Subjects algebra

Solve Rational 77E83B

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1. **Stating the problem:** We are given the equation $$\frac{2x+3}{x-1} = 4$$ and need to solve for $x$. 2. **Formula and rules:** To solve rational equations, multiply both sides by the denominator to eliminate the fraction, but remember to check for values that make the denominator zero (excluded values). 3. **Step 1: Identify excluded values:** The denominator $x-1$ cannot be zero, so $x \neq 1$. 4. **Step 2: Multiply both sides by $x-1$ to clear the fraction:** $$\cancel{\frac{2x+3}{x-1}} \times (x-1) = 4 \times (x-1)$$ which simplifies to $$2x + 3 = 4(x - 1)$$ 5. **Step 3: Expand the right side:** $$2x + 3 = 4x - 4$$ 6. **Step 4: Rearrange terms to isolate $x$:** $$2x + 3 - 4x = -4$$ $$\cancel{2x} + 3 - \cancel{4x} = -4$$ which simplifies to $$-2x + 3 = -4$$ 7. **Step 5: Subtract 3 from both sides:** $$-2x + 3 - 3 = -4 - 3$$ $$-2x = -7$$ 8. **Step 6: Divide both sides by $-2$ to solve for $x$:** $$x = \frac{-7}{-2} = \frac{7}{2}$$ 9. **Step 7: Check the solution against excluded values:** $x = \frac{7}{2} = 3.5$ is not equal to 1, so it is valid. **Final answer:** $$x = \frac{7}{2}$$