1. Solve the inequality: \( \frac{y+3}{5} \geq \frac{y+2}{3}+2 \)
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To solve the inequality \( \frac{y+3}{5} \geq \frac{y+2}{3}+2 \), start by eliminating the fractions. To do this, find a common denominator, which is 15 in this case. Multiply each term by 15: \[ 15 \cdot \frac{y+3}{5} \geq 15 \cdot \left(\frac{y+2}{3}+2\right) \] This simplifies to: \[ 3(y+3) \geq 5(y+2) + 30 \] Now, distribute: \[ 3y + 9 \geq 5y + 10 + 30 \] Combine like terms on the right side: \[ 3y + 9 \geq 5y + 40 \] Next, isolate \( y \) by moving all terms involving \( y \) to one side and constants to the other: \[ 3y - 5y \geq 40 - 9 \\ -2y \geq 31 \] Now, divide by -2, remembering to reverse the inequality sign: \[ y \leq -\frac{31}{2} \] So the solution to the inequality is: \[ y \leq -15.5 \] Happy solving!