For what value of c do the following equations have infinitely many solutions: cx + 3y - (c-3) = 0 and 12x + cy - c = 0?

Step 1: Identify the coefficients of the given linear equations. The general form of a linear equation is Ax + By + C = 0. The given equations are: 1) cx + 3y - (c-3) = 0 2) 12x + cy - c = 0 Comparing these to a_1x + b_1y + c_1 = 0 and a_2x + b_2y + c_2 = 0: From equation 1: a_1 = c, b_1 = 3, c_1 = -(c-3) = 3-c From equation 2: a_2 = 12, b_2 = c, c_2 = -c Step 2: Apply the condition for infinitely many solutions. For a system of linear equations to have infinitely many solutions, the ratios of their coefficients must be equal: (a_1)/(a_2) = (b_1)/(b_2) = (c_1)/(c_2) Substitute the coefficients: (c)/(12) = (3)/(c) = (3-c)/(-c) Step 3: Solve for c using the first two ratios. (c)/(12) = (3)/(c) Cross-multiply: c × c = 12 × 3 c^2 = 36 Take the square root of both sides: c = ± sqrt(36) c = ± 6 So, c can be 6 or -6. Step 4: Solve for c using the second and third ratios. (3)/(c) = (3-c)/(-c) Cross-multiply: 3(-c) = c(3-c) -3c = 3c - c^2 Rearrange the terms to form a quadratic equation: c^2 - 3c - 3c = 0 c^2 - 6c = 0 Factor out c: c(c - 6) = 0 This gives two possible values for c: c = 0 or c - 6 = 0 c = 6 Step 5: Find the value of c that satisfies all conditions. From Step 3, c = 6 or c = -6. From Step 4, c = 0 or c = 6. The value of c that is common to both sets of solutions is c=6. Let's verify with the first and third ratios as well: (c)/(12) = (3-c)/(-c) Cross-multiply: c(-c) = 12(3-c) -c^2 = 36 - 12c Rearrange the terms: c^2 - 12c + 36 = 0 This is a perfect square trinomial, (c-6)^2: (c-6)^2 = 0 c-6 = 0 c = 6 All three conditions consistently yield c=6. The final answer is c=6. Send me the next one 📸