Solve the system using either Gaussian elimination with back-substitution or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTION. If there are an infinite number of solutions, set x3 and solve for x1 and x2. x1-3x3 =-5 3x1 X2 2x34 2x1 + 2x2 + x3 = 7 Need Help? Read It Talk to a Tutor Submit Answer Save Progress Practice Another Version 1 points LarLinAlg8 1.2.033 Solve the system using either Gaussian elimination with back-substitution or Gauss-Jordan elimination. (If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, express x, x2, and x3 in terms of the parameter t.) My Notes Ask Your Teach 2x1+ 4x1-3x2 + 7x3 = 2 8x1 - 9x2 15x3 12 3x33 (x1, x2, x3) -
Accepted Solution
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Answer:x1=-4/5, x2=18/5 and x3=7/5Step-by-step explanation:[tex]\left[\begin{array}{ccc|c}1&0&-3&-5\\3&1&2&4\\2&2&1&7\end{array}\right][/tex]you can do linear combination between the rows:2nd row=R2-3R1 and 3th row=R3-2R1[tex]\left[\begin{array}{ccc|c}1&0&-3&-5\\0&1&11&19\\0&2&7&17\end{array}\right][/tex]3th row=(3R2-R3)/15[tex]\left[\begin{array}{ccc|c}1&0&-3&-5\\0&1&11&19\\0&0&1&\frac{7}{5} \end{array}\right][/tex]1st row=R1+3R3 and R2-11R3[tex]\left[\begin{array}{ccc|c}1&0&0&-4/5 \\0&1&0&18/5 \\0&0&1&7/5\end{array}\right][/tex]x1=-4/5, x2=18/5 and x3=7/5