Solving a Linear System with Two Variables Using the Elimination Method

The solution set of a linear system involving variables x1,…,xn is a list of numbers (s1,…,sn) that makes each equation in the system a true statement when the values s1,…,sn are substituted for x1,…,xn, respectively

In algebra, a linear system consists of a set of linear equations that are to be solved simultaneously

In algebra, a linear system consists of a set of linear equations that are to be solved simultaneously. These equations involve variables such as x1, x2,… xn. The solution set of the linear system is the set of all possible values for these variables that make each equation in the system true.

Let’s take an example linear system with two variables, x1 and x2:

Equation 1: 2×1 + 3×2 = 7
Equation 2: 4×1 – 2×2 = 10

To find the solution set for this system, we need to find the values of x1 and x2 that satisfy both equations.

We can solve this system using various methods, such as substitution or elimination. For this example, let’s use the elimination method:

First, we’ll multiply Equation 1 by 2 to eliminate the x1 term:

Equation 1: 4×1 + 6×2 = 14
Equation 2: 4×1 – 2×2 = 10

Next, we’ll subtract Equation 2 from Equation 1 to eliminate the x1 term:

(4×1 + 6×2) – (4×1 – 2×2) = 14 – 10

This simplifies to:

8×2 = 4

Divide both sides of the equation by 8:

x2 = 0.5

Now that we have the value of x2, we can substitute it back into either of the original equations to solve for x1. Let’s use Equation 1:

2×1 + 3(0.5) = 7

2×1 + 1.5 = 7

Subtract 1.5 from both sides of the equation:

2×1 = 5.5

Divide both sides of the equation by 2:

x1 = 2.75

Therefore, the solution set for this linear system is (2.75, 0.5), which means that these values of x1 and x2 satisfy both Equation 1 and Equation 2.

Please note that this is just an example, and the steps and methods may vary depending on the specific linear system you are working with.

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