Open Resources for Community College Algebra

Example 2.4.2.

Solve for \(x\) in \(3x=3x+4\text{.}\)

To solve this equation, we need to move all terms containing \(x\) to one side of the equals sign:

Notice that \(x\) is no longer present in the equation. What value can we substitute into \(x\) everywhere that you see \(x\) in the equation to make \(0=4\) true? Nothing! There is no \(x\) to substitute a value into and change the equation from being \(0=4\text{.}\) We say this equation has no solution. Or, the equation has an empty solution set. We can write this as \(\emptyset\text{,}\) or \(\{\text{ }\}\text{,}\) which are the symbols for the empty set (not to be confused with the number \(0\text{.}\))

The equation \(0=4\) is known a false statement. It is false no matter what \(x\) is. It indicates there is no solution to the original equation.

Example 2.4.3.

Solve for \(x\) in \(2x+1=2x+1\text{.}\)

We will move all terms containing \(x\) to one side of the equals sign:

At this point, \(x\) is no longer present in the equation. What value can we substitute into \(x\) to make \(1=1\) true? Any number! This means that all real numbers are solutions to the equation \(2x+1=2x+1\text{.}\) We say this equation's solution set contains all real numbers. We can write this set using set-builder notation as \(\{x\mid x\text{ is a real number}\}\) or using interval notation as \((-\infty,\infty)\text{.}\)

The equation \(1=1\) is unambiguously true, since it is true no matter what \(x\) is. It indicates that all real numbers are solutions to the original linear equation.

Example 2.4.6.

Solve for \(t\) in the inequality \(4t+5\gt 4t+2\text{.}\)

To solve for \(t\text{,}\) we will first subtract \(4t\) from each side to get all terms containing \(t\) on one side:

Notice that again, the variable \(t\) is no longer contained in the inequality. We then need to consider which values of \(t\) make the inequality true. The answer is that all values of \(t\) make \(5\gt 2\text{,}\) which we know is a very strange sentence. So our solution set is all real numbers, which we can write as \(\{t\mid t\text{ is a real number}\}\text{,}\) or as \((-\infty,\infty)\text{.}\)

Example 2.4.7.

Solve for \(x\) in the inequality \(-5x+1\le -5x\text{.}\)

To solve for \(x\text{,}\) we will first add \(5x\) to each side to get all terms containing \(x\) on one side:

Once more, the variable \(x\) is absent. So we can ask ourselves, “For which values of \(x\) is \(1\le 0\) true?” The answer is none, and so there is no solution to this inequality. We can write the solution set using \(\emptyset\text{.}\)

Example 2.4.10.

Solve for \(a\) in \(\frac{2}{3}(a+1)-\frac{5}{6}=\frac{2}{3}a\text{.}\)

To solve this equation for \(a\text{,}\) we recall the technique of multiplying each side of the equation by the LCD of all fractions. Here, this means that we will multiply each side by \(6\) as our first step. After that, we'll be able to simplify each side of the equation and continue solving for \(a\text{:}\)

The statement \(-1=0\) is false, so the equation has no solution. We can write the solution set as the empty set, \(\emptyset\text{.}\)

Example 2.4.11.

Solve for \(x\) in the equation \(3(x+2)-8=(5x+4)-2(x+1)\text{.}\)

To solve for \(x\text{,}\) we will first need to simplify the left side and right side of the equation as much as possible by distributing and combining like terms:

From here, we'll want to subtract \(3x\) from each side:

As the equation \(-2=2\) is not true for any value of \(x\text{,}\) there is no solution to this equation. We can write the solution set as the empty set, \(\emptyset\text{.}\)

Example 2.4.12.

Solve for \(z\) in the inequality \(\frac{3z}{5}+\frac{1}{2}\le\left(\frac{z}{10}+\frac{3}{4}\right)+\left(\frac{z}{2}-\frac{1}{4}\right)\text{.}\)

To solve for \(z\text{,}\) we will first need to multiply each side of the inequality by the LCD, which is \(20\text{.}\) After that, we'll finish solving by putting all terms containing a variable on one side of the inequality:

As the equation \(10\le10\) is true for all values of \(z\text{,}\) all real numbers are solutions to this inequality. Thus the solution set is \(\{z\mid z\text{ is a real number}\}\text{,}\) or \((-\infty,\infty)\) in interval notation.