Sure! Let's solve the two equations one by one, then plug in \( x = 3 \) and \( x = -2 \) to see if they hold true.
Equation 1: \( 3x + 3 = 7x + 3 \)
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Start with the equation:
\( 3x + 3 = 7x + 3 \) -
Subtract \( 3 \) from both sides:
\( 3x = 7x \) -
Subtract \( 3x \) from both sides:
\( 0 = 4x \) -
Divide by \( 4 \):
\( x = 0 \)
So, the solution for Equation 1 is \( x = 0 \).
Equation 2: \( 3(4x + 8) = 4(3x + 6) \)
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Start with the equation:
\( 3(4x + 8) = 4(3x + 6) \) -
Distribute on both sides:
Left side: \( 3 \times 4x + 3 \times 8 = 12x + 24 \)
Right side: \( 4 \times 3x + 4 \times 6 = 12x + 24 \) -
Now the equation looks like this:
\( 12x + 24 = 12x + 24 \) -
Since both sides are equal, it means:
This equation is true for all values of \( x \).
So, Equation 2 has infinitely many solutions.
Plugging in Values
Now let's check if \( x = 3 \) and \( x = -2 \) work for these equations.
For Equation 1:
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If \( x = 3 \):
\( 3(3) + 3 = 9 + 3 = 12\)
\( 7(3) + 3 = 21 + 3 = 24\)
Not equal, so \( x=3 \) is not a solution. -
If \( x = -2 \):
\( 3(-2) + 3 = -6 + 3 = -3\)
\( 7(-2) + 3 = -14 + 3 = -11\)
Not equal, so \( x=-2 \) is not a solution.
For Equation 2:
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If \( x = 3 \):
\( 3(4(3) + 8) = 3(12 + 8) = 3(20) = 60\)
\( 4(3(3) + 6) = 4(9 + 6) = 4(15) = 60\)
Equal, so \( x=3 \) is a solution. -
If \( x = -2 \):
\( 3(4(-2) + 8) = 3(-8 + 8) = 3(0) = 0\)
\( 4(3(-2) + 6) = 4(-6 + 6) = 4(0) = 0\)
Equal, so \( x=-2 \) is also a solution.
Summary
- Equation 1 has one solution, \( x = 0 \).
- Equation 2 has infinitely many solutions (it’s true for all \( x \)).
- We checked specific values \( x = 3 \) and \( x = -2 \). Only Equation 2 has those as valid solutions, while Equation 1 does not.
This shows that different equations can have different numbers of solutions.
3 answers