91Ó°ÊÓ

A system of linear equations consists of multiple linear equations that are addressed together, with the aim of finding a common solution for the variables involved. These equations often represent intersecting lines, and the solution is the point(s) where these lines meet. In our example, we have:

• \(-2k_{1} + 5k_{2} = 2\)
• \(4k_{1} + 7k_{2} = 3\)

These equations represent a system because they share the variables \(k_1\) and \(k_2\). Solving the system means finding the values of \(k_1\) and \(k_2\) that satisfy both equations at the same time. To solve them, we can use different methods, including graphing, substitution, and elimination. This problem uses the elimination method.

The elimination method is used to solve a system of equations by eliminating one of the variables, making it easier to solve for the other. In this exercise, we start by eliminating \(k_2\).

To do this, both equations are manipulated so that the coefficients of \(k_2\) are equal. The first equation is multiplied by 7, resulting in:

• \(-14k_{1} + 35k_{2} = 14\)

The second equation is multiplied by 5:

• \(20k_{1} + 35k_{2} = 15\)

Now, subtract one equation from the other:

• \(20k_{1} + 35k_{2} - (-14k_{1} + 35k_{2}) = 15 - 14\)
• \(34k_{1} = 1\)

This leaves us with a significantly simpler equation involving only \(k_1\), allowing easy computation of its value.

In mathematics, vectors have both magnitude and direction. They are depicted in terms of their components on coordinate axes, often as linear combinations of unit vectors. For example, a vector can be expressed in terms of \(\mathbf{i}\) and \(\mathbf{j}\) components:

• \(a\mathbf{i} + b\mathbf{j}\)

Here, \(a\) and \(b\) are the vector components along the \(\mathbf{i}\) and \(\mathbf{j}\) directions, respectively. In the exercise, the expressions \(( -2k_{1} + 5k_{2}) \mathbf{i} + (4k_{1} + 7k_{2}) \mathbf{j} \) and \(2 \mathbf{i} + 3 \mathbf{j}\) represent vectors.

Equating the coefficients of \(\mathbf{i}\) and \(\mathbf{j}\) from both side of the given vector equation allows us to derive a system of linear equations, which we must solve to find the correct proportions \(k_1\) and \(k_2\) for the vectors \(\mathbf{b}\) and \(\mathbf{c}\). This showcases how working with vector components can lead to insightful problem-solving techniques within vector equations.

Solving linear equations is fundamental in algebra. It involves finding the values of variables that make the equations true. For linear systems, solutions can be determined exactly by symbolic manipulations.

• Isolate the variable: Move all terms to one side of the equation to isolate the variable term on the other.
• Simplify: Simplify the equation by performing the same operation on both sides until the variable stands alone.
• Verification: Substitute the values back into the original equations to verify that they satisfy all equations in the system.

In our exercise, upon solving the simplified equation from the elimination method, we find that \(k_1 = \frac{1}{34}\). This solution is verified by substituting back into one of the original equations; solving for \(k_2\) involves additional substitution and manipulation. Ensuring all solutions satisfy every equation solidifies our understanding and provides accurate results.