Question
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Answer
The solutions are (1, 3) and (−4, −2).
Solution
We start with the given equations:
(1) y = x + 2
(2) x + x·y = 4
Substitute equation (1) into (2) by replacing y with (x + 2):
x + x(x + 2) = 4
Now, simplify the equation:
x + x² + 2x = 4
x² + 3x = 4
Subtract 4 from both sides to set the equation to 0:
x² + 3x − 4 = 0
Next, solve the quadratic equation x² + 3x − 4 = 0. We can use the quadratic formula:
x = [−b ± √(b² − 4ac)] / (2a)
Here, a = 1, b = 3, and c = −4. Compute the discriminant:
b² − 4ac = 3² − 4(1)(−4) = 9 + 16 = 25
Then,
x = [−3 ± √25] / 2
x = [−3 ± 5] / 2
This gives two solutions:
x = (−3 + 5) / 2 = 2 / 2 = 1
x = (−3 − 5) / 2 = (−8) / 2 = −4
Now, substitute each value of x back into equation (1) to find y.
For x = 1:
y = 1 + 2 = 3
For x = −4:
y = −4 + 2 = −2
Thus, the solutions to the system of equations are:
(x, y) = (1, 3) and (x, y) = (−4, −2).
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Simplify this solution The Deep Dive
To substitute \( y = x + 2 \) into the equation \( x + x y = 4 \), we first rewrite the second equation: Replace \( y \) with \( x + 2 \): \[ x + x(x + 2) = 4 \] This expands to: \[ x + x^2 + 2x = 4 \] Combining like terms gives: \[ x^2 + 3x - 4 = 0 \] Next, you can factor or use the quadratic formula to find the values of \( x \). Would you like to continue with that?
