Consider the pair of linear equations below.
4x+6y=12
2x+3y=6
Part A: What is the relationship, if any, between 12 and 6?
Part B: Do the two equations have one solution, no solution, or infinitely many solution? Explain.
Part C: How can you verify your answers to Part A and B by solving algebraically?
Eq1: 4x + 6y = 12.
m1 = -A/B = -4/6 = -2/3.
Y-int. = C/B = 12/6 = 2.
Eq2: 2x + 3y = 6.
m2 = -2/3.
Y-int. = 6/3 = 2.
Since the slopes and Y-intercepts are both equal, the Eqs represent the same line. Therefore, we have an infinite number of solutions.
Multiply both sides of Eq2 by 2 and compare the Eqs.
If the slopes were not equal, there will be one solution.
If the slopes were equal with unequal Y-intercepts, we'll have no solutions.
In this case, we have two parallel lines.
Part A: The relationship between 12 and 6 is that 12 is twice the value of 6.
Part B: The two equations have infinitely many solutions. This can be explained by the fact that both equations are multiples of each other. The second equation is obtained by multiplying the first equation by 0.5.
Part C: To verify the answers algebraically, we can solve the equations.
Equation 1: 4x + 6y = 12
Equation 2: 2x + 3y = 6
To solve them, we can use the method of substitution or elimination.
Using the method of substitution:
From Equation 2, we can isolate x: x = (6 - 3y) / 2
Substitute this value of x into Equation 1:
4 * ((6 - 3y) / 2) + 6y = 12
Simplifying:
(12 - 6y) + 6y = 12
12 - 6y + 6y = 12
12 = 12
This equation is always true, meaning that any value of y will satisfy it. Therefore, there are infinitely many solutions.
Using the method of elimination:
Multiply Equation 2 by 2 to make the coefficients of x in both equations the same:
4x + 6y = 12
4x + 6y = 12
Since the equations are identical, they represent the same line. Therefore, there are infinitely many solutions.
Both methods confirm that the equations have infinitely many solutions, supporting the answer in Part B.
Part A: To determine the relationship between 12 and 6, we can compare the given numbers. In this case, since both 12 and 6 are equal, we can say that the relationship between them is that they are equal (12 = 6).
Part B: To determine if the two equations have one solution, no solution, or infinitely many solutions, we need to solve the system of equations. Let's solve them algebraically:
Equation 1: 4x + 6y = 12
Equation 2: 2x + 3y = 6
We can solve this system of equations using the method of substitution or elimination.
Using the method of elimination:
Multiply Equation 2 by 2 to make the coefficients of x the same.
Equation 2 becomes:
4x + 6y = 12
Now, we have two equations with the same coefficients of x:
4x + 6y = 12 (Equation 1)
4x + 6y = 12 (Equation 2)
Since both equations are identical, we can say that the system of equations has infinitely many solutions.
Part C: We can verify our answers to Part A and B by solving algebraically. From our previous solution, we determined that the system has infinitely many solutions. To verify this, we can choose any pair of values for x and y that satisfy the original equations.
Let's choose x = 0 and y = 2:
Substituting these values into the original equations:
Equation 1: 4(0) + 6(2) = 12
Simplifying, we get: 0 + 12 = 12, which is true.
Equation 2: 2(0) + 3(2) = 6
Simplifying, we get: 0 + 6 = 6, which is also true.
Since both equations are true, our chosen values of x = 0 and y = 2 satisfy both equations, confirming that the system has infinitely many solutions.