The following five problems represent the standardized core of US Math 70 intermediate algebra quizzes. Every solution maps the formal mechanical steps required by testing committees.
Problem 1: Solving Rational Equations with Domain Restrictions
Solve for x: (3 / (x - 2)) + (1 / (x + 3)) = 4 / (x² + x - 6)
Step 1: Factor all denominators to establish domain limitations.
x² + x - 6 = (x - 2)(x + 3)
Domain restriction: x ≠ 2 and x ≠ -3.
Step 2: Multiply the entire equation by the lowest common denominator, (x - 2)(x + 3):
3(x + 3) + 1(x - 2) = 4
Step 3: Expand and combine linear terms:
3x + 9 + x - 2 = 4
4x + 7 = 4
4x = -3
x = -3/4
Step 4: Cross-check against domain restrictions. Because -3/4 is neither 2 nor -3, the solution is valid.
Final Answer: x = -3/4
Problem 2: Non-Linear Inequality and Sign Analysis
Solve and express in interval notation: (2x - 5) / (x + 4) ≤ 1
Step 1: Move all terms to the left side to establish zero on the right:
(2x - 5) / (x + 4) - 1 ≤ 0
Step 2: Find a common denominator and combine:
[(2x - 5) - (x + 4)] / (x + 4) ≤ 0
(2x - 5 - x - 4) / (x + 4) ≤ 0
(x - 9) / (x + 4) ≤ 0
Step 3: Identify critical values where numerator and denominator equal zero:
Numerator zero: x = 9
Denominator zero: x = -4 (vertical asymptote, strictly excluded from solution)
Step 4: Test sign intervals across (-∞, -4), (-4, 9], and [9, ∞):
Test x = -5: (-5 - 9) / (-5 + 4) = (-14) / (-1) = 14 > 0 (False)
Test x = 0: (0 - 9) / (0 + 4) = -9/4 ≤ 0 (True)
Test x = 10: (10 - 9) / (10 + 4) = 1/14 > 0 (False)
Final Answer: (-4, 9]
Problem 3: Radical Equations and Extraneous Roots
Solve for y: √(2y + 15) - y = 6
Step 1: Isolate the radical term on one side of the equation:
√(2y + 15) = y + 6
Step 2: Square both sides to eliminate the radical:
2y + 15 = (y + 6)²
2y + 15 = y² + 12y + 36
Step 3: Set the quadratic equation to zero:
y² + 10y + 21 = 0
Step 4: Factor the quadratic polynomial:
(y + 7)(y + 3) = 0
Potential solutions: y = -7, y = -3
Step 5: Test both values in the original radical equation:
Check y = -7: √(2(-7) + 15) - (-7) = √1 + 7 = 1 + 7 = 8 ≠ 6 (Extraneous root, discard)
Check y = -3: √(2(-3) + 15) - (-3) = √9 + 3 = 3 + 3 = 6 (True)
Final Answer: y = -3
Problem 4: Quadratic Function Vertex and Axis Form
Convert f(x) = 3x² - 12x + 7 to vertex form f(x) = a(x - h)² + k using completing the square.
Step 1: Factor the leading coefficient 3 from the variable terms:
f(x) = 3(x² - 4x) + 7
Step 2: Complete the square inside the parentheses. Take half of -4 (-2), square it (4), and balance outside:
f(x) = 3(x² - 4x + 4 - 4) + 7
f(x) = 3(x² - 4x + 4) - (3 · 4) + 7
f(x) = 3(x - 2)² - 12 + 7
f(x) = 3(x - 2)² - 5
Final Answer: f(x) = 3(x - 2)² - 5 (Vertex: (2, -5))
Problem 5: 3x3 Linear System Elimination
Solve the system:
(1) x + 2y - z = 4
(2) 2x + y + z = 7
(3) x - 3y + 2z = -1
Step 1: Eliminate z by adding equation (1) to equation (2):
(x + 2y - z) + (2x + y + z) = 4 + 7
3x + 3y = 11 → Equation (4)
Step 2: Eliminate z using equations (2) and (3). Multiply equation (2) by -2 and add to (3):
-4x - 2y - 2z = -14
x - 3y + 2z = -1
-3x - 5y = -15 → Equation (5)
Step 3: Add Equation (4) and Equation (5) directly:
(3x + 3y) + (-3x - 5y) = 11 + (-15)
-2y = -4
y = 2
Step 4: Substitute y = 2 back into Equation (4):
3x + 3(2) = 11
3x + 6 = 11
3x = 5
x = 5/3
Step 5: Substitute x = 5/3 and y = 2 into Equation (1) to find z:
5/3 + 2(2) - z = 4
5/3 + 4 - z = 4
z = 5/3
Final Answer: (x, y, z) = (5/3, 2, 5/3)
