\dfrac{4x}{30} = 2.

  • Step 2: Eliminate the fraction by multiplying both sides of the equation by 30.

  • Step 3: Simplify the equation to solve for xx.

Now, let's work through each step:

Step 1: The equation is written as 4x30=2\dfrac{4x}{30} = 2.

Step 2: Multiply both sides of the equation by 30 to eliminate the fraction:
30Γ—4x30=2Γ—30 30 \times \dfrac{4x}{30} = 2 \times 30

This simplifies to:
4x=60 4x = 60

Step 3: Solve for xx by dividing both sides by 4:
x=604=15 x = \dfrac{60}{4} = 15

Therefore, the solution to the problem is x=15 x = 15 .

Checking choices, the correct answer is:

x=15 x = 15

Answer:

x=15 x=15

Video Solution
Exercise #3

Solve the equation

7x+5.5=19.5 7x+5.5=19.5

Step-by-Step Solution

To solve the given equation 7x+5.5=19.5 7x + 5.5 = 19.5 , we'll follow these steps:

  • Step 1: Eliminate the constant term from the left side by subtracting 5.5 from both sides of the equation.
  • Step 2: Simplify the equation after subtraction to isolate the term with x x .
  • Step 3: Use division to solve for x x .

Now, let's work through each step:

Step 1: Subtract 5.5 from both sides.

We have:
7x+5.5βˆ’5.5=19.5βˆ’5.5 7x + 5.5 - 5.5 = 19.5 - 5.5

This simplifies to:
7x=14 7x = 14

Step 2: Divide both sides by 7 to solve for x x .

So, we divide by 7:
7x7=147 \frac{7x}{7} = \frac{14}{7}

This simplifies to:
x=2 x = 2

Therefore, the solution to the problem is x=2 x = 2 .

Answer:

x=2 x=2

Video Solution
Exercise #4

Solve the equation

20:4x=5 20:4x=5

Step-by-Step Solution

To solve the exercise, we first rewrite the entire division as a fraction:

204x=5 \frac{20}{4x}=5

Actually, we didn't have to do this step, but it's more convenient for the rest of the process.

To get rid of the fraction, we multiply both sides of the equation by the denominator, 4X.

20=5*4X

20=20X

Now we can reduce both sides of the equation by 20 and we will arrive at the result of:

X=1

Answer:

x=1 x=1

Video Solution
Exercise #5

Solve the equation

8xβ‹…10=80 8x\cdot10=80

Step-by-Step Solution

To solve this linear equation, we need to isolate the variable x x . Here are the steps to follow:

  • Step 1: Simplify the equation by dividing both sides by 10. This gives us:

8xβ‹…1010=8010 \frac{8x \cdot 10}{10} = \frac{80}{10}

This simplifies to:

8x=8 8x = 8

  • Step 2: Now, isolate x x by dividing both sides by 8:

8x8=88 \frac{8x}{8} = \frac{8}{8}

This simplifies to:

x=1 x = 1

Therefore, the solution to the equation 8xβ‹…10=80 8x \cdot 10 = 80 is

x=1 x = 1 .

Answer:

x=1 x=1

Video Solution

Frequently Asked Questions

How do you solve a linear equation with one variable step by step?

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To solve a linear equation with one variable: 1) Identify the variable to isolate, 2) Use inverse operations to move terms to opposite sides, 3) Simplify by combining like terms, 4) Verify your solution by substituting back into the original equation. Always perform the same operation on both sides to maintain equality.

What is the difference between solving X - 1 = 5 and 2X = 8?

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For X - 1 = 5, you add 1 to both sides to get X = 6. For 2X = 8, you divide both sides by 2 to get X = 4. The first uses addition/subtraction to isolate the variable, while the second uses multiplication/division to eliminate the coefficient.

How do you check if your solution to a linear equation is correct?

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Substitute your solution back into the original equation in place of the variable. If both sides of the equation are equal after performing the calculations, your solution is correct. For example, if X = 6 solves X - 1 = 5, then 6 - 1 = 5 gives us 5 = 5, which is true.

What operations can you use to isolate variables in linear equations?

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You can use four basic operations to isolate variables: β€’ Addition (to eliminate subtraction) β€’ Subtraction (to eliminate addition) β€’ Multiplication (to eliminate division/fractions) β€’ Division (to eliminate multiplication/coefficients). Always apply the same operation to both sides of the equation.

How do you solve equations with fractions like (1/3)x = 5?

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To solve equations with fractions, multiply both sides by the denominator to eliminate the fraction. For (1/3)x = 5, multiply both sides by 3: 3 Γ— (1/3)x = 3 Γ— 5, which simplifies to x = 15. Always verify by substituting back: (1/3) Γ— 15 = 5 βœ“

What does it mean when a linear equation has one solution?

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A linear equation with one variable has exactly one solution because there's only one value that makes both sides equal. This is a fundamental property of first-degree equations. When you find this value and substitute it back, you get a true statement like 5 = 5.

How do you solve multi-step linear equations like 2x + 3 = 5?

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For multi-step equations, work backwards using order of operations: 1) First, subtract 3 from both sides: 2x = 2, 2) Then divide both sides by 2: x = 1. Always handle addition/subtraction before multiplication/division when isolating the variable.

What are common mistakes when solving linear equations with one variable?

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Common mistakes include: β€’ Forgetting to apply operations to both sides β€’ Making arithmetic errors during calculations β€’ Not verifying the solution in the original equation β€’ Confusing addition/subtraction with multiplication/division operations β€’ Incorrectly handling negative signs and coefficients

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