Linear Equations Practice Problems - One Variable Solutions

Master solving linear equations with one variable through step-by-step practice problems. Learn isolation techniques, distributive property, and fraction solutions.

πŸ“šMaster Linear Equations with One Variable - Practice Now
  • Solve equations using the distributive property with parentheses
  • Master isolating variables by transposing terms correctly
  • Handle linear equations with fractions and mixed numbers
  • Apply like terms grouping and simplification techniques
  • Determine domain restrictions when variables appear in denominators
  • Check solutions by substitution and verify equivalence

Understanding Linear Equations (One Variable)

Complete explanation with examples

What is an equation with one unknown?

Equations are algebraic expressions containing numbers and unknowns. It is important to differentiate these two groups: the numbers are fixed values while the unknowns, as their name indicates, represent unknown values (at least at the beginning), and in most cases we are asked to find out what this value is.
For example:

A1 - First-degree equations with one unknown

What do we do with the equations?

When we are given an exercise that contains an equation with an unknown, our goal is to solve the equation, that is, to find a solution to the equation. What does it mean to find the solution to an equation? The idea is to find the value of the unknown with the goal of making both sides of the equation equal.

When we have equations that have the same solution, they will be called equivalent equations.

When first degree equations include fractions, and the unknown is in the denominator, it is important to keep in mind the domain of the function


Detailed explanation

Practice Linear Equations (One Variable)

Test your knowledge with 41 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

What are the basic steps to solve linear equations with one variable?

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Follow these steps: 1) Distribute and expand parentheses, 2) Group like terms (variables on one side, numbers on the other), 3) Combine like terms, 4) Isolate the variable by dividing both sides by its coefficient. Remember that when moving terms across the equals sign, their signs change.

How do I solve linear equations with parentheses and distributive property?

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First multiply each term inside the parentheses by the number outside. For example, in 12(2x-3) = -4(3-4x), distribute to get 24x - 36 = -12 + 16x. Then group like terms and solve normally.

What does it mean to transpose terms in linear equations?

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Transposing means moving terms from one side of the equation to the other. When you move a term, its sign changes - if it was positive, it becomes negative and vice versa. This helps group variables on one side and constants on the other.

How do I handle mixed numbers in linear equations?

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Convert mixed numbers to improper fractions first. For example, 3Β½ becomes 7/2 and 4β…“ becomes 13/3. Then solve the equation normally, dividing by the coefficient fraction using multiplication by its reciprocal.

When do linear equations have domain restrictions?

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Domain restrictions occur when variables appear in denominators. The equation is undefined when the denominator equals zero. Find these values by setting the denominator equal to zero and solving - these values must be excluded from the domain.

What are equivalent equations in algebra?

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Equivalent equations are different equations that have the same solution. They're created by performing the same operation on both sides of an equation, such as adding, subtracting, multiplying, or dividing by the same non-zero number.

How do I check if my solution to a linear equation is correct?

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Substitute your answer back into the original equation. Replace the variable with your solution value and calculate both sides. If both sides equal the same number, your solution is correct.

What's the difference between solving ax + b = c and solving equations with variables on both sides?

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Simple equations like ax + b = c require basic isolation steps. Equations with variables on both sides (like 3x + 4 = x + 10) need an extra step to collect all variable terms on one side before isolating the variable.

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