7. For the function defined by:
f(x)= {x^2, x<=1}
{2x+1, x>1}
a. evaluate f(0)
8. Solve the following system of equations algebraically. Verify your solution either graphically or by using matrices.
3x-y=0
5x+2y=22
9. Solve the following system of equations algebraically. Verify your solution either graphically or by using matrices
8x-2y=5
-12x+3y=7
Well, if 8x-2y=5 what does x equal and y equal? (same with all of them)
#7 since 0<1,
f(0) = 0^2 = 0
#8 double the 1st and your system becomes
6x-2y=0
5x+2y=22
Now eliminate y by adding the two equations.
#9 If you try elimination here, you get:
24x-6y=15
-24x+6y=14
Can you see that there is no solution to this system?
7. To evaluate f(0) for the given function, we need to substitute 0 in place of x in the function.
f(0) = 0^2, for x <= 1
Simplifying, we get f(0) = 0.
Therefore, f(0) equals 0.
8. To solve the system of equations algebraically:
Equation 1: 3x - y = 0
Equation 2: 5x + 2y = 22
We can solve this system using the method of substitution or elimination. Let's use the method of elimination to solve it.
Multiply both sides of Equation 1 by 2, and multiply both sides of Equation 2 by -1 to make the coefficient of y equal in magnitude but opposite in sign:
Equation 1: 6x - 2y = 0
Equation 2: -5x - 2y = -22
Now, add the two equations together to eliminate the y term:
(6x - 2y) + (-5x - 2y) = 0 + (-22)
Combining like terms, we get:
6x - 5x - 2y - 2y = -22
Simplifying, we have:
x - 4y = -22
Next, we can solve this new equation alongside Equation 2 using the method of elimination:
Equation 3: x - 4y = -22
Equation 2: 5x + 2y = 22
Multiply Equation 3 by 5 and Equation 2 by 1 to make the coefficient of x equal in magnitude:
Equation 3: 5(x - 4y) = 5(-22)
Equation 2: 1(5x + 2y) = 1(22)
Simplifying, we have:
5x - 20y = -110
5x + 2y = 22
Now, subtract Equation 2 from Equation 3 to eliminate the x term:
(5x - 20y) - (5x + 2y) = -110 - 22
Combining like terms, we get:
5x - 5x - 20y - 2y = -110 - 22
Simplifying, we have:
-22y = -132
Dividing both sides of the equation by -22, we find:
y = 6
Now, substituting this value of y back into Equation 1, we get:
3x - 6 = 0
Adding 6 to both sides of the equation, we have:
3x = 6
Dividing both sides of the equation by 3, we find:
x = 2
Therefore, the solution to the system of equations is x = 2 and y = 6.
To verify the solution graphically, you can plot both equations on a graph and see where they intersect. The point of intersection represents the solution. Alternatively, you can use matrices to verify the solution by putting the coefficients of the equations into a matrix and performing matrix operations.
9. To solve the system of equations algebraically:
Equation 1: 8x - 2y = 5
Equation 2: -12x + 3y = 7
Again, we can solve this system using the method of substitution or elimination. Let's use the method of elimination to solve it.
Multiply Equation 1 by 6 and Equation 2 by 4 to make the coefficient of x equal in magnitude but opposite in sign:
Equation 1: 48x - 12y = 30
Equation 2: -48x + 12y = 28
Now, add the two equations together to eliminate the x term:
(48x - 12y) + (-48x + 12y) = 30 + 28
Combining like terms, we get:
48x - 48x - 12y + 12y = 58
Simplifying, we have:
0 = 58
This equation is not true, which means that the system of equations is inconsistent. There is no possible solution that satisfies both equations simultaneously.
To verify this graphically, plot both equations on a graph. If the lines representing the equations are parallel, they will never intersect, indicating no solution exists. Alternatively, you can use matrices to verify the solution by putting the coefficients of the equations into a matrix and performing matrix operations.