Algebraic Technique Practice
Algebraic Technique Practice. Practice questions to deepen your understanding of algebraic technique. Online math practice with full solutions and step-by-step explanations.
Algebraic Technique Practice — assess your proficiency in basic algebraic operations. Practice computation skills, expression simplification, and equation solving. Suitable for middle school and high school students.
Solve:
\(\frac{3}{x}=6\)
Multiply both sides by x
\(3=6x\)
Divide both sides by 6
and get
x=0.5
Solve:
\(\frac{2}{x}=\frac{4}{x+1}\)
Cross-multiply: 2(x+1) = 4x, that is 2x+2 = 4x. Rearranging: 2 = 2x, therefore x = 1.
How do you factor the expression \(2x^2+7x+3\)
We look for two numbers whose product is 2·3=6 and whose sum is 7.
These are 6 and 1.
Factor: 2x² + 6x + x + 3 = 2x(x+3) + 1(x+3) = (2x+1)(x+3).
What is the result of simplifying the fraction \(\frac{x^2 - 4}{x - 2} \) ?
The numerator x² - 4 is a difference of squares = (x-2)(x+2).
We cancel (x-2) and get x+2.
What is the result of adding \(\frac{2}{x} + \frac{3}{x}\)
When fractions have the same denominator, we add the numerators:
\( 2/x + 3/x = (2+3)/x = 5/x\)
What is the result of the multiplication \(\frac{x}{3} ·\frac{6}{x^2}\) ?
Multiply numerator by numerator and denominator by denominator
\(\frac{x}{3} ·\frac{6}{x^2}=\frac{6x}{3x^2}=\frac{2}{x}\)
Simplify: 6/3 = 2 andx/x² = 1/x,
Therefore the result is \(\frac{2}{x}\)
What is the common factor of the expression \(6x^3-9x^2\)
The greatest common factor is \(3x^2\)
Because: \(6x^3=3x^2 · 2x\)
and \(9x^2=3x^2 · 3\)
Therefore \(6x³ - 9x² = 3x²(2x - 3)\)
What is the result of factoring the expression x² - 9 into factors?
This is the difference of squares formula a² - b² = (a-b)(a+b).
Here x² - 9 = x² - 3²
Therefore the result is (x-3)(x+3)
What are the domain conditions for the equation: \(\frac{5}{x-3} = \frac{2}{x+1}\)
The denominators cannot be zero, therefore \(x-3 \neq 0\) and \(x+1 \neq 0\)
That is \(x \neq 3\) and \(x \neq -1\)
Solve: \(x^4 - 5x^2 + 4 = 0\)
Substitute \(t=x^2\), giving \(t^2-5t+4=0\)
Solutions: \(t=1\) or \(t=4\)
Therefore \(x^2=1\) or \(x^2=4\), that is \(x=\pm 1\) or \(x=\pm 2\)
How many real solutions does the equation \(x^4 = 16\) have?
\(x^4 = 16\) gives \(x^2 = 4\) (taking the positive root)
From \(x^2=4\) we get \(x=2\) or \(x=-2\)
Total: two real solutions
Solve: \(\sqrt{x} = 4\)
Square both sides: \((\sqrt{x})^2 = 4^2\)
Therefore \(x = 16\)
Solve: \(\sqrt{x+5} = 3\)
Square both sides: \(x+5 = 9\), therefore \(x = 4\)
Check: \(\sqrt{4+5} = \sqrt{9} = 3\) ✓
Why must you verify the solution in irrational equations?
When squaring, different equations become identical
(for example \(x=3\) and \(x=-3\) both give \(x^2=9\))
Therefore, solutions may arise that do not satisfy the original equation
What is the domain of the equation \(\sqrt{2x-6} = x-3\)?
The expression under the radical must be non-negative: \(2x-6 \geq 0\)
Therefore \(2x \geq 6\), that is \(x \geq 3\)
Solve: \(\sqrt{x^2} = 5\)
\(\sqrt{x^2} = |x|\), therefore \(|x| = 5\)
This means \(x = 5\) or \(x = -5\)