Pi Power Worksheet
TEKS 8.8(C) Target Practice
Pi Equation Showdown
Solve one-variable linear equations containing \(\pi\) and variables on both sides.
Constant \(\pi \approx 3.1416\)
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Date:
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Architect's Strategy: Eliminate the Pi (\(\pi\)) First!
Since \(\pi\) is a common factor in every term, divide the entire equation by \(\pi\) to make it a standard linear equation.
1. Original Equation \(10\pi x + 5\pi = 4\pi x + 23\pi\)
2. Divide both sides by \(\pi\) \(10x + 5 = 4x + 23\)
3. Solve standard equation \(6x = 18 \implies x = 3\)
PROBLEM 1 Positive Integers
\(9\pi x = 4\pi x + 35\pi\)
Show all algebraic steps
x =
PROBLEM 2 Variables & Constants
\(14\pi h + 8\pi = 6\pi h + 24\pi\)
Show all algebraic steps
h =
PROBLEM 3 Decimals
\(3.5\pi y - 10\pi = 1.5\pi y + 6\pi\)
Show all algebraic steps
y =
PROBLEM 4 Rational Coefficients
\(\frac{2}{3}\pi w + 12\pi = -\frac{1}{3}\pi w + 18\pi\)
Show all algebraic steps
w =
PROBLEM 5 Negative Values
\(5\pi z - 15\pi = -3\pi z + 49\pi\)
Show all algebraic steps
z =
PROBLEM 6 Real-World Cylinder Modeling
Two cylinder containers have equal volumes. Cylinder A has a volume of \(12\pi r\) cubic units. Cylinder B has a volume of \(4\pi r + 40\pi\) cubic units. Set up the equation and solve for the radius \(r\).
Show equation setup & steps
r =
TEKS 8.8(C) • One-Variable Equations with Variables on Both Sides Page 1 of 1
Pi Power Answer Key
TEKS 8.8(C) Answer Key
Pi Equation Showdown [ANSWER KEY]
Teacher guide with complete step-by-step algebraic solutions and simplified models.
Status VERIFIED
Subject: 8th Grade Math
Focus: TEKS 8.8(C)
Unit: Linear Equations
Instructional Tip: Addressing Common Pitfalls
Students often incorrectly try to subtract or approximate \(\pi\) (using 3.14) before simplifying, leading to complex decimal arithmetic. Remind them that dividing every term by \(\pi\) instantly yields simple, beautiful whole-number or basic rational equations.
PROBLEM 1 Positive Integers
\(9\pi x = 4\pi x + 35\pi\)
1. Divide both sides by \(\pi\):
\(9x = 4x + 35\)
2. Subtract \(4x\) from both sides:
\(5x = 35\)
3. Divide by \(5\): \(x = 7\)
Solution Verified
x =
7
PROBLEM 2 Variables & Constants
\(14\pi h + 8\pi = 6\pi h + 24\pi\)
1. Divide both sides by \(\pi\):
\(14h + 8 = 6h + 24\)
2. Subtract \(6h\) and \(8\):
\(8h = 16\)
3. Divide by \(8\): \(h = 2\)
Solution Verified
h =
2
PROBLEM 3 Decimals
\(3.5\pi y - 10\pi = 1.5\pi y + 6\pi\)
1. Divide both sides by \(\pi\):
\(3.5y - 10 = 1.5y + 6\)
2. Subtract \(1.5y\) and add \(10\):
\(2y = 16\)
3. Divide by \(2\): \(y = 8\)
Solution Verified
y =
8
PROBLEM 4 Rational Coefficients
\(\frac{2}{3}\pi w + 12\pi = -\frac{1}{3}\pi w + 18\pi\)
1. Divide both sides by \(\pi\):
\(\frac{2}{3}w + 12 = -\frac{1}{3}w + 18\)
2. Add \(\frac{1}{3}w\) to both sides:
\(1w + 12 = 18\)
3. Subtract \(12\): \(w = 6\)
Solution Verified
w =
6
PROBLEM 5 Negative Values
\(5\pi z - 15\pi = -3\pi z + 49\pi\)
1. Divide both sides by \(\pi\):
\(5z - 15 = -3z + 49\)
2. Add \(3z\) and add \(15\):
\(8z = 64\)
3. Divide by \(8\): \(z = 8\)
Solution Verified
z =
8
PROBLEM 6 Real-World Cylinder Modeling
Two cylinder containers have equal volumes. Cylinder A has a volume of \(12\pi r\). Cylinder B has \(4\pi r + 40\pi\). Solve for radius \(r\).
1. Equalize Volumes: \(12\pi r = 4\pi r + 40\pi\)
2. Divide by \(\pi\): \(12r = 4r + 40\)
3. Subtract \(4r\): \(8r = 40\)
4. Divide by 8: \(r = 5\)
Solution Verified
r =
5
TEKS 8.8(C) ANSWER KEY • One-Variable Equations with Variables on Both Sides Page 1 of 1
