1.1.5 Solving Simultaneous Equations

In this topic, we will learn how to:

solving by substitution, a pair of simultaneous equations, of which one is linear and one is quadratic

To solve a simultaneous equation make one of the unknowns in the linear equation subject of the formula. Substitute this into the quadratic, then solve for the other unknown.

Let’s take a look at the following examples.

1. Use the method of substitution to solve the following equations, simultaneously,
$x + y + 1 = 0\ \ \ \ \ \ x^{2} + y^{2} = 25$ The first step is to make $x$ the subject of the formula in the linear equation,
$x + y + 1 = 0$ $\textcolor{#2192ff}{x = -y -1}$ Substitute $\textcolor{#2192ff}{x = -y -1}$ into the second equation,
$\textcolor{#2192ff}{x}^{2} + y^{2} = 25$ $(\textcolor{#2192ff}{-y - 1})^{2} + y^{2} = 25$ Expand the parentheses,
$\textcolor{#2192ff}{y^{2} + 2y + 1} + y^{2} = 25$ Group like terms,
$y^{2} + y^{2} + 2y + 1 - 25 = 0$ Simplify,
$2y^{2} + 2y - 24 = 0$ Divide the whole equation by $2$ to further simplify it,
$y^{2} + y - 12 = 0$ Solve the quadratic equation using your preferred method, in this case, we will factorise,
$(y + 4)(y - 3) = 0$ $y + 4 = 0\ \ \ \ \ \ y - 3 = 0$ Solve for $y$ in both equations, $y = -4\ \ \ \ \ \ y = 3$ Substitute these values of $y$ into the original linear equation to find the values of x. At $y = -4,$
$x + \textcolor{#2192ff}{y} + 1 = 0$ $x + (\textcolor{#2192ff}{-4}) + 1 = 0$ $x -4 + 1 = 0$ $x - 3 = 0$ $x = 3$ Therefore, at $y = -4$ , $x = 3$ . We can represent this as a set of coordinates,
$(3, -4)$ We do the same for $y = 3$ . At $y = 3$ ,
$x + \textcolor{#0f0}{y} + 1 = 0$ $x + (\textcolor{#0f0}{3}) + 1 = 0$ $x + 3 + 1 = 0$ $x + 4 = 0$ $x = -4$ Therefore, at $y = 3$ , $x = -4$ . We can represent this as a set of coordinates,
$(-4, 3)$ As a result, the final solution for this question is,
$(3, -4), (-4, 3)$ 2. Solve simultaneously by substitution the following equations,
$2x + 3y = 7\ \ \ \ \ \ 3x^{2} = 4 + 4xy$ The first step is to make $x$ the subject of the formula in the linear equation,
$2x + 3y = 7$ $2x = 7 - 3y$ $x = \frac{7}{2} -\frac{3y}{2}$ Substitute $\textcolor{#2192ff}{x = \frac{7}{2} -\frac{3y}{2}}$ into the second equation,
$3\textcolor{#2192ff}{x}^{2} = 4 + 4\textcolor{#2192ff}{x}y$ $3\left(\textcolor{#2192ff}{\frac{7}{2} -\frac{3y}{2}}\right)^{2} = 4 + 4\left(\textcolor{#2192ff}{\frac{7}{2} -\frac{3y}{2}}\right)y$ Expand the parentheses,
$3\left(\textcolor{#2192ff}{\frac{49}{4} - \frac{21y}{2} +\frac{9y^{2}}{4}}\right) = 4 + 14y - 6y^{2}$ $\frac{147}{4} - \frac{63y}{2} + \frac{27y^{2}}{4} = 4 + 14y - 6y^{2}$ Group like terms,
$\frac{147}{4} - 4 - \frac{63y}{2} - 14y + \frac{27y^{2}}{4} + 6y^{2} = 0$ Simplify,
$\frac{131}{4} - \frac{91y}{2} + \frac{51y^{2}}{4} = 0$ Multiply through by $4$ to get to rid of the denominators,
$131 -182y + 51y^{2} = 0$ Write the equation in the form $ax^{2} + bx + c = 0$ ,
$51y^{2} - 182y + 131 = 0$ Solve the quadratic equation using your preferred method, in this case we will factorise,
$(51y - 131)(y - 1) = 0$ $51y - 131 = 0\ \ \ \ \ \ y - 1 = 0$ Solve for $y$ in both equations,
$y = \frac{131}{51}\ \ \ \ \ \ y = 1$ Substitute these values of $y$ into the original linear equation to find the values of x. At $y = \frac{131}{51}$ ,
$2x + 3\textcolor{#2192ff}{y} = 7$ $2x + 3\left(\textcolor{#2192ff}{\frac{131}{51}}\right) = 7$ $2x + \frac{131}{17} = 7$ $2x = 7 - \frac{131}{17}$ $2x = -\frac{12}{17}$ $x = -\frac{6}{17}$ Therefore, at $y = \frac{131}{51}$ , $x = -\frac{6}{17}$ . We can represent this as a set of coordinates,
$\left(-\frac{6}{17}, \frac{131}{51}\right)$ We do the same for $y = 1$ . At $y = 1$ ,
$2x + 3\textcolor{#0f0}{y} = 7$ $2x + 3\left(\textcolor{#0f0}{1}\right) = 7$ $2x + 3 = 7$ $2x = 7 - 3$ $2x = 4$ $x = 2$ Therefore, at $y = 1$ , $x = 2$ . We can represent this as a set of coordinates,
$(2, 1)$ As a result, the final solution for this question is,
$\left(-\frac{6}{17}, \frac{131}{51}\right), (2, 1)$

GCE_AS_Level_Quadratics__Solving_Simultaneous_Equations Download