Quadratics – Solving simultaneous equations (one linear and one quadratic)

Given one linear and one quadratic equation:

1. Rearrange the linear equation to make \(x=\) or \(y=\).
2. Substitute into the quadratic to form a single equation.
3. Solve the quadratic (factor or quadratic formula).
4. Substitute values back into the linear equation to find the other variable.

Solve:

\[ y = x + 1 \qquad\text{and}\qquad y = x^2 - 3 \]

Step 1: Substitute \(y = x + 1\) into \(y = x^2 - 3\):

\[ x + 1 = x^2 - 3 \Rightarrow x^2 - x - 4 = 0 \]

Step 2: Solve the quadratic:

\[ x = \frac{1 \pm \sqrt{1 + 16}}{2} = \frac{1 \pm \sqrt{17}}{2} \]

Step 3: Substitute into \(y = x + 1\):

\[ y = \frac{1 \pm \sqrt{17}}{2} + 1 = \frac{3 \pm \sqrt{17}}{2} \]

Solve:

\[ 2x + y = 4 \qquad\text{and}\qquad y = x^2 + x - 6 \]

Step 1: Rearrange the linear equation:

\[ y = 4 - 2x \]

Step 2: Substitute into the quadratic:

\[ 4 - 2x = x^2 + x - 6 \Rightarrow x^2 + 3x - 10 = 0 \]

Step 3: Solve:

\[ (x + 5)(x - 2) = 0 \Rightarrow x = -5,\ x = 2 \]

Step 4: Substitute into \(y = 4 - 2x\):

For \(x = 2\): \(y = 0\) For \(x = -5\): \(y = 14\)

• Always solve the quadratic fully — expect two solutions.
• If answers look messy (with square roots), they are usually correct.
• Factor if possible, otherwise use the quadratic formula.
• Solutions show points where the line and the curve intersect.
• If no real solutions, the line does not meet the curve.

Solve each pair:

1. \(y = 3x + 2\), \(y = x^2 - 5x + 4\)
2. \(x + y = 1\), \(y = x^2 - 4\)
3. \(2y - x = 1\), \(y = x^2 + 3x\)

Give answers in coordinate form: \((x, y)\).