Linear equation in two variables grade IX -Worksheet 2 with solutions

Self Assessment Test 5

Based on Chapter 5

Time: 1 hour

M.M. 25

Note: Q.1-2 carry 1 mark each, Q.3-5 carry 2 marks each, Q.6-8 carry 3 marks each and Q.9-10 carry 4 marks each.

1. Question: Simplify \(3a^2b \times 2ab^2\).

   Solution: \(3a^2b \times 2ab^2 = 3 \times 2 \times a^2 \times a \times b \times b^2 = 6a^3b^3\).

2. Question: Factorize \(x^2 - 6x + 9\).

   Solution: \(x^2 - 6x + 9 = (x-3)(x-3) = (x-3)^2\).

3. Question: Solve the quadratic equation \(x^2 - 5x + 6 = 0\).

   Solution: Factoring the equation, we get \((x-2)(x-3) = 0\). Therefore, \(x = 2\) or \(x = 3\).

4. Question: Simplify \(\frac{a^2 - b^2}{a^2 + b^2}\) and express in the form \(a + b\).

   Solution: \(\frac{a^2 - b^2}{a^2 + b^2} = \frac{(a+b)(a-b)}{a^2 + b^2} = \frac{a+b}{a^2/b^2 + 1} = a + b - \frac{2b^2}{a^2 + b^2}\).

5. Question: Solve for \(x\): \(3x + 4 = 2x - 5\).

   Solution: Subtracting \(2x\) from both sides, we get \(x + 4 = -5\). Subtracting 4 from both sides, we get \(x = -9\).

6. Question: If \(f(x) = 3x^2 - 4x + 5\), find \(f(2)\).

   Solution: Substituting \(x = 2\) into \(f(x)\), we get \(f(2) = 3(2)^2 - 4(2) + 5 = 3(4) - 8 + 5 = 12 - 8 + 5 = 9\).

7. Question: Find the zeros of the quadratic function \(f(x) = x^2 - 4\).

   Solution: The zeros are the x-values that make \(f(x) = 0\), so we solve the equation \(x^2 - 4 = 0\). Factoring, we get \((x-2)(x+2) = 0\), so \(x = 2\) or \(x = -2\).

8. Question: Simplify \(\frac{3x^3 - 2x^2 + 4x - 3}{x - 1}\).

   Solution: Using polynomial division, we find that \(\frac{3x^3 - 2x^2 + 4x - 3}{x - 1} = 3x^2 + x + 5 - \frac{8}{x - 1}\).

9. Question: Factorize \(2x^2 - 5x - 3\).

   Solution: \(2x^2 - 5x - 3 = (2x+1)(x-3)\).

10. Question: Solve the system of equations: \(x + y = 5\) and \(x - y = 1\).

    Solution: Adding the two equations, we get \(2x = 6\), or \(x = 3\). Substituting into \(x + y = 5\), we get \(3 + y = 5\), or \(y = 2\). Therefore, the solution is \(x = 3\), \(y = 2\).