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Algebra 2 9-12 / Week 04 / Thursday
4/6
Week 04 · Rational Expressions and Equations

Thursday

Equations with x on the bottom
// Plant food in the tank
⏱ about 20 min

Thursday: Equations With x on the Bottom

"The new trays want 4 percent," Comet says, reading the label. "How much water do we add to the 20-liter tank?"

Wren writes 200/(x + 20) = 4 on the wall. "A rational equation. Multiply both sides by the denominator."

"200 = 4(x + 20), so x = 30," Comet says. "Thirty liters, like Monday's table."

"Now a strange one," Wren says. "(x² - 4)/(x - 2) = 5. Cross-multiply and solve."

Comet works. "x² - 4 = 5x - 10. So x² - 5x + 6 = 0. x = 2 or x = 3."

Nova pulses over the x = 2. "Would you like a hint? Put 2 back into the original denominator."

"Two minus two. Zero. The fraction has no value there," Comet says. "So x = 2 is not a solution at all."

"Extraneous," Wren says. "The algebra made it, the equation rejects it."

Solving a rational equation

  1. Note every value that makes a denominator zero. Those values are not allowed.
  2. Multiply both sides by the denominator, or cross-multiply when there is one fraction on each side.
  3. Solve the polynomial equation that is left. It may be linear or quadratic.
  4. Check each answer in the original equation. Throw out any value from step 1.

The tank: (200)/(x + 20) = 4 becomes 200 = 4(x + 20), so x = 30. The crew adds 30 liters of water.

The strange one: (x² - 4)/(x - 2) = 5 becomes x² - 4 = 5(x - 2). Solving gives x = 2 or x = 3.

But x = 2 makes the denominator x - 2 equal to zero. It is extraneous. The only solution is x = 3.

The graph of y = 2/(x - 1): two branches that never touch the dashed line x = 1.

The picture shows y = 2/(x - 1). At x = 1 the denominator is zero, so the graph has no point there. An equation cannot be solved by a value its own denominator forbids.

Why an extraneous solution appears

  1. Start with (x² - 4)/(x - 2) = 5. The value x = 2 is not allowed, because the denominator would be zero.
  2. Multiply both sides by (x - 2). The new equation x² - 4 = 5(x - 2) has no denominator at all.
  3. Multiplying by (x - 2) multiplies both sides by 0 when x = 2. Any equation is true when both sides are 0.
  4. So x = 2 becomes a solution of the new equation, even though it was never a solution of the old one.
  5. Solving the new equation gives x = 2 and x = 3. The check sends x = 2 away, and x = 3 stays.

That is the whole story. Clearing a denominator can turn a forbidden value into a solution of the new equation. The check at the end is not optional.

SOLVE THE EQUATIONS
  • Read the question.
  • Tap your answer.
Solve the tank equation (200)/(x + 20) = 4. How many liters of water?
Solve (x² - 4)/(x - 2) = 5.
Solve (6)/(x) = 3.
Solve (x + 1)/(x - 2) = 3.
WHICH VALUE IS EXTRANEOUS?
  • Read the question.
  • Tap your answer.
Cross-multiplying (x² - 4)/(x - 2) = 5 gives the values x = 3 and x = 2. Which one is extraneous?
Cross-multiplying (x² - 9)/(x - 3) = 6 gives the values x = 3. Which one is extraneous?
(x² - 9)/(x - 3) = 6 gives only x = 3, and x = 3 is extraneous. How many solutions does the equation have?
WHY X = 2 APPEARS, IN ORDER
  • ?(x² - 4)/(x - 2) = 5, and x = 2 is not allowed
  • ?Multiply both sides by (x - 2): x² - 4 = 5(x - 2)
  • ?At x = 2 both sides were multiplied by 0, so both sides are 0
  • ?x = 2 solves the new equation but not the old one
  • ?Solve, check, and keep only x = 3
WHY THIS EXERCISEA derivation is a chain. Each link shows why the check at the end is part of the method, not an extra.
A value the algebra produces that makes a denominator zero is called this. Type one word.
Before solving, note every value that makes this part of the fraction zero. Type one word.
Solve (200)/(x + 20) = 4. How many liters of water does the crew add? Type the number.
StatementTrue or false?
Multiplying both sides of an equation by (x - 2) can add a solution at x = 2.?
(3 × 3 - 4) ÷ (3 - 2) = 5?
x = 2 is a solution of (x² - 4)/(x - 2) = 5.?
200 ÷ (30 + 20) = 4?
A rational equation can have no solution at all.?
WHY THIS EXERCISEChecking by substitution is the step that separates real solutions from extraneous ones.
Try it
Write (x² - 1)/(x - 1) = 2 on a card. Cross-multiply and solve. Then check your value in the original.
Did the equation have a solution? Write one sentence explaining what happened.

Sharp thinking. Tomorrow you write rational equations from Glass House jobs: mixing, and two people filling one tank.

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