Solving Systems of Linear Equations: 5x+2y=-3 and x+5y=4
In this article, we will learn how to solve a system of linear equations using two methods: Substitution and Elimination. The system of equations we will be solving is:
Equation 1: 5x + 2y = -3 Equation 2: x + 5y = 4
Method 1: Substitution
To solve this system using substitution, we need to express one variable in terms of the other in one of the equations. Let's solve Equation 2 for x:
x = 4 - 5y
Now, substitute this expression for x into Equation 1:
5(4 - 5y) + 2y = -3
Expand and simplify the equation:
20 - 25y + 2y = -3 20 - 23y = -3
Add 3 to both sides:
23y = 23
Divide by 23:
y = 1
Now, substitute y back into one of the original equations to find x:
x = 4 - 5(1) x = -1
Therefore, the solution to the system is x = -1 and y = 1.
Method 2: Elimination
To solve this system using elimination, we need to make the coefficients of one variable (either x or y) the same in both equations. Let's multiply Equation 1 by 5 and Equation 2 by 2 to make the coefficients of y the same:
Equation 1: 25x + 10y = -15 Equation 2: 2x + 10y = 8
Now, subtract Equation 2 from Equation 1 to eliminate the y variable:
(25x - 2x) + (10y - 10y) = -15 - 8 23x = -23
Divide by 23:
x = -1
Now, substitute x back into one of the original equations to find y:
-1 + 5y = 4 5y = 5
Divide by 5:
y = 1
Therefore, the solution to the system is x = -1 and y = 1, which is the same as the solution we found using substitution.
Conclusion
In this article, we solved a system of linear equations using two methods: Substitution and Elimination. Both methods yielded the same solution, x = -1 and y = 1. Solving systems of linear equations is an important skill in algebra and is used in many real-world applications.
