Plug more than one point into the original equation and check to show that the equations can have more than one solution. Explain your work. 7x+3=7x+3, 3(4x+8)=4(3x+6)
For the equation 7x+3=7x+3, we can plug in any real number for x and the equation will be true. Let's try plugging in x=0:
7(0) + 3 = 7(0) + 3
0 + 3 = 0 + 3
3 = 3
This shows that the equation is true for x=0. Let's try another value, x=1:
7(1) + 3 = 7(1) + 3
7 + 3 = 7 + 3
10 = 10
Again, the equation is true for x=1. We can continue plugging in different values for x, and in every case, the equation will be true. This shows that the equation 7x+3=7x+3 has infinitely many solutions. No matter what value we choose for x, the equation will always be true.
Now let's consider the equation 3(4x+8)=4(3x+6):
We can simplify this equation by distributing the coefficients:
12x + 24 = 12x + 24
Subtracting 12x from both sides gives:
12x - 12x + 24 = 12x - 12x + 24
24 = 24
The equation 3(4x+8)=4(3x+6) is true for any value of x. This means that there are infinitely many solutions for this equation as well.
To check if the equations have more than one solution, we can plug in different values into the original equations and see if they satisfy the equation. Let's start with the first equation:
1. 7x + 3 = 7x + 3
No matter what value you choose for 'x,' both sides of the equation will always be equal. This means that any value of 'x' is a solution to this equation. So, this equation has infinitely many solutions.
Now, let's move on to the second equation:
2. 3(4x + 8) = 4(3x + 6)
To determine if there is more than one solution for this equation, we'll simplify both sides:
On the left side of the equation, we can distribute the 3:
12x + 24 = 4(3x + 6)
Now, let's distribute the 4 on the right side:
12x + 24 = 12x + 24
As you can see, both sides of the equation simplify to the same expression. This means that any value of 'x' will satisfy the equation, just like the first equation. Therefore, the second equation also has infinitely many solutions.
In conclusion, by plugging in different values for 'x' into both equations, we can see that they both have multiple solutions.
Sure! Let's start with the first equation: 7x+3=7x+3.
To demonstrate that this equation can have more than one solution, we can choose any two different values for x and plug them into the equation to see if they produce a true statement.
Let's pick x=0 and x=1. We can substitute these values into the equation and check if both sides of the equation are equal.
For x=0:
7(0) + 3 = 7(0) + 3
0 + 3 = 0 + 3
3 = 3 (True)
For x=1:
7(1) + 3 = 7(1) + 3
7 + 3 = 7 + 3
10 = 10 (True)
As you can observe, both equations yield true statements, which means that the equation 7x+3=7x+3 has more than one solution. In fact, it has infinitely many solutions since any value of x will satisfy this equation.
Now, let's move on to the second equation: 3(4x+8)=4(3x+6).
Similarly, we can choose two different values of x and substitute them into the equation.
Let's pick x=0 and x=1 again.
For x=0:
3(4(0)+8) = 4(3(0)+6)
3(8) = 4(6)
24 = 24 (True)
For x=1:
3(4(1)+8) = 4(3(1)+6)
3(12) = 4(9)
36 = 36 (True)
Just like the first equation, both equations yield true statements for different values of x. Therefore, the equation 3(4x+8)=4(3x+6) also has more than one solution.
To summarize, by selecting different values of x and plugging them into the equations, we have demonstrated that both 7x+3=7x+3 and 3(4x+8)=4(3x+6) have more than one solution. This shows that the equations are not restrictive and can hold true for various values of x.