To solve the given system of equations using the substitution method, we have the following equations:
- \( y = 2x - 3 \)
- \( y = x^2 + 6x - 3 \)
Since both equations are equal to \( y \), we can set them equal to each other:
\[ 2x - 3 = x^2 + 6x - 3 \]
Next, we can simplify this equation. First, add 3 to both sides to eliminate -3:
\[ 2x = x^2 + 6x \]
Now, subtract \( 2x \) from both sides:
\[ 0 = x^2 + 6x - 2x \]
This simplifies to:
\[ 0 = x^2 + 4x \]
Now, we can factor the right side:
\[ 0 = x(x + 4) \]
Setting each factor equal to zero gives us the possible solutions for \( x \):
- \( x = 0 \)
- \( x + 4 = 0 \) which gives \( x = -4 \)
Now that we have the values for \( x \), we can substitute these values back into the first equation, \( y = 2x - 3 \), to find the corresponding values of \( y \).
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For \( x = 0 \): \[ y = 2(0) - 3 = -3 \] So, one solution is \( (0, -3) \).
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For \( x = -4 \): \[ y = 2(-4) - 3 = -8 - 3 = -11 \] So, another solution is \( (-4, -11) \).
Thus, the solutions to the system of equations are:
\[ (0, -3) \quad \text{and} \quad (-4, -11) \]
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