System Of Linear Equations When We Have Two Variables
5x 2y 194x 3y 6a System Of Linear Equations Is When We Have Two
Provide a comprehensive explanation of what a system of linear equations is, including the types of solutions that can result. Then, demonstrate how to solve the following system of equations using two different methods: substitution and elimination. Verify that both methods lead to the same solution. The system of equations is:
- 5x + 2y = 19
- 4x - 3y = 6
Include detailed steps for each method, explain the reasoning behind each step, and interpret the solution in context. Ensure your presentation is clear, logical, and includes any necessary calculations or graphing considerations.
Paper For Above instruction
A system of linear equations consists of two or more equations that share variables and are considered simultaneously to find common solutions. These solutions can be a single unique point, infinitely many points, or no solution at all, depending on whether the equations are consistent and independent, dependent, or inconsistent (Layton, 2020). Solving systems of equations is fundamental in various fields such as engineering, economics, and sciences because it helps model real-world problems where multiple conditions interact.
The given system involves two equations:
- 5x + 2y = 19
- 4x - 3y = 6
To solve this system, I will employ two methods: substitution and elimination. Both should yield the same solution, confirming the correctness and consistency of the methods.
Method 1: Substitution
In the substitution method, we solve one of the equations for one variable in terms of the other and then substitute this expression into the other equation. Starting with the first equation:
5x + 2y = 19
Rearranged to isolate y:
2y = 19 - 5x
Therefore:
y = (19 - 5x) / 2
Substitute y into the second equation:
4x - 3 * [(19 - 5x) / 2] = 6
Multiply through by 2 to clear the denominator:
2 * 4x - 3(19 - 5x) = 12
8x - 57 + 15x = 12
Combine like terms:
23x - 57 = 12
Add 57 to both sides:
23x = 69
Divide both sides by 23:
x = 3
Now substitute x = 3 back into the expression for y:
y = (19 - 5(3)) / 2 = (19 - 15) / 2 = 4 / 2 = 2
Hence, the solution using substitution method is:
x = 3, y = 2
Method 2: Elimination
The elimination method involves aligning coefficients to eliminate one variable. To do this, I will multiply the equations to match the coefficients of either x or y. Let's eliminate y by making the coefficients of y opposites.
First, multiply the first equation by 3 and the second by 2:
First equation multiplied by 3:
15x + 6y = 57
Second equation multiplied by 2:
8x - 6y = 12
Now, add the two equations:
(15x + 6y) + (8x - 6y) = 57 + 12
23x = 69
Divide both sides by 23:
x = 3
Substitute x = 3 into the original first equation to find y:
5(3) + 2y = 19
15 + 2y = 19
2y = 19 - 15 = 4
y = 2
Again, the solution is:
x = 3, y = 2
Conclusion
Both the substitution and elimination methods led to the same solution, demonstrating the consistency and reliability of solving systems of equations through different strategies. The solution reveals that when x is 3, y is 2 in the system. Graphically, this point (3, 2) would be the intersection point of the two lines represented by the equations, confirming the solution visually.
References
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