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Solve the system of equations: x + y = 4, xy = 16.

Mathematics

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6 steps

To solve the system of equations:

x+y=4(1)x + y = 4 \quad (1) xy=16(2)xy = 16 \quad (2)

  1. Express yy in terms of xx from equation (1).

    From x+y=4x + y = 4, we get:

    y=4−xy = 4 - x

  2. Substitute this expression for yy into equation (2).

    x(4−x)=16x(4 - x) = 16

  3. Expand and rearrange the equation into a standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.

    4x−x2=164x - x^2 = 16 x2−4x+16=0x^2 - 4x + 16 = 0

  4. Solve the quadratic equation using the quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

    For x2−4x+16=0x^2 - 4x + 16 = 0, we have a=1a=1, b=−4b=-4, c=16c=16.

    x=−(−4)±(−4)2−4(1)(16)2(1)x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(16)}}{2(1)} x=4±16−642x = \frac{4 \pm \sqrt{16 - 64}}{2} x=4±−482x = \frac{4 \pm \sqrt{-48}}{2}

  5. Simplify the square root of the negative number.

    Recall that −1=i\sqrt{-1} = i and 48=16×3=43\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}.

    x=4±43i2x = \frac{4 \pm 4\sqrt{3}i}{2} x=2±23ix = 2 \pm 2\sqrt{3}i

    This gives two possible values for xx:

    x1=2+23ix_1 = 2 + 2\sqrt{3}i x2=2−23ix_2 = 2 - 2\sqrt{3}i

  6. Find the corresponding values for yy using y=4−xy = 4 - x.

    For x1=2+23ix_1 = 2 + 2\sqrt{3}i:

    y1=4−(2+23i)y_1 = 4 - (2 + 2\sqrt{3}i) y1=4−2−23iy_1 = 4 - 2 - 2\sqrt{3}i y1=2−23iy_1 = 2 - 2\sqrt{3}i

    For x2=2−23ix_2 = 2 - 2\sqrt{3}i:

    y2=4−(2−23i)y_2 = 4 - (2 - 2\sqrt{3}i) y2=4−2+23iy_2 = 4 - 2 + 2\sqrt{3}i y2=2+23iy_2 = 2 + 2\sqrt{3}i

    The solutions for the system of equations are:

    (x,y)=(2+23i,2−23i) and (2−23i,2+23i)\boxed{(x, y) = (2 + 2\sqrt{3}i, 2 - 2\sqrt{3}i) \text{ and } (2 - 2\sqrt{3}i, 2 + 2\sqrt{3}i)}
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