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Solve the system of equations:
x+3y-z=9
2x+9y+4z=12
x+4y+z=7

Sagot :

LammettHash

[i] … … … x + 3y - z = 9

[ii] … … … 2x + 9y + 4z = 12

[iii] … … … x + 4y + z = 7

Eliminate z by combining …

• … 4 times equation [i] and equation [ii] :

4 (x + 3y - z) + (2x + 9y + 4z) = 4•9 + 12

(4x + 12y - 4z) + (2x + 9y + 4z) = 36 + 12

6x + 21y = 48

2x + 7y = 16

• … equation [i] and equation [iii] :

(x + 3y - z) + (x + 4y + z) = 9 + 7

2x + 7y = 16

Since we ended up with 2 copies of the same equation, we have infinitely many solutions for x, y, and z. That is, we have infinitely many choices for x and y that satisfy 2x + 7y = 16, and consequently infinitely many choices for z to satisfy any of the 3 original equations.

We can parameterize the solution by letting, for instance, x = t; then the solution set is

x = t

2x + 7y = 16   ⇒   y = (16 - 2t)/7

x + 3y - z = 9   ⇒   z = t + 3 (16 - 2t)/7 - 9   ⇒   z = (t - 15)/7

where t is any real number.

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