Quiz Math 70 Answer Key Revealed: Full Step-by-Step Solutions and Proof

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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)