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Verify that each system of equations has the indicated solution. $$\left\\{\begin{array}{l} 3 x-z=3 \\ 2 x+y-2 z=0 \\ 3 x-2 y+z=7 \end{array}\right.$$ \(\text { Solution: } x-1, y=-2, z=0\)

Short Answer

Expert verified
After substituting the given values into the equations, it could be observed that they hold true in all three equations. Hence \(x=1, y=-2, z=0\) is indeed a solution of the given system of equations.

Step by step solution

01

Understand the Given System of Equations

The exercise provides a system of three linear equations. \[\begin{array}{l} 3 x-z=3 \ 2 x+y-2 z=0 \ 3 x-2 y+z=7 \end{array}\] Also, a solution set is provided as \(x=1, y=-2, z=0\).
02

Substitute the Values into the Equations

Substitute \(x=1, y=-2, z=0\) into the equations. Let's start with the first equation: \(3x - z = 3\)\[3(1) - 0 = 3\] which simplifies to `3 = 3`. Similarly, substitute into the second equation: \(2x+y-2z=0\)\[2(1) - 2*(-2) - 2*0 = 0\] which simplifies to `6 = 6`, and the third equation: \(3x - 2y + z = 7\)\[3(1) -2*(-2) + 0 = 7\] that simplifies to `7 = 7`.
03

Verify the Solution

By substituting the given values into the equations, we find that they hold true in each of them. Therefore, we can confirm that \(x=1, y=-2, z=0\) is a valid solution for this system of equations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Equations
Linear equations are a foundational concept in algebra which depict relationships between two or more variables, where each variable is raised to the power of one. Imagine them as straight lines on a graph, hence the term 'linear'.

An example of a linear equation is: \(3x - z = 3\). In a system of linear equations, multiple linear equations are combined, and we seek values for the variables that satisfy all equations simultaneously. For instance, the provided exercise includes a system where three equations must intersect at a common point in three-dimensional space. This point is described by the solution set \(x=1, y=-2, z=0\).

When solving such systems, understanding the structure and components of each linear equation is crucial to identify patterns and apply solution methods effectively.
Substitution Method
The substitution method is a technique often employed in solving systems of equations, particularly when one can solve for a variable easily and 'substitute' it into the other equations. The goal is to reduce the system to fewer equations with fewer variables, making it simpler to handle.

Here's how it typically goes: you first solve one of the equations for one variable in terms of the others, and then replace that variable in the other equations with its equivalent expression. In the context of our exercise, you can see we bypass this initial step—the solution set is already given—and directly substitute the values of \(x, y,\) and \(z\) into each equation to verify the solution.

In situations where the solution is not given, you would solve one of the equations for a variable and proceed with the substitution into the remaining equations, simplifying your work progressively.
Algebraic Solution
An algebraic solution in the context of a system of equations is the set of values for the variables that satisfy all equations simultaneously. To arrive at this solution, mathematicians use methods like substitution, elimination, and sometimes more advanced methods like matrix operations.

The step-by-step solution presented here shows how the given set \(x=1, y=-2, z=0\) is an algebraic solution. When substituted into each equation, this set maintains the balance of the equations, meaning the left-hand side equals the right-hand side for each one. This verification process is essential to confirm that the proposed values indeed form an algebraic solution to the system.

In summary, an algebraic solution is not just a guess; it's a precise answer that comes from comprehensive and logical reasoning, typically following one of the structured methods of solving systems of equations.

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Most popular questions from this chapter

In this set of exercises, you will use the method of solving linear systems using matrices to study real-world problems. A gardener, is mixing organic fertilizers consisting of bone meal, cottonseed meal, and poultry manure. The percentages of nitrogen (N), phosphorus (P), and potassium (K) in each fertilizer are given in the table below. $$\begin{array}{lccc}\hline & \begin{array}{c}\text { Nitrogen } \\\\(\%)\end{array} & \begin{array}{c}\text { Phosphorus } \\\\(\%)\end{array} & \begin{array}{c}\text { Potassium } \\\\(\%)\end{array} \\\\\hline \text { Bone meal } & 4 & 12 & 0 \\\\\text { Cottonseed meal } & 6 & 2 & 1 \\\\\text { Poultry manure } & 4 & 4 & 2\end{array}If Mr. Greene wants to produce a 10 -pound mix containing \(5 \%\) nitrogen content and \(6 \%\) phosphorus content, how many pounds of each fertilizer should he use?

The sum of money invested in two savings accounts is \(\$ 1000 .\) If both accounts pay \(4 \%\) interest compounded annually, is it possible to earn a total of \(\$ 50\) in interest in the first year? (a) Explain your answer in words. (b) Explain your answer using a system of equations.

The sum of the squares of two positive integers is \(74 .\) If the squares of the integers differ by 24 find the integers.

A farmer has 90 acres available for planting corn and soybeans. The cost of seed per acre is \(\$ 4\) for corn and \(\$ 6\) for soybeans. To harvest the crops, the farmer will need to hire some temporary help. It will cost the farmer \(\$ 20\) per acre to harvest the corn and \(\$ 10\) per acre to harvest the soybeans. The farmer has \(\$ 480\) available for seed and \(\$ 1400\) available for labor. His profit is \(\$ 120\) per acre of corn and \(\$ 150\) per acre of soybeans. How many acres of each crop should the farmer plant to maximize the profit?

Let \(A=\left[\begin{array}{ll}1 & -1 \\ 1 & -1\end{array}\right]\) and \(B=\left[\begin{array}{ll}1 & 1 \\ 1 & 1\end{array}\right] .\) What is the prod- uct \(A B ?\) Is it true that if \(A\) and \(B\) are matrices such that \(A B\) is defined and all the entries of \(A B\) are zero, then either all the entries of \(A\) must be zero or all the entries of \(B\) must be zero? Explain.

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