To get exactly two or three solutions to a linear system, we need to create a scenario in which the two linear equations intersect at two or three distinct points.
One way to achieve this is by introducing a third equation that shares the same solution as the original system of equations at one or two points. We can do this by adding a parallel line to either one of the original equations.
For example, suppose the original system of equations is:
Equation 1: x + 2y = 15
Equation 2: 3x - 2y = 3
By rearranging Equation 1, we can write it as:
Equation 1 (rearranged): x = 15 - 2y
Now, let's introduce a third equation that shares the same y-intercept as the original system:
Equation 3: x = 15
By adding this equation to the original system, we now have three equations:
Equation 1 (rearranged): x = 15 - 2y
Equation 2: 3x - 2y = 3
Equation 3: x = 15
Now let's solve this system:
1. Substitute Equation 1 (rearranged) into Equation 2:
3(15 - 2y) - 2y = 3
45 - 6y - 2y = 3
45 - 8y = 3
-8y = 3 - 45
-8y = -42
y = -42/-8
y = 5.25
2. Substitute the value of y back into Equation 1 (rearranged):
x = 15 - 2(5.25)
x = 15 - 10.5
x = 4.5
Thus, we have found one solution to this system of linear equations, which is x = 4.5 and y = 5.25.
Now, let's substitute Equation 3 into the original system of equations:
Equation 1 (rearranged): 15 = 15 - 2y
0 = -2y
y = 0
Substituting y = 0 into Equation 1 (rearranged), we get:
x = 15 - 2(0)
x = 15
Therefore, another solution to this system of equations is x = 15 and y = 0.
Hence, by introducing the third equation, we obtained two distinct solutions to the linear system.
Note: If a third equation shares the same solution with the original system at two points instead of one, we would end up with three solutions.
The solution to a system of two linear equations is x = 3 weeks; y = 9 feet How could you get exactly two or three solutions to this linear system?
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