Solve the following systems of linear equations and determine if they are consistent and independent: 4(vii) 2x + 5y = 17, 5x + 3y = 14; 4(viii) 3x - 5y - 10 = 0, 6x - 10y = 20

Here are the solutions for the circled problems: 4(vii) The given system of equations is: 1) 2x + 5y = 17 2) 5x + 3y = 14 Step 1: Multiply equation (1) by 5 and equation (2) by 2 to eliminate x. 5(2x + 5y) = 5(17) 10x + 25y = 85 (3) 2(5x + 3y) = 2(14) 10x + 6y = 28 (4) Step 2: Subtract equation (4) from equation (3). (10x + 25y) - (10x + 6y) = 85 - 28 19y = 57 y = (57)/(19) y = 3 Step 3: Substitute the value of y=3 into equation (1). 2x + 5(3) = 17 2x + 15 = 17 2x = 17 - 15 2x = 2 x = 1 Step 4: The solution to the system is (x, y) = (1, 3). Step 5: To determine if the system is consistent and independent, we compare the ratios of the coefficients. For a system a_1x + b_1y = c_1 and a_2x + b_2y = c_2: Here, a_1 = 2, b_1 = 5 and a_2 = 5, b_2 = 3. (a_1)/(a_2) = (2)/(5) (b_1)/(b_2) = (5)/(3) Since (a_1)/(a_2) ≠ (b_1)/(b_2) ((2)/(5) ≠ (5)/(3)), the system has a unique solution. Therefore, the system is consistent and independent. The solution is x=1, y=3. The system is consistent and independent. 4(viii) The given system of equations is: 1) 3x - 5y - 10 = 0 3x - 5y = 10 2) 6x - 10y = 20 Step 1: To determine if the system is consistent and independent, we compare the ratios of the coefficients. For a system a_1x + b_1y = c_1 and a_2x + b_2y = c_2: Here, a_1 = 3, b_1 = -5, c_1 = 10 and a_2 = 6, b_2 = -10, c_2 = 20. Step 2: Calculate the ratios of the coefficients. (a_1)/(a_2) = (3)/(6) = (1)/(2) (b_1)/(b_2) = (-5)/(-10) = (1)/(2) (c_1)/(c_2) = (10)/(20) = (1)/(2) Step 3: Compare the ratios. Since (a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2), the system has infinitely many solutions. Step 4: A system with infinitely many solutions is consistent but dependent. It is not independent because the two equations represent the same line. Therefore, the system is consistent but not independent. The system has infinitely many solutions and is consistent but dependent. Send me the next one 📸