Solve the following system using elimination: The preceding { should cover all 3 lines, but it doesn't work in this interface. i.e it should be one big bracket covering 3 lines. I omitted them on the next steps and result. see attached Answer for clarification.{3 x + 2 y + 4 z = 7 | (equation 1){4 x + 3 y + 5 z = 9 | (equation 2){7 x + 4 y + 7 z = 5 | (equation 3)
Swap equation 1 with equation 3:{7 x + 4 y + 7 z = 5 | (equation 1)4 x + 3 y + 5 z = 9 | (equation 2)3 x + 2 y + 4 z = 7 | (equation 3)
Subtract 4/7 × (equation 1) from equation 2:{7 x + 4 y + 7 z = 5 | (equation 1)0 x+(5 y)/7 + z = 43/7 | (equation 2)3 x + 2 y + 4 z = 7 | (equation 3)
Multiply equation 2 by 7:{7 x + 4 y + 7 z = 5 | (equation 1)0 x+5 y + 7 z = 43 | (equation 2)3 x + 2 y + 4 z = 7 | (equation 3)
Subtract 3/7 × (equation 1) from equation 3:{7 x + 4 y + 7 z = 5 | (equation 1)0 x+5 y + 7 z = 43 | (equation 2)0 x+(2 y)/7 + z = 34/7 | (equation 3)
Multiply equation 3 by 7:{7 x + 4 y + 7 z = 5 | (equation 1)0 x+5 y + 7 z = 43 | (equation 2)0 x+2 y + 7 z = 34 | (equation 3)
Subtract 2/5 × (equation 2) from equation 3:{7 x + 4 y + 7 z = 5 | (equation 1)0 x+5 y + 7 z = 43 | (equation 2)0 x+0 y+(21 z)/5 = 84/5 | (equation 3)
Multiply equation 3 by 5/21:{7 x + 4 y + 7 z = 5 | (equation 1)0 x+5 y + 7 z = 43 | (equation 2)0 x+0 y+z = 4 | (equation 3)
Subtract 7 × (equation 3) from equation 2:{7 x + 4 y + 7 z = 5 | (equation 1)0 x+5 y+0 z = 15 | (equation 2)0 x+0 y+z = 4 | (equation 3)
Divide equation 2 by 5:{7 x + 4 y + 7 z = 5 | (equation 1)0 x+y+0 z = 3 | (equation 2)0 x+0 y+z = 4 | (equation 3)
Subtract 4 × (equation 2) from equation 1:{7 x + 0 y+7 z = -7 | (equation 1)0 x+y+0 z = 3 | (equation 2)0 x+0 y+z = 4 | (equation 3)
Subtract 7 × (equation 3) from equation 1:{7 x+0 y+0 z = -35 | (equation 1)0 x+y+0 z = 3 | (equation 2)0 x+0 y+z = 4 | (equation 3)
Divide equation 1 by 7:{x+0 y+0 z = -5 | (equation 1)0 x+y+0 z = 3 | (equation 2)0 x+0 y+z = 4 | (equation 3)
Collect results: {x = -5
y = 3
z = 4
