Solve the following system of linear equations: 2a + 3b + 4c = 11, a + 5b + 7c = 15, a + 11b + 13 = 25

Here's the solution to the system of equations. The given system of equations is: 1) 2a + 3b + 4c = 11 2) a + 5b + 7c = 15 3) a + 11b + 13 = 25 Step 1: Simplify Equation (3). Subtract 13 from both sides of Equation (3): a + 11b = 25 - 13 a + 11b = 12 (Equation 3 simplified) Step 2: Express a in terms of b from the simplified Equation (3). a = 12 - 11b (Equation 4) Step 3: Substitute Equation (4) into Equation (1) and Equation (2) to eliminate a. Substitute a = 12 - 11b into Equation (1): 2(12 - 11b) + 3b + 4c = 11 24 - 22b + 3b + 4c = 11 -19b + 4c = 11 - 24 -19b + 4c = -13 (Equation 5) Substitute a = 12 - 11b into Equation (2): (12 - 11b) + 5b + 7c = 15 12 - 6b + 7c = 15 -6b + 7c = 15 - 12 -6b + 7c = 3 (Equation 6) Step 4: Solve the system of two equations (Equation 5 and Equation 6) for b and c. We have: 5) -19b + 4c = -13 6) -6b + 7c = 3 Multiply Equation (5) by 7 and Equation (6) by 4 to eliminate c: 7(-19b + 4c) = 7(-13) -133b + 28c = -91 (Equation 7) 4(-6b + 7c) = 4(3) -24b + 28c = 12 (Equation 8) Subtract Equation (8) from Equation (7): (-133b + 28c) - (-24b + 28c) = -91 - 12 -133b + 24b = -103 -109b = -103 b = (-103)/(-109) b = (103)/(109) Step 5: Substitute the value of b into Equation (6) to find c. -6((103)/(109)) + 7c = 3 -(618)/(109) + 7c = 3 7c = 3 + (618)/(109) 7c = (3 × 109)/(109) + (618)/(109) 7c = (327 + 618)/(109) 7c = (945)/(109) c = (945)/(109 × 7) c = (135)/(109) Step 6: Substitute the value of b into Equation (4) to find a. a = 12 - 11b a = 12 - 11((103)/(109)) a = 12 - (1133)/(109) a = (12 × 109)/(109) - (1133)/(109) a = (1308 - 1133)/(109) a = (175)/(109) The solution to the system of equations is: a = (175)/(109) b = (103)/(109) c = (135)/(109) Last free one today — make it count tomorrow, or type /upgrade for unlimited.