Algebra · real student question

Simplify and decide whether these three expressions in S are equal: ((0.6S - 25 - 150*0.04 - 100*0.04)*0.75 - 30*0.08)/30, ((0.6S - 25 - 150*0.04)*0.75 - 30*0.08)/(30 + 100/40), and (0.45S - 25.65)/32.5.

Question

Simplify each expression and decide whether the claimed chain of equalities is correct:

(0.6S−25−150⋅0.04−100⋅0.04)⋅0.75−30⋅0.0830=(0.6S−25−150⋅0.04)⋅0.75−30⋅0.0830+100/40=0.45S−25.6532.5\frac{(0.6S-25-150\cdot 0.04-100\cdot 0.04)\cdot 0.75-30\cdot 0.08}{30}=\frac{(0.6S-25-150\cdot 0.04)\cdot 0.75-30\cdot 0.08}{30+100/40}=\frac{0.45S-25.65}{32.5}

Step-by-step solution

  1. Evaluate the shared numerical constants first. 150⋅0.04=6150\cdot 0.04=6, 100⋅0.04=4100\cdot 0.04=4, 30⋅0.08=2.430\cdot 0.08=2.4, and 100/40=2.5100/40=2.5. Reducing every product to a number before touching SS keeps the three expressions comparable.

  2. Simplify the first expression. Its bracket is 0.6S−25−6−4=0.6S−350.6S-25-6-4=0.6S-35. Then (0.6S−35)(0.75)−2.4=0.45S−26.25−2.4=0.45S−28.65(0.6S-35)(0.75)-2.4=0.45S-26.25-2.4=0.45S-28.65, and dividing by 3030 gives 0.015S−0.955=3S200−1912000.015S-0.955=\frac{3S}{200}-\frac{191}{200}.

  3. Simplify the second expression. Here only 150⋅0.04=6150\cdot 0.04=6 is subtracted, so the bracket is 0.6S−310.6S-31. Then (0.6S−31)(0.75)−2.4=0.45S−23.25−2.4=0.45S−25.65(0.6S-31)(0.75)-2.4=0.45S-23.25-2.4=0.45S-25.65, and the denominator is 30+2.5=32.530+2.5=32.5, giving 0.45S−25.6532.5\frac{0.45S-25.65}{32.5}.

  4. Compare with the third expression. The second expression has already reduced to exactly the third one, 0.45S−25.6532.5=9S650−513650≈0.0138462S−0.789231\frac{0.45S-25.65}{32.5}=\frac{9S}{650}-\frac{513}{650}\approx 0.0138462S-0.789231. So the second equality in the chain is correct.

  5. Compare the first with the others. 3S200−191200\frac{3S}{200}-\frac{191}{200} and 9S650−513650\frac{9S}{650}-\frac{513}{650} have different slopes (0.0150.015 versus 0.01384620.0138462) and different intercepts, so they are not the same expression. The first equality in the chain is false: dropping the extra −100⋅0.04-100\cdot 0.04 and changing the denominator from 3030 to 32.532.5 are not compensating changes.

  6. Find where they do coincide. Setting 3S200−191200=9S650−513650\frac{3S}{200}-\frac{191}{200}=\frac{9S}{650}-\frac{513}{650} and multiplying by 1300013000 gives 195S−12415=180S−10260195S-12415=180S-10260, so 15S=215515S=2155 and S=4313≈143.67S=\frac{431}{3}\approx 143.67. The two sides agree at that single value only.

  7. Numerical check at S = 100 and at the crossing point. At S=100S=100, expression 1 gives 0.015(100)−0.955=0.5450.015(100)-0.955=0.545 while expressions 2 and 3 both give 0.0138462(100)−0.789231=0.5953850.0138462(100)-0.789231=0.595385 - different, as predicted. At S=4313S=\frac{431}{3} both sides give exactly 1.21.2, confirming that single crossing point.

Answer

Only the last two are equal: 3S200−191200≠9S650−513650 except at S=4313\text{Only the last two are equal: }\tfrac{3S}{200}-\tfrac{191}{200}\ne\tfrac{9S}{650}-\tfrac{513}{650}\ \text{except at } S=\tfrac{431}{3}

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