Name: Gauss Method Exercises
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Chapters

What Are Linear Equations?
Steps to Solve a Linear Equation
Linear Equation Exercises

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What Are Linear Equations?

Linear equations or first-degree equations are of the type $\text{[math]}$ , with $\text{[math]}$ , or any other equation that, when operated on, transposed, and simplified, adopts this expression.

Steps to Solve a Linear Equation

In general, to solve a linear equation or first-degree equation we must follow these steps:

1 Remove parentheses

That is, if there are expressions like

$\text{[math]}$

Then we expand using the distributive property, which is $\text{[math]}$ , and the law of signs will also be important.

$\text{[math]}$

2 Remove denominators

In case there are fractional terms in the expression, we must identify the different denominators present, calculate the least common multiple (LCM) of these, and multiply the equation by the LCM. Instead of the LCM, you can also calculate the product of all denominators, although the first method is more recommended since it gives a smaller or more simplified number. For example:

$\text{[math]}$

We multiply the first fraction by $\text{[math]}$

$\text{[math]}$

Here again we might need to remove parentheses to simplify

$\text{[math]}$

3 Group the terms with x on one side and the independent terms on the other

Once we have completed step 1 and step 2, we will have the sum and subtraction of terms with x and independent terms on both sides of the equation. What follows is to gather the $\text{[math]}$ terms on one side and the independent terms on the other. For this, remember that if a $\text{[math]}$ is being added on one side of the equation, for example, I can move it to the other side with the inverse operation, that is, it would become $\text{[math]}$ on the other side.

$\text{[math]}$

4 Combine like terms

Now that we have terms with $\text{[math]}$ together, we add or subtract them depending on the case. Similarly with the independent terms, for example:

$\text{[math]}$

5 Solve for the unknown

If there is a coefficient accompanying the variable $\text{[math]}$ , since it is multiplying, we will move it to the other side with the inverse operation, that is, dividing. This is what we call solving for the variable.

$\text{[math]}$

Linear Equation Exercises

Score:

$\text{[math]}$

Solution

We solve for the unknown by dividing both sides by $\text{[math]}$ . Also, as a practical method, we can say that the $\text{[math]}$ that is multiplying on the first side moves to the second side as a divisor.

$\text{[math]}$

Solution

We group like terms by adding $\text{[math]}$ and $\text{[math]}$ to both sides, obtaining an equivalent equation.

$\text{[math]}$

In practice, we usually say that if a term is adding $\text{[math]}$ on one side, it moves to the other side subtracting $\text{[math]}$ , and if it was subtracting $\text{[math]}$ , it moves to the other side adding $\text{[math]}$ . We add:

$\text{[math]}$

Solution

We use the distributive property to operate on the parentheses, that is, multiply each algebraic term inside the parentheses by $\text{[math]}$ . Thus, on the left side we have: $\text{[math]}$

$\text{[math]}$

We group like terms; the x that is adding moves to the other side subtracting, and the $\text{[math]}$ that is subtracting moves to the other side adding. We add:

$\text{[math]}$

We solve for the unknown; the $\text{[math]}$ that is multiplying moves to the other side dividing:

$\text{[math]}$

Solution

To remove the denominators, we need to find the least common multiple of $\text{[math]}$ and $\text{[math]}$

$\text{[math]}$

We multiply both sides of the equation by the LCM, in this case $\text{[math]}$ , and obtain:

$\text{[math]}$

We multiply using the distributive property to solve the parentheses, group and add like terms:

$\text{[math]}$

We solve for the unknown:

$\text{[math]}$

Solution

We multiply $\text{[math]}$ by each term inside the parentheses (distributive property) to solve the parentheses and simplify:

$\text{[math]}$

We group and add like terms:

$\text{[math]}$

Solution

In this case it's convenient to first develop the operation $\text{[math]}$ . When solving it, we can change the bracket to a parenthesis.

$\text{[math]}$

We operate the terms inside the parentheses by −1 to remove the negative sign and the parentheses from the equation:

$\text{[math]}$

To remove the denominators, we need to find the least common multiple of $\text{[math]}$ and $\text{[math]}$ .

$\text{[math]}$ $\text{[math]}$

$\text{[math]}$

Using the distributive property to develop the parentheses, we multiply the first parenthesis by $\text{[math]}$ and the second by $\text{[math]}$ :

$\text{[math]}$

We group like terms:

$\text{[math]}$

We add:

$\text{[math]}$

We divide both sides by: $\text{[math]}$

$\text{[math]}$

Solution

Using the distributive property to develop the parentheses, we multiply the first parenthesis by $\text{[math]}$ and the second by $\text{[math]}$ .

$\text{[math]}$

We group like terms:

$\text{[math]}$

We add like terms and solve:

$\text{[math]}$

Solution

Using the distributive property to solve the parentheses, we multiply the first parenthesis by $\text{[math]}$ and the second by $\text{[math]}$ .

$\text{[math]}$

We group like terms:

$\text{[math]}$

We add like terms and solve:

$\text{[math]}$

Solution

To remove the denominators, we need to find the least common multiple of $\text{[math]}$ , $\text{[math]}$ and $\text{[math]}$

$\text{[math]}$

We divide the common denominator by each denominator and multiply the result by the corresponding numerator:

$\text{[math]}$

Using the distributive property to operate on the parentheses, we multiply the first by $\text{[math]}$ , the second by $\text{[math]}$ and the third by $\text{[math]}$ .

$\text{[math]}$

We group like terms:

$\text{[math]}$

We add like terms and solve:

$\text{[math]}$

Solution

Using the distributive property to develop the parentheses, we multiply the first parenthesis by $\text{[math]}$ , the second by $\text{[math]}$ and the third by $\text{[math]}$ .

It's necessary to remember that when we multiply a whole number by a fraction, we solve it by multiplying the whole number by the numerator of the fraction and the denominator remains the same.

$\text{[math]}$

We apply the distributive property to develop the parentheses in the numerators:

$\text{[math]}$

To remove the denominators, we need to find the least common multiple of $\text{[math]}$ , $\text{[math]}$ and $\text{[math]}$ .

$\text{[math]}$

We divide the common denominator by each denominator and multiply the result by the corresponding numerator.

$\text{[math]}$

Using the distributive property to develop the parentheses, we multiply the first parenthesis by $\text{[math]}$ and simplify carefully with attention to sign changes.

$\text{[math]}$

We group like terms:

$\text{[math]}$

We solve for the unknown:

$\text{[math]}$

Solution

For the equality between the two fractions to hold, the product of the extremes must equal the product of the means.

Or if preferred, we can also find the LCM which is $\text{[math]}$ because both binomials are irreducible. Then we divide the LCM by each denominator and multiply the result by the corresponding numerator.

$\text{[math]}$

Using the distributive property to develop the parentheses, we multiply the first parenthesis by $\text{[math]}$ and the second by $\text{[math]}$ .

$\text{[math]}$

We group like terms:

$\text{[math]}$

We solve for the unknown:

$\text{[math]}$

Solution

In this case it's convenient to first develop the operation $\text{[math]}$ . When solving it, we can change the bracket to a parenthesis.

$\text{[math]}$

We operate the terms inside the parentheses by −1 to remove the negative sign and the parentheses from the equation:

$\text{[math]}$

To remove the denominators, we need to find the least common multiple of $\text{[math]}$ and $\text{[math]}$ .

$\text{[math]}$ $\text{[math]}$

$\text{[math]}$

Using the distributive property to develop the parentheses, we multiply the first parenthesis by $\text{[math]}$ and the second by $\text{[math]}$ :

$\text{[math]}$

We group like terms:

$\text{[math]}$

We add:

$\text{[math]}$

We divide both sides by: $\text{[math]}$

$\text{[math]}$

Solution

To remove the denominators, we need to find the least common multiple of $\text{[math]}$ , $\text{[math]}$ , $\text{[math]}$ and $\text{[math]}$ .

$\text{[math]}$

We divide the common denominator by each denominator and multiply the result by the corresponding numerator:

$\text{[math]}$

Using the distributive property to develop the parentheses, we multiply the first parenthesis by $\text{[math]}$ , the second by $\text{[math]}$ , and the third by $\text{[math]}$ .

$\text{[math]}$

We group like terms:

$\text{[math]}$

We add like terms and solve:

$\text{[math]}$

Solution

For the equality between the two fractions to hold, the product of the extremes must equal the product of the means.

$\text{[math]}$

Using the distributive property to develop the parentheses, we multiply the first parenthesis by $\text{[math]}$ and the second by $\text{[math]}$ .

$\text{[math]}$

We add like terms:

$\text{[math]}$

We solve for the unknown:

$\text{[math]}$

Solution

We operate the terms inside the parentheses by $\text{[math]}$ to remove the negative sign and the parentheses from the equation; now we can substitute the bracket with a parenthesis.

$\text{[math]}$

We use the distributive property to develop the parentheses.

It's necessary to remember that when we multiply one fraction by another, we must multiply numerator by numerator and denominator by denominator.

$\text{[math]}$

To remove the denominators, we need to find the least common multiple of $\text{[math]}$ and $\text{[math]}$ .

$\text{[math]}$ $\text{[math]}$

$\text{[math]}$

We group like terms:

$\text{[math]}$

We add and solve:

$\text{[math]}$

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Agostina Babbo is an English and Italian to Spanish translator and writer, specializing in product localization, legal content for tech, and team sports—particularly handball and e-sports. With a degree in Public Translation from the University of Buenos Aires and a Master's in Translation and New Technologies from ISTRAD/Universidad de Madrid, she brings both linguistic expertise and technical insight to her work.

First-Degree Equations

What Are Linear Equations?

Steps to Solve a Linear Equation

Linear Equation Exercises

Linear Equations

Gauss Method Exercises

Cancel reply

First-Degree Equations

What Are Linear Equations?

Steps to Solve a Linear Equation

Linear Equation Exercises

Theory

Linear Equations

Other Exercises

Gauss Method Exercises

Cancel reply