Answer :
Answer:
one solution
Explanation:
a1/a2 is not equals to b1/b2
Explanation:
To determine how many solutions the system of equations \( 2x + 3y = 6 \) and \( 3x + 4y = -12 \) has, we can analyze the equations together.
First, let's express \( y \) from the first equation:
\[ 2x + 3y = 6 \]
\[ 3y = 6 - 2x \]
\[ y = \frac{6 - 2x}{3} \]
\[ y = 2 - \frac{2x}{3} \]
Now substitute \( y \) in the second equation:
\[ 3x + 4\left(2 - \frac{2x}{3}\right) = -12 \]
Expand and simplify:
\[ 3x + 8 - \frac{8x}{3} = -12 \]
To eliminate the fraction, multiply through by 3:
\[ 9x + 24 - 8x = -36 \]
\[ x + 24 = -36 \]
\[ x = -60 \]
Now substitute \( x = -60 \) back into the expression for \( y \):
\[ y = 2 - \frac{2(-60)}{3} \]
\[ y = 2 + 40 \]
\[ y = 42 \]
Thus, the solution to the system is \( x = -60 \) and \( y = 42 \).
To confirm there is exactly one solution, note that both equations are linear and have different slopes (\( -\frac{2}{3} \) for the first and \( -\frac{3}{4} \) for the second), indicating they intersect at exactly one point.
Therefore, the system of equations \( 2x + 3y = 6 \) and \( 3x + 4y = -12 \) has 1
solution.
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