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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.
For example, in the given exercise, we used elimination to remove the variable \(z\). This allowed us to simplify the system, eventually making it easier to solve for \(x\) and \(y\). The process can significantly cut down on the complexity of multi-variable problems.
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).
In our exercise, after isolating \(z\) in one equation, we substituted this expression into the remaining equations. This turns a system of equations into a single-variable equation, simplifying the process.
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:
  • 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.
In the given system, we had to solve for each variable (\(x, y, \) and \(z\)) step by step, ensuring we isolated the variable efficiently using elimination and substitution methods.
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.
In our exercise, we used algebraic manipulation several times: clearing fractions, combining like terms, and rearranging equations. This sequential manipulation ensures we can solve for one variable at a time, leading to the final solution. By understanding these basic principles, you can tackle and simplify even complex systems of equations.

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Most popular questions from this chapter

Three retired couples each require an additional annual income of \(\$ 2000\) per year. As their financial consultant, you recommend that they invest some money in Treasury bills that yield \(7 \%\), some money in corporate bonds that yield \(9 \%,\) and some money in "junk bonds" that yield \(11 \%\). Prepare a table for each couple showing the various ways that their goals can be achieved: (a) If the first couple has \(\$ 20,000\) to invest. (b) If the second couple has \(\$ 25,000\) to invest. (c) If the third couple has \(\$ 30,000\) to invest. (d) What advice would you give each couple regarding the amount to invest and the choices available?

Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. What is the amount that results if \(\$ 2700\) is invested at \(3.6 \%\) compounded monthly for 3 years?

Solve each system of equations. If the system has no solution, state that it is inconsistent. $$ \left\\{\begin{aligned} 2 x-y &=-1 \\ x+\frac{1}{2} y &=\frac{3}{2} \end{aligned}\right. $$

IS-LM Model in Economics In economics, the IS curve is a linear equation that represents all combinations of income \(Y\) and interest rates \(r\) that maintain an equilibrium in the market for goods in the economy. The LM curve is a linear equation that represents all combinations of income \(Y\) and interest rates \(r\) that maintain an equilibrium in the market for money in the economy. In an economy, suppose that the equilibrium level of income (in millions of dollars) and interest rates satisfy the system of equations $$ \left\\{\begin{array}{l} 0.05 Y-1000 r=10 \\ 0.05 Y+800 r=100 \end{array}\right. $$ Find the equilibrium level of income and interest rates.

Solve each system of equations. If the system has no solution, state that it is inconsistent. $$ \left\\{\begin{array}{r} x-y+z=-4 \\ 2 x-3 y+4 z=-15 \\ 5 x+y-2 z=12 \end{array}\right. $$

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