Step 1: Find the common denominator The denominator is $x(2x + 3)$. Rewrite: \[ \frac{4(2x + 3) - 5x}{x(2x + 3)} = 3. \] Step 2: Simplify the numerator \[ \frac{8x + 12 - 5x}{x(2x + 3)} = 3 \implies \frac{3x + 12}{x(2x + 3)} = 3. \] Step 3: Cross-multiply \[ 3x + 12 = 3x(2x + 3) \implies 3x + 12 = 6x^2 + 9x. \] Rearrange: \[ 6x^2 + 6x - 12 = 0. \] Simplify: \[ x^2 + x - 2 = 0. \] Step 4: Solve the quadratic equation Factorize: \[ x^2 + x - 2 = (x + 2)(x - 1) = 0. \] \[ x = -2 \quad \text{or} \quad x = 1. \] Step 5: Check solutions For $x = -2$, $2x + 3 = -1$, so the denominator is undefined. Therefore, $x = -2$ is not valid. Correct Answer: $x = 1$.
A shopkeeper sells an item at a 20 % discount on the marked price and still makes a 25 % profit. If the marked price is 500 rupees, what is the cost price of the item?
In the adjoining figure, \( AP = 1 \, \text{cm}, \ BP = 2 \, \text{cm}, \ AQ = 1.5 \, \text{cm}, \ AC = 4.5 \, \text{cm} \) Prove that \( \triangle APQ \sim \triangle ABC \).
Hence, find the length of \( PQ \), if \( BC = 3.6 \, \text{cm} \).
In the adjoining figure, $\triangle CAB$ is a right triangle, right angled at A and $AD \perp BC$. Prove that $\triangle ADB \sim \triangle CDA$. Further, if $BC = 10$ cm and $CD = 2$ cm, find the length of AD.