L2 · Converting to an equivalent integer equation
“I can convert an equation with rational coefficients into an equivalent equation with integer coefficients by multiplying EVERY term by the LCM of the denominators (or by a power of 10 for decimals), then solve the integer equation and verify the solution in the original.”
Solve (1/2)x + 3 = 5. (This is the curriculum's own example of an equivalent equation.) Step 1: The only denominator is 2, so the clearing multiplier is 2. Step 2: Multiply EVERY term by 2 - the x-term, the +3, and the 5: (1/2)x x 2 = x; 3 x 2 = 6; 5 x 2 = 10. The equivalent integer equation is x + 6 = 10. Step 3: Solve the integer equation: x = 4. Step 4: Verify in the ORIGINAL (the converted equation's answer must satisfy the original - that is what "equivalent" promises): (1/2)(4) + 3 = 2 + 3 = 5. Say it plainly: scaling a balance by the same factor on both pans changes nothing about what balances it. But the stamp must land on every term - a term that escapes the multiplier breaks the equivalence.
Why must the +3 be multiplied by 2 as well, when it has no fraction in it?
Convert and solve (1/3)x + 2 = 6 -> multiply by 3: x + 6 = 18, so x = 12 (check: (1/3)(12) + 2 = 4 + 2 = 6)
"(1/4)x + 5 = 9, multiplied through by 4, becomes x + 20 = ___" -> 36
10 scored questions with a 3-rung hint ladder. Mastery needs 8/10 plus both gate questions.