Problem 12 - Understanding Elementary Algebra with Geometry

Question

$$\text { In Exercises } 1-14, \text { solve the system of equations using the elimination method.}$$ $$\left\\{\begin{array}{l} 6 u-w=2 \\ 2 u-3 w=2 \end{array}\right.$$

Step-by-Step Solution

Answer

u = \frac{1}{4}, w = -\frac{1}{2}

Step 1 - Analyze the System of Equations

First, consider the given system of equations: 1) 6u - w = 2 2) 2u - 3w = 2

Step 2 - Eliminate Variable by Aligning Coefficients

To eliminate one of the variables, we can align the coefficients of either variable. Let's eliminate the variable \(u\) by making sure both equations have the same coefficient for \(u\). We need to multiply the second equation by 3 so that the coefficient for \(u\) in both equations is 6.1) \(6u - w = 2\)3*(2) \( \rightarrow 6u - 9w = 6\)

Step 3 - Subtract Equations to Eliminate the Variable

Subtract the second equation from the first to eliminate \(u\):(6u - w) - (6u - 9w) = 2 - 6This simplifies to:- w + 9w = -48w = -4

Step 4 - Solve for Remaining Variable

Now, solve for \(w\):8w = -4w = \frac{-4}{8} w = -\frac{1}{2}

Step 5 - Substitute Back to Find the Second Variable

Substitute \(w = -\frac{1}{2}\) back into the first equation to solve for \(u\):6u - (-\frac{1}{2}) = 26u + \frac{1}{2} = 26u = 2 - \frac{1}{2}6u = \frac{4}{2} - \frac{1}{2}6u = \frac{3}{2}u = \frac{3}{2} * \frac{1}{6}u = \frac{1}{4}

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of variables. Finding the values of these variables which satisfy all equations simultaneously is the main goal when dealing with a system of equations.
For example, consider the given system of equations: \[\begin{cases}{6u - w = 2} \ {2u - 3w = 2}\right.\] Here, we have two equations involving the variables \(u\) and \(w\).
To solve this system using the elimination method, we need to find a common coefficient for one of the variables so that we can eliminate it.
This will simplify our calculations and help us isolate and solve for one of the variables.

Linear Equations

Linear equations are equations of the first degree, meaning each term is either a constant or the product of a constant and a single variable. Linear equations have the general form \ax + by = c\.
In our example, both equations \(6u - w = 2\) and \(2u - 3w = 2\) are linear.
These equations graph as straight lines, and the solution to the system is the point where these lines intersect.
Solving a system of linear equations involves finding this intersection point.

Variable Elimination

Variable elimination, or the elimination method, is a technique for solving systems of equations. The main idea is to eliminate one variable by aligning its coefficients across the equations.
In our example, we multiplied the second equation by 3 to get: \[6u - w = 2 \] \[6u - 9w = 6\] We can then subtract the second equation from the first, effectively eliminating \(u\).
This leaves us with a simpler equation involving only \(w\).
The next step is to solve this simplified equation.

Solving Equations

After eliminating one variable, we need to solve for the remaining variable. From the previous step, we have: \[8w = -4\] Solving for \(w\), we get: \w = - \frac{1}{2}\.
Substituting this value back into one of the original equations allows us to solve for the second variable: \[6u - (-\frac{1}{2}) = 2\] Simplifying, we obtain: \[6u + \frac{1}{2} = 2\] \[6u = \frac{3}{2}\] \u = \frac{1}{4}\.
So the solution to the system of equations is \u = \frac{1}{4}\ and \w = - \frac{1}{2}\. This means the values \(u\) and \(w\) satisfy both original equations.