Explore detailed, fully explained solutions to intermediate algebra problems featured in Sample 3 on AnalyzeMath.com. This page includes clear step-by-step answers to questions involving absolute value equations, quadratic equations, systems of equations, factoring, and functions — all designed to help students build a solid understanding of algebra.
Write \(1.5 \times 10^{-5}\) in standard form.
Solution
\(1.5 \times 10^{-5} = \frac{1.5}{10^5} = \frac{1.5}{100000} = 0.000015\)
Evaluate: \(30 - |-x + 6|\) for \(x = 10\)
Solution
Substitute \(x = 10\) in the expression:
\(30 - |-(10) + 6| = 30 - |-10 + 6| = 30 - |-4| = 30 - 4 = 26\)
Evaluate: \(2xy^3 + x - 2y\) for \(x = 2\) and \(y = -2\)
Solution
Substitute \(x = 2\), \(y = -2\):
\(2(2)(-2)^3 + 2 - 2(-2) = 4(-8) + 2 + 4 = -26\)
What is the slope of the line perpendicular to the line \(y = -4\)?
Solution
The line \(y = -4\) is horizontal. A line perpendicular to it is vertical and has an undefined slope.
Write an equation of the line with slope 2 and x-intercept \((-4, 0)\).
Solution
Use point-slope form: \(y - b = m(x - a)\)
Using point \((-4, 0)\) and \(m = 2\):
\(y - 0 = 2(x - (-4)) \Rightarrow y = 2x + 8\)
Solve the equation: \(-3(-x + 5) + 20 = -10(x - 3) + 4\)
Solution
Expand both sides:
\(3x - 15 + 20 = -10x + 30 + 4\)
\(3x + 5 = -10x + 34\)
\(13x = 29 \Rightarrow x = \frac{29}{13}\)
Solve the inequality: \(4(x - 6) + 4 < 8(x - 4)\)
Solution
Expand:
\(4x - 24 + 4 < 8x - 32\)
\(4x - 20 < 8x - 32\)
\(-4x < -12 \Rightarrow x > 3\)
Solve the equation: \(3(x - 2)^2 - 12 = 0\)
Solution
\(3(x - 2)^2 = 12 \Rightarrow (x - 2)^2 = 4\)
\(x - 2 = \pm 2 \Rightarrow x = 0, 4\)
The solutions are \(x = 0\) and \(x = 4\)
Solve the equation: \(\frac{x}{3} + \frac{2}{7} = \frac{x}{7} - 5\)
Solution
Multiply all terms by LCD = 21:
\(21\left(\frac{x}{3} + \frac{2}{7}\right) = 21\left(\frac{x}{7} - 5\right)\)
\(7x + 6 = 3x - 105\)
\(4x = -111 \Rightarrow x = \frac{-111}{4}\)
Line \(L\) passes through \((2, 7)\) and is perpendicular to \(x + y = 0\). Find the intersection of \(L\) and \(x + y = 0\).
Solution
Slope of \(x + y = 0\) is -1 (rewritten as \(y = -x\)).
Slope of line \(L\) is 1 (since perpendicular → negative reciprocal).
Let point of intersection be \((a, b)\).
Using slope formula: \(\frac{7 - b}{2 - a} = 1\)
Also, point lies on \(x + y = 0 \Rightarrow a + b = 0 \Rightarrow a = -b\)
Substitute into slope equation: \(7 - b = 2 - (-b) = 2 + b\)
\(7 - b = 2 + b \Rightarrow 2b = 5 \Rightarrow b = \frac{5}{2}, a = -\frac{5}{2}\)
The intersection point is \(\left(-\frac{5}{2}, \frac{5}{2}\right)\)
Find the point of intersection of \(x + 2y = 4\) and \(-x - 3y = -7\)
Solution
Add the equations:
\((x + 2y) + (-x - 3y) = 4 + (-7)\)
\(-y = -3 \Rightarrow y = 3\)
Substitute into \(x + 2y = 4 \Rightarrow x + 6 = 4 \Rightarrow x = -2\)
The intersection point is \((-2, 3)\)
How many solutions does the system \(2x - 3y = 4\) and \(4x - 6y = -7\) have?
Solution
Multiply first equation by 2:
\(4x - 6y = 8\)
Subtract: \((4x - 6y) - (4x - 6y) = 8 - (-7)\)
\(0 = 15\) → contradiction → No solution.
Find the value of \(A\) such that the system \(Ax + 6y = 0\) and \(2x - 7y = 3\) has no solution.
Solution
Solve first equation for \(y\): \(y = -\frac{A}{6}x\)
Substitute into second equation:
\(2x - 7\left(-\frac{A}{6}x\right) = 3 \Rightarrow 2x + \frac{7A}{6}x = 3\)
\(x\left(2 + \frac{7A}{6}\right) = 3\)
No solution if denominator = 0:
\(2 + \frac{7A}{6} = 0 \Rightarrow A = -\frac{12}{7}\)
Solve \( |2x - 4| - 2 = 6 \).
Solution
Add 2 to both sides: \( |2x - 4| = 8 \)
Split into two cases:
Case 1: \( 2x - 4 = 8 \Rightarrow 2x = 12 \Rightarrow x = 6 \)
Case 2: \( 2x - 4 = -8 \Rightarrow 2x = -4 \Rightarrow x = -2 \)
The solutions are \(x = -2\) and \(x = 6\)
How many solutions does the equation \(2x^2 + 3x = 8\) have?
Solution
Rearrange the equation: \(2x^2 + 3x - 8 = 0\)
Use the discriminant: \(D = b^2 - 4ac = 3^2 - 4(2)(-8) = 9 + 64 = 73\)
Since the discriminant is positive, the equation has two real solutions
Solve the equation \(3x^2 + 6x - 1 = 8\)
Solution
Rearrange: \(3x^2 + 6x - 9 = 0\)
Divide by 3: \(x^2 + 2x - 3 = 0\)
Factor: \((x + 3)(x - 1) = 0 \Rightarrow x = -3, x = 1\)
The solutions are \(x = -3\) and \(x = 1\)
Solve the system of equations: \(2x + 5y = 18\) and \(-3x - y = -1\)
Solution
Multiply the second equation by 5: \(-15x - 5y = -5\)
Add to first equation:
\((2x + 5y) + (-15x - 5y) = 18 + (-5) \Rightarrow -13x = 13 \Rightarrow x = -1\)
Substitute \(x = -1\) into \(2x + 5y = 18\):
\(2(-1) + 5y = 18 \Rightarrow -2 + 5y = 18 \Rightarrow 5y = 20 \Rightarrow y = 4\)
The solution is \(x = -1\), \(y = 4\)
What is the range of function \(f = \{(2,3), (1,4), (5,4), (0,3)\}\)?
Solution
Range is the set of output values (second values): \(\{3, 4, 4, 3\}\)
Remove duplicates: \(\{3, 4\}\)
The range is \(\{3, 4\}\)
Factor the expression \(2x^2 + 3x + 1\)
Solution
Find factors of \(2x^2 + 3x + 1\):
\(2x^2 + 2x + x + 1 = 2x(x + 1) + 1(x + 1) = (2x + 1)(x + 1)\)
Factored form: \((2x + 1)(x + 1)\)
Factor the expression \(10x^2 + 20x - 80\)
Solution
Factor out the GCF: \(10(x^2 + 2x - 8)\)
Factor the quadratic: \(x^2 + 4x - 2x - 8 = x(x + 4) -2(x + 4) = (x - 2)(x + 4)\)
Factored form: \(10(x - 2)(x + 4)\)