ion **Substitution** means replacing a variable with an equal expression. The **subject** of an equation is the variable isolated on one side; in $x = y - 1$, $x$ is the subject. Use the linear equation to replace one variable in the quadratic equation. This gives you one quadratic equation in one variable. A quadratic in one variable is in standard form when it is written as $ax^2 + bx + c = 0$, where $a \neq 0$. **Factorising** means rewriting an expression as a product of brackets. The **roots** of an equation are its solutions. 1. Solve the simultaneous equations: $$ \begin{aligned} x^2 + y^2 &= 25 \\ x &= y - 1 \end{aligned} $$ 2. Substitute $x = y - 1$ into $x^2 + y^2 = 25$. $$ (y - 1)^2 + y^2 = 25 $$ 3. Expand the bracket. $$ y^2 - 2y + 1 + y^2 = 25 $$ 4. Rearrange into standard form, then divide by 2. $$ \begin{aligned} 2y^2 - 2y - 24 &= 0 \\ y^2 - y - 12 &= 0 \end{aligned} $$ 5. Factorise. $$ (y - 4)(y + 3) = 0 $$ 6. Therefore $y = 4$ or $y = -3$. 7. Use $x = y - 1$ to find the matching $x$-values: if $y = 4$, then $x = 3$; if $y = -3$, then $x = -4$. 8. The solutions are $(3, 4)$ and $(-4, -3)$. Writing separate lists of $x$-values and $y$-values is not enough unless the pairings are clear. Always give coordinate pairs. ## 3. When the linear equation needs rearranging Sometimes the linear equation is not already written as $x = \dots$ or $y = \dots$. Rearrange it first, choosing the variable that looks easiest to substitute. 1. Solve: $$ \begin{aligned} x^2 + y^2 &= 25 \\ 3x + y &= 5 \end{aligned} $$ 2. Make $y$ the subject of the linear equation. $$ y = 5 - 3x $$ 3. Substitute into the quadratic equation and simplify. $$ \begin{aligned} x^2 + (5 - 3x)^2 &= 25 \\ x^2 + 25 - 30x + 9x^2 &= 25 \\ 10x^2 - 30x &= 0 \end{aligned} $$ 4. Factorise and solve. $$ 10x(x - 3) = 0 $$ 5. So $x = 0$ or $x = 3$. 6. Use $y = 5 - 3x$: if $x = 0$, then $y = 5$; if $x = 3$, then $y = -4$. 7. The solutions are $(0, 5)$ and $(3, -4)$. When substituting $y = 5 - 3x$, write $(5 - 3x)^2$. Do not treat it as $5 - 3x^2$. ## 4. When the quadratic does not factorise nicely If factorising is awkward, use the **quadratic formula**. A **coefficient** is the number multiplying a term; in $5x^2 - 16x - 2 = 0$, the coefficients are $a = 5$, $b = -16$ and $c = -2$. For $ax^2 + bx + c = 0$: $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ The expression $b^2 - 4ac$ is the **discriminant**. If it is positive, you get two real roots; if it is zero, you get one repeated root. If the discriminant $b^2 - 4ac$ is negative, the square root is not a real number. That means the line does not meet the curve in real coordinate pairs. 1. Solve, giving answers to 3 significant figures: $$ \begin{aligned} x^2 + y^2 &= 18 \\ 2x + y &= 4 \end{aligned} $$ 2. Make $y$ the subject. $$ y = 4 - 2x $$ 3. Substitute and collect terms. $$ \begin{aligned} x^2 + (4 - 2x)^2 &= 18 \\ x^2 + 16 - 16x + 4x^2 &= 18 \\ 5x^2 - 16x - 2 &= 0 \end{aligned} $$ 4. Use the quadratic formula. $$ x = \frac{16 \pm \sqrt{296}}{10} $$ 5. Calculator values give $x \approx 3.320\ldots$ or $x \approx -0.120\ldots$. 6. Use $y = 4 - 2x$ with the unrounded values: $y \approx -2.640\ldots$ or $y \approx 4.240\ldots$. 7. To 3 significant figures, the solutions are $(3.32, -2.64)$ and $(-0.120, 4.24)$.