HELP PLEASE,
Solve using the elimination method. Show your work. If the system has no solution or an infinite number of solutions, state this.
-2x Ð 8y = 22
-6x Ð y = 20
At first I tried to multiply by 8 but that did not work.
I assume the illegible sign is a + sign, since the - sign shows clearly.
-2x + 8y = 22 ....(1)
-6x + y = 20 ....(2)
First, we determine if the system has 0 solution, one solution or many solutions.
Calculate the determinant of the left-hand-side:
Δ=(-2)(1)-(-6)(8)=46
Since Δ≠0, the system has one solution. Multiplying 3*(-1) and add to (2) gives
-23y=-46
solving for y gives
y=2
Substitute y=2 into either (1) or (2) to solve for x.
Then substitute both x and y in the remaining equation for a check.
To solve the given system of equations using the elimination method, we need to eliminate one variable by adding or subtracting the equations. In this case, we can eliminate the variable "y" by adjusting the second equation.
The given system of equations is:
-2x - 8y = 22 ---(Equation 1)
-6x - y = 20 ---(Equation 2)
To eliminate the variable "y", multiply Equation 2 by -8, which will make the coefficient of "y" in both equations equal:
-8 * (-6x - y) = -8 * 20
48x + 8y = -160 ---(Equation 3)
Now, we have the equations:
-2x - 8y = 22 ---(Equation 1)
48x + 8y = -160 ---(Equation 3)
Next, add Equation 1 and Equation 3 to eliminate the variable "y":
(-2x - 8y) + (48x + 8y) = 22 + (-160)
-2x + 48x - 8y + 8y = -138
46x = -138
Divide both sides of the equation by 46 to solve for "x":
46x/46 = -138/46
x = -3
Now that we have the value of "x", substitute it back into one of the original equations to solve for "y". Let's use Equation 1:
-2(-3) - 8y = 22
6 - 8y = 22
-8y = 22 - 6
-8y = 16
Divide both sides of the equation by -8 to solve for "y":
-8y/-8 = 16/-8
y = -2
Therefore, the solution to the given system of equations is x = -3 and y = -2.
In this case, the system has a unique solution, and it is (-3, -2).