Solve the system of equations by Elimination OR Substitution. Explain why you chose one method over the other.
3x – 2y = 8
2x + 5y = -1
I would only use substitution if I can isolate one of the variables without having to go into fractions.
In this case I would use elimination.
multiply the first by 5 and the second by 2, then add them :
15x - 10y = 40
4x + 10y = -2
19x = 38
x = 2
now sub that back into one of the originals. Take it from here.
To solve the system of equations:
3x - 2y = 8 (Equation 1)
2x + 5y = -1 (Equation 2)
There are two common methods to solve systems of equations: Elimination and Substitution. Let's explore both methods and determine which one is more suitable for this particular system.
1. Elimination Method:
The goal of the elimination method is to eliminate one variable by adding or subtracting the equations. To do this, we want to manipulate one or both equations so that when we add or subtract them, one variable will cancel out.
Looking at the given system, we can notice that the coefficients of "y" in both equations (2 and -5) are convenient for elimination. In order to cancel out the "y" variable, we need the coefficients to be equal in absolute value but opposite in sign.
To achieve this, we could multiply Equation 1 by 5 and Equation 2 by 2:
5(3x - 2y) = 5(8) -> 15x - 10y = 40 (Equation 3)
2(2x + 5y) = 2(-1) -> 4x + 10y = -2 (Equation 4)
Now, when we add Equation 3 and Equation 4, the "y" variable will cancel out:
(15x - 10y) + (4x + 10y) = 40 + (-2)
19x = 38
Dividing both sides of the equation by 19, we can solve for "x":
x = 38/19
x = 2
Substituting this value back into Equation 1 or Equation 2, we can find "y":
3(2) - 2y = 8
6 - 2y = 8
-2y = 8 - 6
-2y = 2
y = 2/(-2)
y = -1
Therefore, the solution to the system of equations is x = 2 and y = -1.
2. Substitution Method:
The goal of the substitution method is to solve one equation for one variable and substitute it into the other equation. This allows us to solve for the remaining variable.
Looking at the given system, both equations are already solved for one variable. We can solve Equation 1 for "x" and Equation 2 for "y". Let's solve Equation 1 for "x":
3x - 2y = 8 -> 3x = 2y + 8 -> x = (2y + 8)/3
Now, we substitute this value of "x" into Equation 2:
2((2y + 8)/3) + 5y = -1
By simplifying the equation and solving for "y", we can attain the value of "y".
If we go through these calculations, we would eventually find the value of "y" as -1 and then substitute it back into Equation 1 to find "x" as 2.
In this example, both the elimination and substitution methods are applicable. However, the coefficients in Equation 2 are simpler to use for elimination, making it the preferable method. By quickly identifying the most efficient method, we can save time in solving the system.