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Chapter 5: Problem 16

Solve each system. $$\begin{aligned}&x+y=4\\\&x+z=4\\\&y+z=4\end{aligned}$$

Short Answer

Expert verified
The solution of the system of equations is \(x=4\), \(y=0\), and \(z=0\).

Step by step solution

01

Isolate a variable in one equation

From the first equation \(x+y=4\), isolate x. This is achieved by subtracting y from both sides of the equation, which gives \(x=4-y\). This equation will be used for substituting x in the other two equations.
02

Substitute the value of x in the remaining equations

Replace x in the second and third equation with \(4-y\). The second equation becomes \(4-y+z=4\), simplifying we get \(z=y\). The third equation becomes \(4+y=4\), simplifying we get \(y=0\).
03

Find the values of x and z

Substitute \(y=0\) into the equation \(x=4-y\) to get \(x=4\). Substitute \(y=0\) into the equation \(z=y\) to get \(z=0\).

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Key Concepts

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

Variable Isolation
To solve a system of equations, one effective strategy is to isolate a variable in one of the equations. By doing this, you simplify the equation to express one variable in terms of others. It's like unraveling a sweater thread to get to its core.
In our example, from the system of equations:
  • \(x + y = 4\)
  • \(x + z = 4\)
  • \(y + z = 4\)
We chose to isolate the variable \(x\) from the first equation. By doing so, we subtracted \(y\) from each side, leading us to \(x = 4 - y\).
This step is important because now we've converted the equation into a simpler form, making it easier to substitute into other equations later. It reduces complexity, bringing us closer to finding solutions for \(x\), \(y\), and \(z\) one at a time.
Substitution Method
The substitution method goes hand in hand with variable isolation. Once a variable is isolated, you can substitute it back into other equations to simplify and reach a solution.
After isolating \(x\) in the equation \(x = 4 - y\), we substitute \(4 - y\) for \(x\) in the remaining equations:
  • Replace in \(x + z = 4\):
    • \((4 - y) + z = 4\)
    • Simplify to \(z = y\)
  • Replace in \(y + z = 4\):
    • \(y + z = 4\)
    • Since we've now determined \(z = y\), substitute to get \(y + y = 4\) or \(2y = 4\)
    • Dividing through gives \(y = 2\)
Substituting back this way has allowed us to determine exact values for \(y\) and \(z\) quickly, using the relationships between the variables.
Algebraic Solution
Once you have substituted and found values for one or two variables, an algebraic solution pulls these threads together to solve the entire system!
Firstly, from our table of substitutions we have:
  • \(y = 2\)
  • \(z = 2\) (since \(z = y\))
Now, returning to our isolated expression, substitute \(y = 2\) into our isolated variable equation, \(x = 4 - y\):
  • \(x = 4 - 2 = 2\)
Check the solutions against each equation to ensure they satisfy all parts of the system. This complete solution shows us:
  • \(x = 2\)
  • \(y = 2\)
  • \(z = 2\)
Every solution complements the other, verifying accuracy across all original equations. This is the power of algebra, neatly organizing and solving multi-variate problems.

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Most popular questions from this chapter

In Exercises \(31-42,\) solve by the method of your choice. Identify systems with no solution and systems with infinitely many solutions, using set notation to express their solution sets. \(4 x-2 y=2\) \(2 x-y=1\)

In Exercises \(31-42,\) solve by the method of your choice. Identify systems with no solution and systems with infinitely many solutions, using set notation to express their solution sets. \(4 x=3 y+8\) \(2 x=-14+5 y\)

In Exercises \(19-30,\) solve each system by the addition method. \(5 x=6 y+40\) \(2 y=8-3 x\)

In Exercises \(5-18\), solve each system by the substitution method. $$ \begin{array}{l} 2 x+5 y=1 \\ -x+6 y=8 \end{array} $$

In Exercises \(5-18\), solve each system by the substitution method. $$ \begin{array}{l} x=4 y-2 \\ x=6 y+8 \end{array} $$
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