First Degree Equations

A first-degree equation is a first-degree polynomial set equal to zero. In general, an equation is of the first degree if it can be written in the canonical form:

\[ ax + b = 0 \quad \text{where} \quad a \neq 0 \]

The part to the left of the equality sign is called the first member, while the part to the right is called the second member. 

Table of Contents

• How to solve a first-degree equation
• First principle of equivalence
• Second principle of equivalence
• Exercises
• Common mistakes to avoid
• Geometric meaning

Solving a first-degree equation means finding the value that, when substituted for the unknown \( x \), satisfies the equation. This means that the value (solution to the equation) must make the equality true. The solving process involves some steps, called equivalence principles for equations.

The first principle of equivalence states that, by adding or subtracting the same quantity or algebraic expression to both sides of an equation, the set of solutions does not change.

Thanks to this principle, we can subtract the quantity \( -b \) from both sides:

\[ ax = -b \]

Note that adding or subtracting a quantity to both sides is equivalent to "moving" it from one side to the other, as long as its sign is changed. For now, we have moved \( b \) to the second member, changing its sign to \( -b \).

The second principle of equivalence states that, by multiplying or dividing both sides of an equation by the same non-zero number, the set of solutions does not change. 

Applying this principle to the equivalent equation \( ax = -b \), we divide both sides by \( a \neq 0 \), yielding:

\[ x = -\frac{b}{a} \]

It is important to emphasize that \( a \) must be different from zero for the equation to make sense. In fact, if \( a = 0 \), the equation becomes \( 0 \cdot x + b = 0 \), which simplifies to \( b = 0 \), a situation that does not represent an equation in \( x \) and would be impossible if \( b \neq 0 \).

From now on, the goal will be to isolate the variable \( x \) on the first member, or on the second (it doesn't change anything).

Exercise 1. Solve the equation \( 3x - 1 = 0 \).

Solution. Move \( -1 \) to the second member (changing its sign):

\[ 3x = 1 \]

Finally, divide both sides by \( 3 \) to obtain the desired solution:

\[ x = \frac{1}{3} \]

Verification. To verify that it is the correct solution, substitute the found value back into the original equation:

\[ 3 \cdot \frac{1}{3} - 1 = 1 - 1 = 0 \]

Therefore, the solution is correct.

Exercise 2: Solve the first-degree equation \(\displaystyle\frac{1}{2}(x-1)=-x+1\).

Solution. Start by isolating the unknown \( x \) on the first member:

\[ \frac{1}{2}(x - 1) = -x + 1 \implies \frac{x}{2} - \frac{1}{2} = -x + 1 \]

Now add \(x\) to both sides of the equation:

\[ \frac{x}{2} + x - \frac{1}{2} = 1 \]

Simplify by transforming \(x\) into a term with a common denominator:

\[ \frac{x}{2} + \frac{2x}{2} - \frac{1}{2} = 1 \implies \frac{3x}{2} - \frac{1}{2} = 1 \]

Add \(\displaystyle\frac{1}{2}\) to both sides:

\[ \frac{3x}{2} = \frac{3}{2} \]

Now multiply both sides by \(\displaystyle\frac{2}{3}\) to solve for \(x\):

\[ x = 1 \]

Therefore, the solution is \( x = 1 \).

Verification. As before, substitute \( x = 1 \) into the original equation:

\[ \frac{1}{2}(x - 1) = -x + 1 \]

When \( x = 1 \), we get:

\[ \frac{1}{2}(1 - 1) = -1 + 1 \]

Calculate both sides:

\[ \frac{1}{2} \cdot 0 = 0 \quad \text{and} \quad -1 + 1 = 0 \]

Both sides are equal, so the solution is correct.

Exercise 3. Solve the equation \( 5(x - 2) - 3(2x + 1) = 7 - 4x \)

Solution. Apply the distributive property:

\[ 5x - 10 - 6x - 3 = 7 - 4x \]

\[ -x - 13 = 7 - 4x \]

Move the terms with \(x\) to the first member and the constant terms to the second member:

\[ -x + 4x = 7 + 13 \]

\[ 3x = 20 \]

Divide both sides by 3:

\[ x = \frac{20}{3} \]

Verification. Substitute \( x = \frac{20}{3} \) into the original equation:

\[ 5\left(\frac{20}{3} - 2\right) - 3\left(2\cdot\frac{20}{3} + 1\right) = 7 - 4\cdot\frac{20}{3} \]

Calculate the first side:

\[ 5\left(\frac{20}{3} - \frac{6}{3}\right) - 3\left(\frac{40}{3} + \frac{3}{3}\right) = 5\cdot\frac{14}{3} - 3\cdot\frac{43}{3} = \frac{70}{3} - \frac{129}{3} = -\frac{59}{3} \]

Calculate the second side:

\[ 7 - 4\cdot\frac{20}{3} = \frac{21}{3} - \frac{80}{3} = -\frac{59}{3} \]

Both sides are equal, so the solution is verified:

\[ x = \frac{20}{3} \]

When solving first-degree equations, it is important to be aware of some common mistakes:

Sign Change Error: When moving a term from one side of the equation to the other, remember to change its sign. For example, in the equation \(2x + 3 = 5\), when moving the \(3\), you get \(2x = 5 - 3\), not \(2x = 5 + 3\).

Incomplete Distribution: When you have an expression like \(3(x + 2)\), the coefficient \( 3 \) must be multiplied by all the terms inside the parentheses. A common mistake is writing \(3x + 2\) instead of the correct \(3x + 6\).

Fraction Errors: When you have an equation like \(\displaystyle \frac{x}{2} = 3\), to isolate \(x\) you must multiply both sides by 2, yielding \(x = 6\). It is wrong to write \(x = \displaystyle \frac{3}{2}\).

Imprecise Simplification: In an equation like \(2x - x = 5\), do not forget to simplify the like terms before proceeding. The correct form is \(x = 5\).

Missing Verification: Skipping the verification step can result in missing calculation errors. It is always recommended to substitute the found solution back into the original equation to confirm its correctness.

Solving a first-degree equation \( ax + b = 0 \) means finding the value at which the line with the equation

\[ y = ax + b \]

intersects the x-axis (horizontal axis). For example, the line with the equation \( y = 2x - 1 \)

intersects the x-axis at the point \( x = \displaystyle \frac{1}{2} \), as shown in the figure.

We said that the solution of a first-degree equation \( ax + b = 0 \) is the x-coordinate where the line intersects the x-axis. Now, let's ask another question: how can we determine the value of \( x \) for which the line \( y = ax + b \) takes a specific value, say \( y = 2 \)?

To do this, simply substitute \( y = 2 \) into the equation of the line. We get \( 2 = 2x - 1 \iff 2x - 3 = 0 \), so:

\[ x=\frac{3}{2} \]

As shown in the figure, the solution we found corresponds to the ordinate \(y=2\).