Quadratic simultaneous equations

In a nutshell

Simultaneous equations can have one linear equation and one quadratic equation. To solve these simultaneous equations you can use the substitution method. Simultaneous equations with one linear and one quadratic equation can have up to two pairs of solutions.

Solving quadratic simultaneous equations

To solve simultaneous equations with one linear and one quadratic equation, you rearrange the linear equation to create an equation for one of the unknown variables. By substituting this equation into the quadratic and solving the quadratic equation, you can get two values for the unknown. Substituting both these values into the linear equation will give two different values for the remaining unknown variable. 

Example

Solve the simultaneous equations:

​$$2x+y=1 \quad \textcircled{1}\\x^2 + y^2 = 1 \quad \textcircled{2}$$​​

Rearrange the linear equation:

$$y=1-2x \quad \textcircled{1}$$​​

Substitute this expression for $$y$$ into the quadratic:

$$x^2 + (1-2x)^2 = 1 \quad \textcircled{2}$$

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Solve for $$x$$:​

​$$\begin {aligned}x^2 + 1 - 4x + 4x^2 &= 1 \\5x^2-4x&=0 \\x(5x-4)&=0\\x=0,\ x&=\dfrac45\end {aligned}$$

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Substitute both values into the linear equation and solve for values of $$y$$​:

​$$x=0$$:​

​$$\begin {aligned}2(0)+y&=1\\y&=1\end {aligned}$$​​

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​$$x = \dfrac45$$:

$$\begin {aligned}2\bigg(\dfrac45\bigg)+y&=1\\y&=-\dfrac{3}{5}\end {aligned}$$

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Therefore, $$\underline{x=0,\ y=1}$$ or $$\underline{x=\dfrac45,\ y=-\dfrac35}$$.

Note: It's important to pair the solutions correctly.

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