Question

Solve the system of equations. $$\left\\{\begin{array}{r}x+y+z+w=6 \\ 2 x+3 y-w=0 \\ -3 x+4 y+z+2 w=4 \\\ x+2 y-z+w=0\end{array}\right.$$

Short answer

The solution of the system of equations is: \(x = -2\), \(y = z = 1\) and \(w = 6\).

Step-by-step solution

Pair and combine equations

First, pair two equations and eliminate a variable. For instance, add the second and fourth equations \(2x + 3y - w = 0\) and \(x + 2y - z + w = 0\) to cancel out variable \(w\). The result will be \(3x + 5y - z = 0\)

Eliminate another variable

Now, pair a previously unused equation with the derived equation from Step 1, with the aim of eliminating another variable. So, use the equations \(x+y+z+w=6\) and \(3x + 5y - z = 0\). When you subtract the second one from the first one, you will get \(-2x -4y + 2z + w = 6\), and simplifying you will get \(-x -2y + z + w/2 = 3\).

Eliminate variable third variable

Next step consists in eliminating a third variable. Choose another pair of equations that suits best, ideally one of the equations found in previous steps. If we pair the modified equation from Step 2 \(-x - 2y + z + w/2 = 3\) with the third initial equation \(-3x + 4y + z + 2w = 4\), it will be possible to cancel out variable \(z\). The calculation looks like this: \(2(-x -2y + z + w/2) + (-3x + 4y + z + 2w) = 2*3 + 4\). This simplifies to \(-5x = 10\).

Solve for the first variable

After eliminating three variables in the previous steps, we can directly solve for \(x\) in equation \(-5x = 10\). This gives us the value of \(x=10/-5 = -2\).

Substitute \(x\) into the equations

Next, substitute the found value of \(x\) into the original four equations and solve for remaining the variables one by one. This process will lead to the solution for the remaining variables \(y\), \(z\) and \(w\).

Final solution

By substituting \(x = -2\) in the original four equations and solving for the remaining variables, one obtains \(y = z = 1\) and \(w = 6\). This is the final solution of the system.

Key concepts

Linear Equations

A linear equation is a type of algebraic equation where the highest power of the variable involved is 1. In simpler terms, if you have terms like x, y, or z but not x², y², or higher powers, you're dealing with a linear equation. It's called "linear" because if you were to graph these equations, they would form straight lines. Linear equations can have one or more variables.
The system of equations provided in the exercise is a perfect example of a set of linear equations with four variables: x, y, z, and w. Each equation in the given system respects the linear nature as they only involve the variables raised to the first power.
Understanding how to manipulate these equations is a fundamental skill in algebra, which helps to find the values of the variables that make all the equations true simultaneously.

Variable Elimination

Variable elimination is a process used to simplify solving systems of equations by getting rid of variables one at a time. The main idea is to combine two equations in such a way that one of the variables cancels out. This simplifies the system and reduces the number of unknowns.
In our exercise, we used variable elimination by initially combining the second equation (\(2x + 3y - w = 0\)) and the fourth equation (\(x + 2y - z + w = 0\)) to eliminate the variable \(w\). This step-by-step reduction in variables makes it easier to solve the system.
By continuing to strategically pair equations and eliminate additional variables, we ease the complexity of our algebraic manipulations. Ultimately, this method simplifies a multi-variable system to fewer variables that are easier to handle.

Algebraic Solution

The algebraic solution involves manipulating the equations using basic algebraic operations like addition, subtraction, multiplication, or division, to isolate variables and find their values.
In our setup, the algebraic solution came into play when manipulating the equations to either simplify or reveal the values of unknowns with clear straightforward steps. As seen, the steps also involved combining equations and applying arithmetic operations to simplify the system to \(-5x = 10\), allowing us to readily solve for \(x\).
The goal is to be left with equations that directly give the values of one or more variables. This problem-solving edge does not just find solutions; it builds a solid understanding of how equations interact with one another.

Substitution Method

The substitution method is another alternative to solving systems of equations, with an approach different from elimination. However, in our exercise, it acted as a complement to the elimination process.
Once we've reduced the system to simpler equations, substitution makes it straightforward to solve for remaining variables. As mentioned in the solution steps, once we have determined the value for \(x = -2\), we substitute this value back into the other equations. By doing so, we can easily solve for the other variables \(y\), \(z\), and \(w\).
This method involves substituting known values and solving the remaining simplified equations. It provides a chain reaction of solutions that help unravel all the unknowns in the system efficiently.