To solve the system of equations using the substitution method, we start with the given equations:
- \( y = 2x - 3 \)
- \( y = x^2 + 6x - 3 \)
Since we have \( y \) expressed in terms of \( x \) in the first equation, we can substitute \( 2x - 3 \) for \( y \) in the second equation.
Substituting, we have:
\[ 2x - 3 = x^2 + 6x - 3 \]
Now, we will simplify and solve for \( x \):
- Add \( 3 \) to both sides:
\[ 2x = x^2 + 6x \]
- Rearranging gives us:
\[ 0 = x^2 + 6x - 2x \] \[ 0 = x^2 + 4x \]
- Factor the quadratic equation:
\[ 0 = x(x + 4) \]
Setting each factor to zero gives:
\[ x = 0 \quad \text{or} \quad x + 4 = 0 \Rightarrow x = -4 \]
Now, we can find the corresponding \( y \) values by substituting back into the first equation:
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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, the other solution is \( (-4, -11) \).
Thus, the solutions to the system are:
- \( (0, -3) \)
- \( (-4, -11) \)
Considering the possible responses given:
- (1, -1) and (2, 1)
- (-4, -11) and (0, -3)
- (-4, -11)
- (0, -3)
The correct choice is:
(-4, -11) and (0, -3).
1 answer