Chapter 12: Problem 82
Solve each system of equations using any method you wish. \(\left\\{\begin{aligned} 2 x+3 y-z &=-2 \\ 4 x+& 3 z=6 \\ 6 y-2 z &=2 \end{aligned}\right.\)
Short Answer
Expert verified
The solution is x = -3/2, y = 5/3, and z = 4.
Step by step solution
01
Write Down the Equations
List the system of equations for better visibility:1. \(2x + 3y - z = -2\)2. \(4x + 3z = 6\)3. \(6y - 2z = 2\)
02
Eliminate z from Equations 2 and 3
Use substitution or elimination to eliminate variable \(z\). Start with equation 2:\(4x + 3z = 6\).We can rearrange for \(z\):\(3z = 6 - 4x\).\(z = \frac{6 - 4x}{3}\).
03
Substitute z into Equation 1
Substitute \(z\) from Step 2 into Equation 1:\(2x + 3y - \frac{6 - 4x}{3} = -2\).Clear the fraction by multiplying through by 3:\(6x + 9y - (6 - 4x) = -6\).Simplify:\(6x + 9y - 6 + 4x = -6\).Combine like terms:\(10x + 9y - 6 = -6\).\(10x + 9y = 0\).So, equation (4) is:\(10x + 9y = 0\).
04
Substitute z into Equation 3
Use \(z = \frac{6 - 4x}{3}\) in Equation 3:\(6y - 2(\frac{6 - 4x}{3}) = 2\).Clear the fraction by multiplying through by 3:\(18y - 2(6 - 4x) = 6\).Simplify:\(18y - 12 + 8x = 6\).Combine like terms:\(18y + 8x - 12 = 6\).\(8x + 18y = 18\).Divide through by 2:\(4x + 9y = 9\).So, equation (5) is:\(4x + 9y = 9\).
05
Solve the System of Equations (Equations 4 and 5)
Now we solve for \(x\) and \(y\) from the simplified equations:\(10x + 9y = 0\) (Equation 4)\(4x + 9y = 9\) (Equation 5).Subtract Equation 4 from Equation 5 to eliminate \(y\):\((4x + 9y) - (10x + 9y) = 9 - 0\).Simplify:\(-6x = 9\).Solve for \(x\):\(x = -\frac{3}{2}\).
06
Back-Substitute \(x\) to Find \(y\)
Substitute \(x = -\frac{3}{2}\) back into either simplified equation. Here, using Equation 4:\(10(-\frac{3}{2}) + 9y = 0\).Simplify:\(-15 + 9y = 0\).Solve for \(y\):\(9y = 15\).\(y = \frac{5}{3}\).
07
Back-Substitute \(x\) and \(y\) to Find \(z\)
Use \(x = -\frac{3}{2}\) and \(y = \frac{5}{3}\) in the expression for \(z\) from Step 2:\(z = \frac{6 - 4(-\frac{3}{2})}{3}\).Simplify:\(z = \frac{6 + 6}{3}\).\(z = 4\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Elimination Method
When faced with a system of equations, the elimination method can be a powerful tool. This method aims to eliminate one of the variables by adding or subtracting equations. Here's how it works:
- Align the equations so corresponding variables are vertically lined up.
- Adjust coefficients if necessary to make the variables you'd like to eliminate match in opposite signs.
- Add or subtract equations to cancel out one variable.
Substitution Method
The substitution method involves solving one equation for one variable and then substituting that solution into another equation. The steps are straightforward:
- Choose an equation and solve it for one variable.
- Substitute this expression into the other equation(s).
- Solve for the remaining variable(s).
Solving Linear Equations
A linear equation is an equation that forms a straight line when graphed. They have variables that are to the first power and can usually be written in standard form \(Ax + By = C\). Solving these equations involves finding the values of the variables that make the equation true.
The steps to solve linear equations are:
The steps to solve linear equations are:
- Isolate the variable on one side of the equation.
- Simplify both sides of the equation if needed.
- Perform inverse operations to solve for the variable.
Algebraic Manipulation
Algebraic manipulation is the process of rearranging equations to isolate a variable or simplify an expression. It's a fundamental skill in solving equations and systems of equations. Here are key techniques:
- Combine like terms to simplify expressions.
- Use the distributive property to remove parentheses.
- Perform inverse operations to isolate variables.