Solving equations and inequalities: Linear inequalities in one unknown
Solving a linear inequality via equations
You can also solve a linear inequality by
- first replacing the inequality sign by an equal sign,
- then solving this equation, and
- finally, determining the sign of the inequality for point to the left and to the right of the solution of the equation.
Determine the exact solution of the inequality \[-3x + 10 \lt -x {\,+\,}16\] via equations.
\(x \gt -3\)
We follow the following roadmap:
- Get started with the corresponding equation \[-3x + 10 = -x {\,+\,}16\]
- Solve this equation:
- Get the terms with \(x\) on the left-hand side of the equation (by adding \(x\) on both sides):
\(-3x + 10 +x = -x {\,+\,}16 +x\), which simplifies to \(-2x +10 = 16\). - Then move the terms without \(x\) to the right (by adding \(-10\) both sides):
\(-2x +10 - 10 = 16 - 10\), which simplifies to \(-2x = 6\).- Next, divide the left- and right-hand side by the coefficient of \(x\) (which is here \(-2\)); this gives \(x = \;\frac{6}{-2}\).
- So, the solution of the equation is \(x = {-3}\).
- Get the terms with \(x\) on the left-hand side of the equation (by adding \(x\) on both sides):
- Find out whether the solutions are on the number line to the left or to the right of \(-3\).
- First calculate the left- and right-hand sides of the inequality \(-3x + 10 \lt -x {\,+\,}16\) when you substitute a value of \(x\) less than \(-3\). For example, when you fill in \(x=-10\), then you get \(40 \lt 26\) and this is a false statement. Any other value of \(x\) less than \(-3\) may be used too, and you still get a false statement.
- Then calculate the left- and right-hand sides of the inequality \(-3x + 10 \lt -x {\,+\,}16\) when you substitute a value of \(x\) greater than \(-3\). For example, when you fill in \(x=10\), then you get \(-20 \lt 6\) and this is a true statement. Any other value of \(x\) greater than \(-3\) may be used too, and you still get a true statement.
- From these two numeric examples follows that solutions \(x\) of \(-3x + 10 \lt -x {\,+\,}16\) must satisfy \(x \gt -3\).
The points where the inequality holds are shown in green in the number line below. An open circle around \(x=-3\) indicates that we are dealing with an inequality of the type \(\lt\) or \(\gt\), where in this case the point itself is not a solution. A closed circle indicates an inequality of the type \(\le\) or \(\ge\), and then the point marked on the number line is element of the solution set.
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