By Nicole Harms • Updated Aug 30, 2022
Wachiwit/iStock/GettyImages
Solving linear equations is a cornerstone of algebra. Mastering this skill not only builds confidence but also provides a toolkit for tackling a wide range of algebraic problems.
Start by moving every term that contains a variable to the left‑hand side. For example, with the equation
\(5a + 16 = 3a + 22\)
subtract \(3a\) from both sides, yielding
\(2a + 16 = 22\)
Now shift the constants to the right side by adding the opposite of \(+16\), which is \(-16\):
\(2a = 6\)
The variable \(a\) is multiplied by 2. Divide both sides by 2 to solve for \(a\):
\(\frac{2a}{2} = \frac{6}{2}\)
so \(a = 3\).
Substitute \(a = 3\) back into the original equation to confirm:
\(5(3) + 16 = 3(3) + 22\)
Both sides equal 31, confirming the solution is correct.
Consider the equation
\(\frac{5}{4}x + \frac{1}{2} = 2x - \frac{1}{2}\)
Subtract \(2x\) from both sides. To combine with \(\frac{5}{4}x\), express \(2x\) as \(\frac{8}{4}x\):
\(\frac{5}{4}x - \frac{8}{4}x + \frac{1}{2} = -\frac{1}{2}\)
which simplifies to
\(-\frac{3}{4}x + \frac{1}{2} = -\frac{1}{2}\)
Add \(-\frac{1}{2}\) to both sides to move the constant term:
\(-\frac{3}{4}x = -1\)
Divide both sides by \(-\frac{3}{4}\), or multiply by its reciprocal \(-\frac{4}{3}\):
\(x = \frac{4}{3}\)
Plugging \(x = \frac{4}{3}\) into the original equation gives:
\(\frac{5}{4}\times\frac{4}{3} + \frac{1}{2} = 2\times\frac{4}{3} - \frac{1}{2}\)
Both sides evaluate to \(\frac{13}{6}\), confirming the solution.
For an alternative walkthrough, watch the video below.
Tip: Solving by hand, especially with fractions, often yields quicker results than relying on a calculator.
Warning: Always double‑check your work; small errors can easily creep in during the process.
