Solve the system. Then classify the system as consistent and independent, consistent and dependent, or inconsistent.
24.) 4x+5y=3
6x+9y=9
26.) 1/2x -3y=10
1/4x +2y=-2
24. Eq1: 4x + 5y = 3
Eq2: 6x + 9y = 9
Multiply Eq1 by 6 and Eq2 by -4 and add:
24x + 30y = 18
-24x - 36y = -36
Sum: -6y = --18
Y = 3
In Eq1, replace y with 3 and solve for x:
4x + 5*3 = 3
4x = 3-15 = -12
X = -3
26. Use same procedure as #24.
Multiply Eq1 by 2 and Eq2 by -4 and add
the Eqs.
To solve the systems of equations, we can use either substitution or elimination method. Let's start with the first system:
24.)
4x + 5y = 3 ...(Equation 1)
6x + 9y = 9 ...(Equation 2)
To solve the system using the elimination method, we need to eliminate one variable by multiplying either Equation 1 or Equation 2 by a suitable constant.
We can see that if we multiply Equation 1 by 2, we can get the coefficients of 'x' to be the same in both equations. Let's do that:
2(4x + 5y) = 2(3)
8x + 10y = 6 ...(Equation 3)
Now, let's subtract Equation 3 from Equation 2:
(6x + 9y) - (8x + 10y) = 9 - 6
6x + 9y - 8x - 10y = 3
-2x - y = -3 ...(Equation 4)
Now, we have two equations:
-2x - y = -3 ...(Equation 4)
8x + 10y = 6 ...(Equation 3)
If we analyze Equation 4, we can see that we can eliminate 'y' by multiplying it by -10. Let's do that:
-10(-2x - y) = -10(-3)
20x + 10y = 30 ...(Equation 5)
Now, let's add Equation 3 and Equation 5:
(8x + 10y) + (20x + 10y) = 6 + 30
28x + 20y = 36 ...(Equation 6)
Now, we can divide Equation 6 by 4 to simplify it:
(28x + 20y)/4 = 36/4
7x + 5y = 9 ...(Equation 7)
If we compare Equation 7 with Equation 4, we can see that they are the same equation. Therefore, the system is consistent and dependent.
Now, let's move on to the second system:
26.)
(1/2)x - 3y = 10 ...(Equation 1)
(1/4)x + 2y = -2 ...(Equation 2)
To solve this system, we can again use the elimination method. We can see that if we multiply Equation 1 by 4, we can get the coefficients of 'x' to be the same in both equations:
4[(1/2)x - 3y] = 4(10)
2x - 12y = 40 ...(Equation 3)
Now, let's subtract Equation 3 from Equation 2:
[(1/4)x + 2y] - (2x - 12y) = -2 - 40
(1/4)x + 2y - 2x + 12y = -42
(-7/4)x + 14y = -42 ...(Equation 4)
Now, we have two equations:
(-7/4)x + 14y = -42 ...(Equation 4)
(1/4)x + 2y = -2 ...(Equation 2)
If we compare Equation 4 with Equation 2, we can see that they are not the same equation and the coefficients of 'x' are not equal. Therefore, the system is inconsistent.
So, to classify the systems:
24.) The system is consistent and dependent.
26.) The system is inconsistent.