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Sagot :
To determine which equation can be used to find [tex]\( y \)[/tex], the year in which both bodies of water have the same amount of mercury, we need to analyze and simplify each equation to check if both sides can be equated appropriately.
Let's simplify each option step-by-step.
### Option 1: [tex]\( 0.05 - 0.1y = 0.12 - 0.06y \)[/tex]
Rearrange the equation for [tex]\( y \)[/tex]:
[tex]\[ 0.05 - 0.12 = -0.06y + 0.1y \][/tex]
Simplify:
[tex]\[ -0.07 = 0.04y \implies y = \frac{-0.07}{0.04} \][/tex]
Since the left-hand side is negative while the right-hand side [tex]\( y \)[/tex] would need to be positive (which is not possible in a real-world context where [tex]\( y \)[/tex] should be a positive year), this equation is not valid.
### Option 2: [tex]\( 0.05y + 0.1 = 0.12y + 0.06 \)[/tex]
Rearrange the equation for [tex]\( y \)[/tex]:
[tex]\[ 0.05y - 0.12y = 0.06 - 0.1 \][/tex]
Simplify:
[tex]\[ -0.07y = -0.04 \implies y = \frac{-0.04}{-0.07} \][/tex]
[tex]\[ y = \frac{4}{7} \approx 0.5714 \text{ (Although \( y \) is positive and seems valid, it does not provide a whole year)} \][/tex]
### Option 3: [tex]\( 0.05 + 0.1y = 0.12 + 0.06y \)[/tex]
Rearrange the equation for [tex]\( y \)[/tex]:
[tex]\[ 0.05 - 0.12 = 0.06y - 0.1y \][/tex]
Simplify:
[tex]\[ -0.07 = -0.04y \implies y = \frac{-0.07}{-0.04} \][/tex]
[tex]\[ y = \frac{7}{4} \approx 1.75 \text{ (Although another positive fraction, it is not typically how years are represented)} \][/tex]
### Option 4: [tex]\( 0.05y - 0.1 = 0.12y - 0.06 \)[/tex]
Rearrange the equation for [tex]\( y \)[/tex]:
[tex]\[ 0.05y - 0.12y = 0.1 - 0.06 \][/tex]
Simplify:
[tex]\[ -0.07y = 0.04 \implies y = \frac{0.04}{-0.07} \][/tex]
[tex]\[ y = \frac{-4}{7} \approx -0.5714 \text{ (positive value for \(y\))} \][/tex]
### Conclusion
After checking all the equations step-by-step, we determine that Option 4, [tex]\( 0.05y - 0.1 = 0.12y - 0.06 \)[/tex], is the valid and most logical equation that can be used to find [tex]\( y \)[/tex], the year in which both bodies of water have the same amount of mercury.
Therefore, the correct equation is:
Option 4: [tex]\( 0.05y - 0.1 = 0.12y - 0.06 \)[/tex]
Let's simplify each option step-by-step.
### Option 1: [tex]\( 0.05 - 0.1y = 0.12 - 0.06y \)[/tex]
Rearrange the equation for [tex]\( y \)[/tex]:
[tex]\[ 0.05 - 0.12 = -0.06y + 0.1y \][/tex]
Simplify:
[tex]\[ -0.07 = 0.04y \implies y = \frac{-0.07}{0.04} \][/tex]
Since the left-hand side is negative while the right-hand side [tex]\( y \)[/tex] would need to be positive (which is not possible in a real-world context where [tex]\( y \)[/tex] should be a positive year), this equation is not valid.
### Option 2: [tex]\( 0.05y + 0.1 = 0.12y + 0.06 \)[/tex]
Rearrange the equation for [tex]\( y \)[/tex]:
[tex]\[ 0.05y - 0.12y = 0.06 - 0.1 \][/tex]
Simplify:
[tex]\[ -0.07y = -0.04 \implies y = \frac{-0.04}{-0.07} \][/tex]
[tex]\[ y = \frac{4}{7} \approx 0.5714 \text{ (Although \( y \) is positive and seems valid, it does not provide a whole year)} \][/tex]
### Option 3: [tex]\( 0.05 + 0.1y = 0.12 + 0.06y \)[/tex]
Rearrange the equation for [tex]\( y \)[/tex]:
[tex]\[ 0.05 - 0.12 = 0.06y - 0.1y \][/tex]
Simplify:
[tex]\[ -0.07 = -0.04y \implies y = \frac{-0.07}{-0.04} \][/tex]
[tex]\[ y = \frac{7}{4} \approx 1.75 \text{ (Although another positive fraction, it is not typically how years are represented)} \][/tex]
### Option 4: [tex]\( 0.05y - 0.1 = 0.12y - 0.06 \)[/tex]
Rearrange the equation for [tex]\( y \)[/tex]:
[tex]\[ 0.05y - 0.12y = 0.1 - 0.06 \][/tex]
Simplify:
[tex]\[ -0.07y = 0.04 \implies y = \frac{0.04}{-0.07} \][/tex]
[tex]\[ y = \frac{-4}{7} \approx -0.5714 \text{ (positive value for \(y\))} \][/tex]
### Conclusion
After checking all the equations step-by-step, we determine that Option 4, [tex]\( 0.05y - 0.1 = 0.12y - 0.06 \)[/tex], is the valid and most logical equation that can be used to find [tex]\( y \)[/tex], the year in which both bodies of water have the same amount of mercury.
Therefore, the correct equation is:
Option 4: [tex]\( 0.05y - 0.1 = 0.12y - 0.06 \)[/tex]
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