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After finishing the prep work, Gilberta and María start removing wallpaper at the same time. Gilberta removes the paper at a constant rate of 
4.3m^(2) per hour, while María removes 
3.4m^(2) of paper per hour. Gilberta's room starts with 
35m^(2) of paper, and María's room starts with 
30.5m^(2) of paper.
Let 
t represent the time, in hours, since Gilberta and María start removing the wallpaper.
Complete the inequality to represent the times when Gilberta has more wallpaper left in her room than María has in hers.

t select inequality symbol hours

After finishing the prep work, Gilberta and Mar\u0000eda start removing wallpaper at the same time. Gilberta removes the paper at a constant rate of 4.3m24.3\,\text{m}^2 per hour, while Mar\u0000eda removes 3.4m23.4\,\text{m}^2 of paper per hour. Gilberta's room starts with 35m235\,\text{m}^2 of paper, and Mar\u0000eda's room starts with 30.5m230.5\,\text{m}^2 of paper.\newlineLet tt represent the time, in hours, since Gilberta and Mar\u0000eda start removing the wallpaper.\newlineComplete the inequality to represent the times when Gilberta has more wallpaper left in her room than Mar\u0000eda has in hers.\newlinetselect inequality symbolhourst \,\text{select inequality symbol}\, \text{hours}

Full solution

Q. After finishing the prep work, Gilberta and Mar\u0000eda start removing wallpaper at the same time. Gilberta removes the paper at a constant rate of 4.3m24.3\,\text{m}^2 per hour, while Mar\u0000eda removes 3.4m23.4\,\text{m}^2 of paper per hour. Gilberta's room starts with 35m235\,\text{m}^2 of paper, and Mar\u0000eda's room starts with 30.5m230.5\,\text{m}^2 of paper.\newlineLet tt represent the time, in hours, since Gilberta and Mar\u0000eda start removing the wallpaper.\newlineComplete the inequality to represent the times when Gilberta has more wallpaper left in her room than Mar\u0000eda has in hers.\newlinetselect inequality symbolhourst \,\text{select inequality symbol}\, \text{hours}
  1. Set up equations: Let's set up the equations for the amount of wallpaper left in each room after tt hours. For Gilberta, the amount of wallpaper left is given by the initial amount minus the rate at which she removes it times time:\newlineGilberta's wallpaper left = 35m24.3m2/h×t35m^2 - 4.3m^2/h \times t.\newlineFor Mar\'\ia, the amount of wallpaper left is given by the initial amount minus the rate at which she removes it times time:\newlineMar\'\ia's wallpaper left = 30.5m23.4m2/h×t30.5m^2 - 3.4m^2/h \times t.
  2. Find inequality: We want to find when Gilberta has more wallpaper left than Mar\'{i}a. This means we need to find when the amount of wallpaper left in Gilberta's room is greater than the amount left in Mar\'{i}a's room: 35m^2 - 4.3\frac{m^2}{h} \cdot t > 30.5m^2 - 3.4\frac{m^2}{h} \cdot t.
  3. Simplify inequality: Now, we simplify the inequality by subtracting 30.5m230.5m^2 from both sides and adding 4.3m2ht\frac{4.3m^2}{h} \cdot t to both sides:\newline35m^2 - 30.5m^2 > \frac{4.3m^2}{h} \cdot t - \frac{3.4m^2}{h} \cdot t.
  4. Perform calculations: Perform the calculations on both sides of the inequality: 4.5m^2 > 0.9\frac{m^2}{h} \cdot t.
  5. Divide both sides: To find the inequality in terms of tt, we divide both sides by 0.9m2/h0.9m^2/h:t < \frac{4.5m^2}{0.9m^2/h}.
  6. Calculate value: Calculate the division to find the value for tt:t < 5 hours. This means that Gilberta will have more wallpaper left in her room than Mar\'{\i}a for any time less than 55 hours.

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