Solve the system of linear equations and check any solutions algebraically. (If the the system is dependent, express x,y, and z in terms of the parameter a.) {[x-3y+2z=18],[5x-13 y+12 z=100]:}

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Solve the system of linear equations and check any solutions algebraically. (If the the system is dependent, express  x,y, and  z in terms of the parameter a.)  {[x-3y+2z=18],[5x-13 y+12 z=100]:}

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Write Equations: Write down the system of equations. We have the following system of equations: 1) x - 3y + 2z = 18 2) 5x - 13y + 12z = 100Multiply and Prepare: Multiply the first equation by 5 to prepare for elimination with the second equation. 5(x - 3y + 2z) = 5(18) This gives us: 5x - 15y + 10z = 90Eliminate x: Subtract the new equation from the second equation in the system to eliminate x. (5x - 13y + 12z) - (5x - 15y + 10z) = 100 - 90 This simplifies to: 2y + 2z = 10Simplify Equation: Simplify the equation from Step 3. Divide the entire equation by 2 to simplify: y + z = 5Express y in terms of z: Express y in terms of z. Let's use z as our parameter (we can call it a). y = 5 - aSubstitute y into 1st equation: Substitute y = 5 - a into the first original equation. x - 3(5 - a) + 2a = 18 This simplifies to: x - 15 + 3a + 2a = 18Solve for x: Solve for x in terms of a. x + 5a = 18 + 15 x + 5a = 33 x = 33 - 5aFinal Solution: Write the final solution in terms of the parameter a. We have: x = 33 - 5a y = 5 - a z = aCheck Solution: Check the solution by substituting x, y, and z back into the original equations. Substitute into the first equation: (33 - 5a) - 3(5 - a) + 2a = 18 33 - 5a - 15 + 3a + 2a = 18 33 - 15 = 18 18 = 18 (True) Substitute into the second equation: 5(33 - 5a) - 13(5 - a) + 12a = 100 165 - 25a - 65 + 13a + 12a = 100 165 - 65 = 100 (True)

Solve the system of linear equations and check any solutions algebraically. (If the the system is dependent, express $x, y$ , and $z$ in terms of the parameter a.) $\newline$ $\left\{\begin{array}{r} x-3 y+2 z=18 \\ 5 x-13 y+12 z=100 \end{array}\right.$

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