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,
$$\left(-\frac{6}{17}, \frac{131}{51}\right), (2, 1)$$