Solve the system of equations.%0D%0A2x+8y=8%0D%0A2%0D%0A%0D%0A+%0D%0A8%0D%0A%0D%0A=%0D%0A8%0D%0A%0D%0A2x−4y=−16%0D%0A2%0D%0A%0D%0A−%0D%0A4%0D%0A%0D%0A=%0D%0A−%0D%0A16%0D%0A(1 point)%0D%0A(, )
The system of equations is:
2x + 8y = 8 .................(1)
2x - 4y = -16 .................(2)
We can solve this system of equations by either substitution or elimination method.
Using elimination method:
Multiply equation (2) by 2 to make the coefficients of x in both equations the same:
2(2x - 4y) = 2(-16)
4x - 8y = -32 .................(3)
Now, subtract equation (3) from equation (1) to eliminate y:
(2x + 8y) - (4x - 8y) = 8 - (-32)
2x + 8y - 4x + 8y = 8 + 32
-2x + 16y = 40 .................(4)
Next, divide equation (4) by 2 to simplify:
(-2x + 16y)/2 = 40/2
- x + 8y = 20 .................(5)
Now we have two equations:
- x + 8y = 20 .................(5)
2x + 8y = 8 .................(1)
Add equation (5) to equation (1) to eliminate x:
(-x + 8y) + (2x + 8y) = 20 + 8
x + 16y = 28 .................(6)
Now we have two equations:
x + 16y = 28 .................(6)
2x + 8y = 8 .................(1)
We can subtract equation (6) from equation (1) to eliminate x:
(2x + 8y) - (x + 16y) = 8 - 28
x - 8y = -20 .................(7)
Now we have two equations:
x - 8y = -20 .................(7)
x + 16y = 28 .................(6)
Adding equations (7) and (6), we can eliminate x:
(x - 8y) + (x + 16y) = -20 + 28
2x + 8y = 8 .................(1)
This is the same equation as equation (1), which means the system of equations is dependent and has infinitely many solutions. Therefore, there are no unique values of x and y to be determined.
To solve the system of equations,
2x + 8y = 8 ----(1)
2x - 4y = -16 ----(2)
We can use the method of substitution or elimination. Let's use the method of substitution.
From equation (1), we get:
2x = 8 - 8y
=> x = 4 - 4y
Substituting this value of x in equation (2), we have:
2(4 - 4y) - 4y = -16
8 - 8y - 4y = -16
-12y = -24
y = 2
Substituting this value of y in equation (1), we can solve for x:
2x + 8(2) = 8
2x + 16 = 8
2x = -8
x = -4
Therefore, the solution to the system of equations is:
(x, y) = (-4, 2)
To solve the system of equations:
Equation 1: 2x + 8y = 8
Equation 2: 2x - 4y = -16
We can use the method of substitution or elimination to solve this system. Let's use the elimination method.
Step 1: Multiply equation 1 by 2 and equation 2 by 1 to make the coefficients of x in both equations equal:
Equation 1: 4x + 16y = 16
Equation 2: 2x - 4y = -16
Step 2: Subtract equation 2 from equation 1 to eliminate x:
(4x + 16y) - (2x - 4y) = 16 - (-16)
Simplifying:
4x + 16y - 2x + 4y = 16 + 16
2x + 20y = 32
Step 3: Now, we have a new equation: 2x + 20y = 32. Let's divide the equation by 2 to simplify it:
(2x + 20y)/2 = 32/2
x + 10y = 16
Step 4: Now we have two equations:
x + 10y = 16 (Equation 3)
2x - 4y = -16 (Equation 4)
Step 5: Multiply equation 3 by 2:
2(x + 10y) = 2(16)
2x + 20y = 32
This equation is identical to the equation we obtained in step 3, which means the system is dependent.
The solution to this system is not a single point but a whole line. Any (x, y) that satisfies the equation x + 10y = 16 is a solution to this system.
For example, when x = 6, y can be any value. Possible solutions include (6, 0), (6, 1), (6, -2), etc.
Therefore, the solution to the system of equations is (x, y) = (6, y), where y belongs to the set of real numbers.