Expanding the terms in the numerator gives (x² - 7x + 10) + (x² - 7x + 12) = 2x² - 14x + 22. Dividing both sides' numerators by 2 gives (x² - 7x + 11)/(x² - 8x + 15) = (5)/(3).

Cross-multiplying yields 3(x² - 7x + 11) = 5(x² - 8x + 15), which expands to 3x² - 21x + 33 = 5x² - 40x + 75. Rearranging all terms to one side gives the standard form 2x² - 19x + 42 = 0.

We need two numbers whose sum is -19 and whose product is 2 × 42 = 84. These numbers are -12 and -7, so we split -19x as -12x - 7x.

23. The perimeter of an isosceles triangle is $32\text{ cm}$ . If each equal side is $\frac{5}{6}\text{th}$ of the base, find the area of the triangle. (CBSE-2025) 24. Express the equation $\frac{x-2}{x-3} + \frac{x-4}{x-5} = \frac{10}{3}$ ( $x \neq 3, 5$ ) as a quadratic equation in standard form and find the roots of the equation so formed. (CBSE-2025) 25. A 2-digit number is seven times the sum of its digits and two (2) more than 5 times the product of its digits. Find the number.

Answer: Standard form: $2x^2 - 19x + 42 = 0$ ; Roots: $x = \frac{7}{2}$ and $x = 6$

Step-by-step solution

Step 1: Combine fractions on the left-hand side

Taking the common denominator $(x-3)(x-5)$ on the left-hand side, we combine the two rational expressions by cross-multiplying each numerator by the other's denominator.

Step 2: Expand and simplify the numerator and denominator

Expanding the terms in the numerator gives $(x^2 - 7x + 10) + (x^2 - 7x + 12) = 2x^2 - 14x + 22$ . Dividing both sides' numerators by 2 gives $\frac{x^2 - 7x + 11}{x^2 - 8x + 15} = \frac{5}{3}$ .

Step 3: Cross-multiply to obtain the standard quadratic form

Cross-multiplying yields $3(x^2 - 7x + 11) = 5(x^2 - 8x + 15)$ , which expands to $3x^2 - 21x + 33 = 5x^2 - 40x + 75$ . Rearranging all terms to one side gives the standard form $2x^2 - 19x + 42 = 0$ .

Step 4: Split the middle term to factorise

We need two numbers whose sum is $-19$ and whose product is $2 \times 42 = 84$ . These numbers are $-12$ and $-7$ , so we split $-19x$ as $-12x - 7x$ .

Step 5: Factor by grouping and find the roots

Factoring out common terms gives $(2x - 7)(x - 6) = 0$ . Setting each linear factor to zero gives $2x - 7 = 0$ or $x - 6 = 0$ , which yields $x = \frac{7}{2}$ and $x = 6$ . Neither value equals the excluded values 3 or 5.

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