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Chapter 5: Problem 63

Find all solutions of the form \((u, v, w)\). $$ \begin{array}{l} \frac{u-3}{4}+\frac{v+1}{3}+\frac{w-2}{8}=1 \\ \frac{u-3}{2}+\frac{v+1}{2}+\frac{w-2}{4}=0 \\ \frac{u-3}{4}-\frac{v+1}{2}+\frac{w-2}{2}=-6 \end{array} $$

Short Answer

Expert verified
The solution is \((u, v, w) = (-13, 11, 10)\).

Step by step solution

01

Simplify the equations

First, let's simplify each given equation by eliminating the denominators: 1) \( \frac{u-3}{4} + \frac{v+1}{3} + \frac{w-2}{8} = 1 \) Multiply through by 24 to clear the denominators: \( 6(u-3) + 8(v+1) + 3(w-2) = 24 \) Simplify: \( 6u - 18 + 8v + 8 + 3w - 6 = 24 \) \( 6u + 8v + 3w - 16 = 24 \) \( 6u + 8v + 3w = 40 \)2) \( \frac{u-3}{2} + \frac{v+1}{2} + \frac{w-2}{4} = 0 \) Multiply through by 4 to clear the denominators: \( 2(u-3) + 2(v+1) + (w-2) = 0 \) Simplify: \( 2u - 6 + 2v + 2 + w - 2 = 0 \) \( 2u + 2v + w - 6 = 0 \) \( 2u + 2v + w = 6 \)3) \( \frac{u-3}{4} - \frac{v+1}{2} + \frac{w-2}{2} = -6 \) Multiply through by 4 to clear the denominators: \( (u-3) - 2(v+1) + 2(w-2) = -24 \) Simplify: \( u - 3 - 2v - 2 + 2w - 4 = -24 \) \( u - 2v + 2w - 9 = -24 \) \( u - 2v + 2w = -15 \)
02

Write the system of linear equations

Rewrite the three simplified equations as a system of linear equations: 1) \( 6u + 8v + 3w = 40 \) 2) \( 2u + 2v + w = 6 \) 3) \( u - 2v + 2w = -15 \)
03

Solve the second equation for w

From the second equation, solve for \( w \): \( w = 6 - 2u - 2v \)
04

Substitute w into the other equations

Substitute \( w = 6 - 2u - 2v \) into the first and third equations: First equation: \( 6u + 8v + 3(6 - 2u - 2v) = 40\) Simplify: \( 6u + 8v + 18 - 6u - 6v = 40 \)\( 2v = 22 \) \( v = 11 \)Third equation: \( u - 2v + 2(6 - 2u - 2v) = -15 \) Simplify: \( u - 2v + 12 - 4u - 4v = -15 \) \( -3u - 6v = -27 \) \( 3u + 6v = 27 \) \( u + 2v = 9 \) \( u + 2(11) = 9 \) \( u = 9 - 22 \) \( u = -13 \)
05

Solve for w

Substitute \( u = -13 \) and \( v = 11 \) back into the equation for \( w \): \( w = 6 - 2(-13) - 2(11) \) Simplify: \( w = 6 + 26 - 22 = 10 \)
06

Verify the solution

Substitute \( u = -13 \), \( v = 11 \), and \( w = 10 \) back into the original equations to verify the solution: 1) \( \frac{-13-3}{4} + \frac{11+1}{3} + \frac{10-2}{8} = -4 + 4 + 1 = 1 \) 2) \( \frac{-13-3}{2} + \frac{11+1}{2} + \frac{10-2}{4} = -8 + 6 + 2 = 0 \) 3) \( \frac{-13-3}{4} - \frac{11+1}{2} + \frac{10-2}{2} = -4 - 6 + 4 = -6 \)The solution is correct.

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Key Concepts

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

Solving linear equations
Understanding how to solve linear equations is a foundation in mathematics. A linear equation in several variables can be visualized as a straight line in coordinate geometry. To solve these equations, your goal is to find the values of the variables that make the equation true. For example, given the system of equations: \[ \frac{u-3}{4} + \frac{v+1}{3} + \frac{w-2}{8} = 1 \] \[ \frac{u-3}{2} + \frac{v+1}{2} + \frac{w-2}{4} = 0 \] \[ \frac{u-3}{4} - \frac{v+1}{2} + \frac{w-2}{2} = -6 \] The key is to express the equations in a simpler form where solving becomes easier. Knowing how to manipulate and simplify these equations step-by-step ensures that you can systematically find the values that satisfy the equations.
Substitution method
The substitution method is a powerful technique for solving systems of linear equations. Here's how it works: you solve one of the equations for one variable in terms of the others, and then substitute this expression into the remaining equations. To demonstrate, let's look at the second simplified equation: \[ 2u + 2v + w = 6 \] By solving for \( w \), we have: \[ w = 6 - 2u - 2v \] This expression for \( w \) can now be substituted in the remaining two equations, ultimately allowing all variables to be solved step-by-step.
Simplification of equations
Simplifying equations involves eliminating fractions and combining like terms. This step reduces complexity and makes the equations easier to handle. Let's take the first original equation: \[ \frac{u-3}{4} + \frac{v+1}{3} + \frac{w-2}{8} = 1 \] To remove the denominators, multiply through by 24 (the least common multiple of 4, 3, and 8): \[ 24 \left( \frac{u-3}{4} \right) + 24 \left( \frac{v+1}{3} \right) + 24 \left( \frac{w-2}{8} \right) = 24 \] This results in: \[ 6(u-3) + 8(v+1) + 3(w-2) = 24 \] After simplifying, we get: \[ 6u + 8v + 3w = 40 \] This simplification process allows us to work with the equation more easily in subsequent steps.
Verification of solutions
Verifying your solutions is crucial to confirm correctness. After finding values for your variables, substitute them back into the original equations to see if they hold true. For our system, we found \( u = -13 \), \( v = 11 \), and \( w = 10 \). let's verify:For the first equation: \[ \frac{-13-3}{4} + \frac{11+1}{3} + \frac{10-2}{8} = 1 \] \[ -4 + 4 + 1 = 1 \] For the second equation: \[ \frac{-13-3}{2} + \frac{11+1}{2} + \frac{10-2}{4} = 0 \] \[ -8 + 6 + 2 = 0 \] For the third equation: \[ \frac{-13-3}{4} - \frac{11+1}{2} + \frac{10-2}{2} = -6 \] \[ -4 - 6 + 4 = -6 \] All satisfy their respective equations, confirming our solutions are correct.

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Most popular questions from this chapter

Use a graphing utility to approximate the solution(s) to the system of equations. Round the coordinates to 3 decimal places. $$ \begin{array}{l} y=-0.7 x+4 \\ y=\ln x \end{array} $$

The minimum and maximum distances from a point \(P\) to a circle are found using the line determined by the given point and the center of the circle. Given the circle defined by \(x^{2}+y^{2}=9\) and the point \(P(4,5)\), a. Find the point on the circle closest to the point (4,5) . b. Find the point on the circle furthest from the point (4,5) .

Use a graphing utility to approximate the solution(s) to the system of equations. Round the coordinates to 3 decimal places. $$ \begin{array}{l} y=0.2 e^{x} \\ y=-0.6 x^{2}-2 x-3 \end{array} $$

The attending physician in an emergency room treats an unconscious patient suspected of a drug overdose. The physician does not know the initial concentration \(A_{0}\) of the drug in the bloodstream at the time of injection. However, the physician knows that after \(3 \mathrm{hr}\), the drug concentration in the blood is \(0.69 \mu \mathrm{g} / \mathrm{dL}\) and after \(4 \mathrm{hr}\), the concentration is \(0.655 \mu \mathrm{g} / \mathrm{dL}\). The model \(A(t)=A_{0} e^{-k t}\) represents the drug concentration \(A(t)\) (in \(\mu \mathrm{g} / \mathrm{dL}\) ) in the bloodstream \(t\) hours after injection. The value of \(k\) is a constant related to the rate at which the drug is removed by the body. a. Substitute 0.69 for \(A(t)\) and 3 for \(t\) in the model and write the resulting equation. b. Substitute 0.655 for \(A(t)\) and 4 for \(t\) in the model and write the resulting equation. c. Use the system of equations from parts (a) and (b) to solve for \(k .\) Round to 3 decimal places. d. Use the system of equations from parts (a) and (b) to approximate the initial concentration \(A_{0}\) (in \(\mu \mathrm{g} / \mathrm{dL}\) ) at the time of injection. Round to 2 decimal places. e. Determine the concentration of the drug after \(12 \mathrm{hr}\). Round to 2 decimal places.

$$ \begin{array}{l} 2(x+y)=2-y \\ 4 x-1=2-5 y \end{array} $$$$ \begin{array}{l} 5(x+y)=9+2 y \\ 6 y-2=10-7 x \end{array} $$
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