/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Verify that each system of equat... [FREE SOLUTION] | 91Ó°ÊÓ

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Verify that each system of equations has the indicated solution. $$\left\\{\begin{array}{rr} x+2 y-3 z= & 1 \\ 2 x+3 y+z= & -3 \end{array}\right.$$ Solution: \(x=-9-11 z, y=5+7 z\) (for every real number \(z\) )

Short Answer

Expert verified
The solution \(x=-9-11 z, y=5+7 z\) for all real \(z\) is not a valid solution for the given system of linear equations.

Step by step solution

01

Express the solution

First, the solution is expressed as \(x=-9-11 z, y=5+7 z\). This shows that the variables x and y are expressed in terms of z. Since z is any real number, both x and y will change accordingly.
02

Substitute in the first equation

Next, substitute \(x=-9-11 z\) and \(y=5+7 z\) into the first equation. It turns into \(-9-11z + 2(5 + 7z) - 3z = 1\). This simplifies to \(-9 -11z + 10 +14z -3z = 1\). Further simplifying gives \(0 = 1\), which is not a valid statement. Thus, this solution is not valid for the first equation.
03

Substitute in the second equation

Now, let's substitute \(x=-9-11 z\) and \(y=5+7 z\) into the second equation. The equation becomes \(2(-9-11z) + 3(5+7z)+z=-3\). Simplifying, this equation produces \(0=-3\), which again is not valid. Thus the solution does not hold for the second equation either.

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Key Concepts

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

Solution Verification
When we talk about solution verification in the context of a system of equations, it means checking whether a proposed set of values satisfies all the equations in the system. In this exercise, we had a system of two linear equations. We were given a set of solutions for that system, specifically expressing the solutions in terms of a parameter, "z." These proposed solutions were:
  • \(x = -9 - 11z\)
  • \(y = 5 + 7z\)
Here, "z" is any real number. The verification process involves substituting these expressions for \(x\) and \(y\) back into the original set of equations. If after substitution, both equations hold true (i.e., both sides of the equations are equal), then the proposed solutions are verified. However, in this scenario, the substitution resulted in invalid statements such as \(0 = 1\). Thus, the solution does not satisfy the system of equations. This means that the initial expressions for \(x\) and \(y\) are not true solutions.
Linear Equations
Linear equations are algebraic expressions where each term is either a constant or the product of a constant and a single variable. In simplest terms, they graph as straight lines in a coordinate system. A general form of a linear equation in two variables usually looks like \(ax + by = c\).
Linear equations can have multiple variables but importantly, each variable is only raised to the power of one. Let's consider the linear equations from our original exercise:
  • \(x + 2y - 3z = 1\)
  • \(2x + 3y + z = -3\)
These equations include the variables \(x\), \(y\), and \(z\). Each term aligns with our definition of linearity since they don't contain squares, cubes, or any higher powers of the variables or two variables multiplied together. Often, problems involving linear systems, as in this exercise, require us to find where these lines intersect. Solving them in terms of a parameter "z", as was attempted, is common when dealing with systems that have infinitely many solutions or are dependent on a variable.
Real Numbers
Real numbers are the collection of numbers that include both the rational and irrational numbers, essentially encompassing all the numbers that could appear on an infinite number line. Rational numbers are any numbers that can be expressed as the fraction of two integers, and include integers themselves, such as -2, 0, or 4. Irrational numbers are non-repeating, non-terminating decimals – they can't be accurately expressed as simple fractions. Famous examples include \(\pi\) and \(\sqrt{2}\). In our problem, "z" was stated to represent any real number. This means \(z\) could be any number on the number line; positive, negative, whole, fraction, or irrational. This freedom indicates that for every possible value of \(z\), there would ideally be corresponding values of \(x\) and \(y\) that make up the solution set, if the solution is valid for the system. However, not every assignment of real numbers produces a true statement in this case, as our verification exercises showed. Understanding real numbers is essential in interpreting how such continuously infinite possibilities interact within equations.

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