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Sagot :
To solve the equation [tex]\( y + \frac{y^2 - 5}{y^2 - 1} = \frac{y^2 + y + 2}{y + 1} \)[/tex], we need to clear the denominators by multiplying both sides of the equation by the least common denominator (LCD).
1. Identify the denominators present in the equation:
- [tex]\( y^2 - 1 \)[/tex]
- [tex]\( y + 1 \)[/tex]
2. Notice that [tex]\( y^2 - 1 \)[/tex] can be factored:
[tex]\[ y^2 - 1 = (y + 1)(y - 1) \][/tex]
3. Now the denominators are:
- [tex]\( (y + 1)(y - 1) \)[/tex]
- [tex]\( y + 1 \)[/tex]
4. Observe that the LCD must include all factors present in the denominators:
- Since [tex]\( (y + 1) \)[/tex] is a factor of [tex]\( y^2 - 1 \)[/tex], the LCD is [tex]\( y^2 - 1 \)[/tex].
Therefore, Janet should multiply both sides of the equation by [tex]\( y^2 - 1 \)[/tex] to clear the fractions. Thus, the answer is:
[tex]\[ y^2 - 1 \][/tex]
1. Identify the denominators present in the equation:
- [tex]\( y^2 - 1 \)[/tex]
- [tex]\( y + 1 \)[/tex]
2. Notice that [tex]\( y^2 - 1 \)[/tex] can be factored:
[tex]\[ y^2 - 1 = (y + 1)(y - 1) \][/tex]
3. Now the denominators are:
- [tex]\( (y + 1)(y - 1) \)[/tex]
- [tex]\( y + 1 \)[/tex]
4. Observe that the LCD must include all factors present in the denominators:
- Since [tex]\( (y + 1) \)[/tex] is a factor of [tex]\( y^2 - 1 \)[/tex], the LCD is [tex]\( y^2 - 1 \)[/tex].
Therefore, Janet should multiply both sides of the equation by [tex]\( y^2 - 1 \)[/tex] to clear the fractions. Thus, the answer is:
[tex]\[ y^2 - 1 \][/tex]
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