Elimination x2y=2x5, xy=3 \square!
X+y/2=4 x/3+2y=5 by elimination method-Divide both sides by 5 z = 4 Substitute z = 4 in (2)6y 5(4) = 46y = 4 Subtract from both sides6y = 24 Divide both sides by 6 y = 4 Substitute y = 4 and z = 4 in (1) x 2(4) 4 = 3 x 8 4 = 3 x 4 = 3 x = 1 Therefore the solution of the system is x = 1, y = 4 and z = 4 Example 2 2x 4y 6z = 22 3x 8y 5z = 27x y 2z = 2 SolutionStepbystep explanation We will multiply the equation x y = 5 by 3 so that the leading coefficient of x in both equation will be equal Subtract first equation to the second equation Divide both sides by 5 to get the value of y Substitute the value of y to equation x y = 5 Let's check the equation and value of x and y
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`x/3 2y = 5` (ii) From (i), we get `(2x y)/2 = 4` 2x y = 8 y = 8 2x From (ii), we get x 6y = 15 (iii) Substituting y = 8 2x in (iii), we get x 6(8 2x) = 15 `=> x 48 12x = 15` => 11x = 15 48 => 11x = 33 `=> x = (33)/(11) = 3` Putting x = 3 in y = 8 2x we get y = 8 2 x 3 = 8 6 = 2 y = 2 Hence, solution of the given system of equation is x= 3, y = 2Solve the system shown below using the Gauss Jordan Elimination method x 2 y = 4 x – 2 y = 6 Solution Let's write the augmented matrix of the system of equations 1 2 4 1 – 2 6 Now, we do the elementary row operations on this matrix until we arrive in the reduced row echelon form Step 1
Incoming Term: x+y/2=4 x/3+2y=5 by elimination method,
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