1. **State the problem:** Solve the system of equations using Gauss Elimination Method: $$\begin{cases} 4x - 5y + 6z = 12 \\ 7x + 3y - 2z = -5 \\ 5x - 4y + 8z = 10 \end{cases}$$ 2. **Write the augmented matrix:** $$\left[\begin{array}{ccc|c} 4 & -5 & 6 & 12 \\ 7 & 3 & -2 & -5 \\ 5 & -4 & 8 & 10 \end{array}\right]$$ 3. **Eliminate $x$ from rows 2 and 3:** - Multiply row 1 by $\frac{7}{4}$ and subtract from row 2: $$R_2 \to R_2 - \frac{7}{4}R_1$$ $$\left[\begin{array}{ccc|c} 4 & -5 & 6 & 12 \\ 7 & 3 & -2 & -5 \\ 5 & -4 & 8 & 10 \end{array}\right] \to \left[\begin{array}{ccc|c} 4 & -5 & 6 & 12 \\ 0 & \cancel{3} - \cancel{\frac{35}{4}} & -2 - \frac{42}{4} & -5 - 21 \\ 5 & -4 & 8 & 10 \end{array}\right]$$ Simplify row 2: $$0, 3 - \frac{35}{4} = \frac{12}{4} - \frac{35}{4} = -\frac{23}{4}, \quad -2 - \frac{42}{4} = -2 - 10.5 = -12.5, \quad -5 - 21 = -26$$ So row 2 becomes: $$\left[0, -\frac{23}{4}, -\frac{25}{2}, -26\right]$$ - Multiply row 1 by $\frac{5}{4}$ and subtract from row 3: $$R_3 \to R_3 - \frac{5}{4}R_1$$ $$\left[5, -4, 8, 10\right] - \frac{5}{4} \times \left[4, -5, 6, 12\right] = \left[5 - 5, -4 + \frac{25}{4}, 8 - \frac{30}{4}, 10 - 15\right]$$ Simplify: $$0, -4 + 6.25 = 2.25, 8 - 7.5 = 0.5, -5$$ Row 3 becomes: $$\left[0, \frac{9}{4}, \frac{1}{2}, -5\right]$$ 4. **New matrix:** $$\left[\begin{array}{ccc|c} 4 & -5 & 6 & 12 \\ 0 & -\frac{23}{4} & -\frac{25}{2} & -26 \\ 0 & \frac{9}{4} & \frac{1}{2} & -5 \end{array}\right]$$ 5. **Eliminate $y$ from row 3:** Multiply row 2 by $\frac{9}{4}$ and row 3 by $-\frac{23}{4}$ to align coefficients: $$R_3 \to R_3 + \frac{9}{4} \times \frac{4}{23} R_2$$ Or more simply: Multiply row 2 by 9 and row 3 by 23: $$R_2: 0, -\frac{23}{4} \times 9 = -\frac{207}{4}, -\frac{25}{2} \times 9 = -\frac{225}{2}, -26 \times 9 = -234$$ $$R_3: 0, \frac{9}{4} \times 23 = \frac{207}{4}, \frac{1}{2} \times 23 = \frac{23}{2}, -5 \times 23 = -115$$ Add $R_2$ and $R_3$: $$0, 0, -\frac{225}{2} + \frac{23}{2} = -\frac{202}{2} = -101, -234 - 115 = -349$$ Row 3 becomes: $$\left[0, 0, -101, -349\right]$$ 6. **Back substitution:** - From row 3: $$-101z = -349 \implies z = \frac{-349}{-101} = \frac{349}{101}$$ - From row 2: $$-\frac{23}{4}y - \frac{25}{2}z = -26$$ Substitute $z$: $$-\frac{23}{4}y - \frac{25}{2} \times \frac{349}{101} = -26$$ Calculate: $$-\frac{23}{4}y = -26 + \frac{25}{2} \times \frac{349}{101} = -26 + \frac{25 \times 349}{2 \times 101} = -26 + \frac{8725}{202}$$ Convert $-26$ to fraction with denominator 202: $$-26 = -\frac{5252}{202}$$ Sum: $$-\frac{5252}{202} + \frac{8725}{202} = \frac{3473}{202}$$ So: $$-\frac{23}{4}y = \frac{3473}{202} \implies y = \frac{3473}{202} \times \frac{-4}{23} = -\frac{13892}{4646} = -\frac{6946}{2323}$$ - From row 1: $$4x - 5y + 6z = 12$$ Substitute $y$ and $z$: $$4x - 5 \times \left(-\frac{6946}{2323}\right) + 6 \times \frac{349}{101} = 12$$ Calculate: $$4x + \frac{34730}{2323} + \frac{2094}{101} = 12$$ Convert 12 to common denominator (2323*101=234,623): $$12 = \frac{12 \times 234,623}{234,623} = \frac{2,815,476}{234,623}$$ Convert fractions: $$\frac{34730}{2323} = \frac{34730 \times 101}{234,623} = \frac{3,510,730}{234,623}$$ $$\frac{2094}{101} = \frac{2094 \times 2323}{234,623} = \frac{4,864,962}{234,623}$$ Sum fractions: $$\frac{3,510,730 + 4,864,962}{234,623} = \frac{8,375,692}{234,623}$$ Equation: $$4x + \frac{8,375,692}{234,623} = \frac{2,815,476}{234,623}$$ Subtract: $$4x = \frac{2,815,476}{234,623} - \frac{8,375,692}{234,623} = -\frac{5,560,216}{234,623}$$ Divide by 4: $$x = -\frac{5,560,216}{4 \times 234,623} = -\frac{5,560,216}{938,492} = -\frac{1,390,054}{234,623}$$ 7. **Final solution:** $$x = -\frac{1,390,054}{234,623}, \quad y = -\frac{6946}{2323}, \quad z = \frac{349}{101}$$ This completes the Gauss Elimination solution for the first system.