3. $10x + 7y - 2z = 46$\n$3x - 2y + 9z = 22$\n$5x + y - 3z = 28$

Explanation:

Step1: Eliminate (y)

Multiply the first equation (5x + y-3z = 28) by (2) to get (10x+2y - 6z=56). Add it to the second equation (3x - 2y+9z = 22): $$(10x + 2y-6z)+(3x - 2y + 9z)=56 + 22$$ $$13x+3z = 78$$

Step2: Another elimination of (y)

We have the third equation (10x+7y-2z = 46) and the first equation (5x + y-3z = 28) (rewrite as (y=28 - 5x + 3z)). Substitute (y = 28-5x + 3z) into (10x+7y-2z = 46): $$10x+7(28 - 5x + 3z)-2z=46$$ $$10x+196-35x + 21z-2z=46$$ $$- 25x+19z=-150$$

Step3: Solve the system of two - variable equations

From (13x+3z = 78), we get (z = 26-\frac{13x}{3}). Substitute (z = 26-\frac{13x}{3}) into (-25x + 19z=-150): $$-25x+19(26-\frac{13x}{3})=-150$$ $$-25x+\frac{494}{1}-\frac{247x}{3}=-150$$ Multiply through by (3) to clear the fraction: $$-75x + 1482-247x=-450$$ $$-322x=-1932$$ $$x = 6$$

Step4: Find (z)

Substitute (x = 6) into (13x+3z = 78): $$13\times6+3z=78$$ $$78+3z=78$$ $$3z=0$$ $$z = 0$$

Step5: Find (y)

Substitute (x = 6) and (z = 0) into (5x + y-3z = 28): $$5\times6+y-3\times0=28$$ $$30 + y=28$$ $$y=-2$$

Answer:

(x = 6,y=-2,z = 0)