We think you are located in South Africa. Is this correct?

Simultaneous Equations

2.8 Simultaneous equations (EMBFS)

Simultaneous linear equations can be solved using three different methods: substitution, elimination or using a graph to determine where the two lines intersect. For solving systems of simultaneous equations with linear and non-linear equations, we mostly use the substitution method. Graphical solution is useful for showing where the two equations intersect.

In general, to solve for the values of \(n\) unknown variables requires a system of \(n\) independent equations.

An example of a system of simultaneous equations with one linear equation and one quadratic equation is \begin{align*} y - 2x &= -4\\ x^2 + y &= 4 \end{align*}

Solving by substitution (EMBFT)

  • Use the simplest of the two given equations to express one of the variables in terms of the other.

  • Substitute into the second equation. By doing this we reduce the number of equations and the number of variables by one.

  • We now have one equation with one unknown variable which can be solved.

  • Use the solution to substitute back into the first equation to find the value of the other unknown variable.

Worked example 20: Simultaneous equations

Solve for \(x\) and \(y\): \begin{align*} y-2x &= -4 \qquad \ldots (1) \\ x^2 + y &= 4 \qquad \ldots (2) \end{align*}

Make \(y\) the subject of the first equation

\[y = 2x - 4\]

Substitute into the second equation and simplify

\begin{align*} x^2 + \left(2x-4\right) &= 4 \\ x^2 + 2x - 8 &= 0 \end{align*}

Factorise the equation

\begin{align*} \left(x+4\right)\left(x-2\right) &= 0 \\ \therefore x = -4 &\text{ or } x = 2 \end{align*}

Substitute the values of \(x\) back into the first equation to determine the corresponding \(y\)-values

If \(x = -4\): \begin{align*} y &= 2(-4)-4 \\ &= -12 \end{align*}

If \(x = 2\): \begin{align*} y &= 2(2) - 4 \\ &= 0 \end{align*}

Check that the two points satisfy both original equations

Write the final answer

The solution is \(x = -4 \text{ and } y = -12\) or \(x = 2 \text{ and } y = 0\). These are the coordinate pairs for the points of intersection as shown below.

b654f2d9c090834fa2da66b63c289670.png

Solving by elimination (EMBFV)

  • Make one of the variables the subject of both equations.

  • Equate the two equations; by doing this we reduce the number of equations and the number of variables by one.

  • We now have one equation with one unknown variable which can be solved.

  • Use the solution to substitute back into either original equation, to find the corresponding value of the other unknown variable.

Worked example 21: Simultaneous equations

Solve for \(x\) and \(y\): \begin{align*} y &= x^2 - 6x \qquad \ldots (1) \\ y + \frac{1}{2}x - 3 &= 0 \qquad \ldots (2) \end{align*}

Make \(y\) the subject of the second equation

\begin{align*} y + \frac{1}{2}x - 3 &= 0 \\ y &= -\frac{1}{2}x + 3 \end{align*}

Equate the two equations and solve for \(x\)

\begin{align*} x^2 - 6x &= -\frac{1}{2}x + 3 \\ x^2 - 6x + \frac{1}{2}x - 3 &= 0\\ 2x^2 - 12x + x - 6 &= 0\\ 2x^2 -11x - 6 &= 0\\ (2x +1)(x-6) &= 0\\ \text{Therefore } x = -\frac{1}{2}&\text{ or } x = 6 \end{align*}

Substitute the values for \(x\) back into the second equation to calculate the corresponding \(y\)-values

If \(x = -\dfrac{1}{2}\): \begin{align*} y &= -\frac{1}{2} \left( -\frac{1}{2} \right) + 3 \\ \therefore y &= 3\frac{1}{4} \end{align*} This gives the point \(\left(-\dfrac{1}{2};3\dfrac{1}{4}\right)\).

If \(x = 6\): \begin{align*} y &= -\frac{1}{2}(6) + 3 \\ &= -3 + 3 \\ \therefore y &= 0 \end{align*} This gives the point \((6;0)\).

Check that the two points satisfy both original equations

Write the final answer

The solution is \(x = -\dfrac{1}{2} \text{ and } y = 3\dfrac{1}{4}\) or \(x = 6 \text{ and } y = 0\). These are the coordinate pairs for the points of intersection as shown below.

52528403900da2fe061055f5d6d10b18.png

Worked example 22: Simultaneous equations

Solve for \(x\) and \(y\): \begin{align*} y &= \dfrac{5}{x - 2} \qquad \ldots (1)\\ y + 1 &= 2x \qquad \ldots (2) \end{align*}

Make \(y\) the subject of the second equation

\begin{align*} y + 1 &= 2x \\ y &= 2x - 1 \end{align*}

Equate the two equations and solve for \(x\)

\begin{align*} 2x - 1 &= \frac{5}{x - 2} \\ (2x - 1)(x - 2) &= 5 \\ 2x^2 - 5x + 2 &= 5 \\ 2x^2 - 5x - 3 &= 0 \\ (2x + 1)(x - 3) &= 0 \\ \text{Therefore } x = -\frac{1}{2}&\text{ or } x = 3 \end{align*}

Substitute the values for \(x\) back into the second equation to calculate the corresponding \(y\)-values

If \(x -\dfrac{1}{2}\): \begin{align*} y &= 2(-\frac{1}{2}) - 1 \\ \therefore y &= -2 \end{align*} This gives the point \((-\dfrac{1}{2};-2)\).

If \(x = 3\): \begin{align*} y &= 2(3) -1 \\ &= 5 \end{align*} This gives the point \((3;5)\).

Check that the two points satisfy both original equations

Write the final answer

The solution is \(x = -\dfrac{1}{2} \text{ and } y = -2\) or \(x = 3 \text{ and } y = 5\). These are the coordinate pairs for the points of intersection as shown below.

68ed42298fce291b00d3fa2af26547c9.png

Solving graphically (EMBFW)

  • Make \(y\) the subject of each equation.

  • Draw the graph of each equation on the same system of axes.

  • The final solutions to the system of equations are the coordinates of the points where the two graphs intersect.

Worked example 23: Simultaneous equations

Solve graphically for \(x\) and \(y\): \begin{align*} y + x^2 &= 1 \qquad \ldots (1) \\ y - x + 5 &= 0 \qquad \ldots (2) \end{align*}

Make \(y\) the subject of both equations

For the first equation we have

\begin{align*} y + x^2 &= 1 \\ y &= - x^2 + 1 \end{align*}

and for the second equation

\begin{align*} y - x + 5 &= 0 \\ y &= x - 5 \end{align*}

Draw the straight line graph and parabola on the same system of axes

4faaade382907e96b747d6431eae078b.png

Determine where the two graphs intersect

From the diagram we see that the graphs intersect at \((-3;-8)\) and \((2;-3)\).

Check that the two points satisfy both original equations

Write the final answer

The solutions to the system of simultaneous equations are \((-3;-8)\) and \((2;-3)\).

Solving simultaneous equations

Exercise 2.9

Solve the following systems of equations algebraically. Leave your answer in surd form, where appropriate.

\(y+x=5\)

\(y-{x}^{2}+3x-5=0\)

We make \(y\) the subject of each equation:

\begin{align*} y + x &= 5 \\ y &= 5 - x \end{align*} \begin{align*} y - x^{2} + 3x - 5 &= 0 \\ y &= x^{2} - 3x + 5 \end{align*}

Next equate the two equations and solve for \(x\):

\begin{align*} 5 - x &= x^{2} - 3x + 5 \\ 0 &= x^{2} - 2x \\ 0 &= x(x - 2) \\ \text{Therefore } x = 0 &\text{ or } x = 2 \end{align*}

Now we substitute the values for \(x\) back into the first equation to calculate the corresponding \(y\)-values

If \(x = 0\): \begin{align*} y &= 5 - 0 \\ \therefore y &= 5 \end{align*} This gives the point \((0;5)\).

If \(x = 2\): \begin{align*} y &= 5 - 2 \\ &= 3 \end{align*} This gives the point \((2;3)\).

The solution is \(x = 0 \text{ and } y = 5\) or \(x = 2 \text{ and } y = 3\). These are the coordinate pairs for the points of intersection.

\(y = 6 - 5x + x^2\)

\(y-x+1=0\)

We make \(y\) the subject of each equation:

\[y = 6 - 5x + x^2\] \begin{align*} y - x + 1 &= 0 \\ y &= x - 1 \end{align*}

Next equate the two equations and solve for \(x\):

\begin{align*} 6 - 5x + x^2 &= x - 1 \\ x^{2} - 6x + 7 &= 0 \\ x &= \dfrac{-(-6) \pm \sqrt{(-6)^{2} - 4(1)(7)}}{2(1)} \\ x &= \dfrac{6 \pm \sqrt{36 - 28}}{2} \\ x &= \dfrac{6 \pm \sqrt{8}}{2} \\ x &= \dfrac{6 \pm 2\sqrt{2}}{2} \\ x &= 3 \pm \sqrt{2} \end{align*}

Now we substitute the values for \(x\) back into the second equation to calculate the corresponding \(y\)-values

If \(x = 3 - \sqrt{2}\): \begin{align*} y &= 3 - \sqrt{2} - 1 \\ \therefore y &= 2 - \sqrt{2} \end{align*}

If \(x = 3 + \sqrt{2}\): \begin{align*} y &= 3 + \sqrt{2} - 1 \\ &= 2 + \sqrt{2} \end{align*}

The solution is \(x = 3 \pm \sqrt{2} \text{ and } y = 2 \pm \sqrt{2}\)

\(y=\dfrac{2x+2}{4}\)

\(y-2{x}^{2}+3x+5=0\)

We make \(y\) the subject of each equation:

\[y = \dfrac{2x+2}{4}\] \begin{align*} y - 2{x}^{2} + 3x + 5 &= 0 \\ y &= 2x^{2} - 3x - 5 \end{align*}

Next equate the two equations and solve for \(x\):

\begin{align*} \dfrac{2x+2}{4} &= 2x^{2} - 3x - 5 \\ 2x + 2 &= 8x^{2} - 12x - 20 \\ 0 &= 8x^{2} - 14x - 22 \\ 0 &= 4x^{2} - 7x - 11 \\ 0 &= (4x - 11)(x + 1) \\ \text{Therefore } x = -1 &\text{ or } x = \frac{11}{4} \end{align*}

Now we substitute the values for \(x\) back into the first equation to calculate the corresponding \(y\)-values

If \(x = -1\): \begin{align*} y &= \dfrac{2(-1) + 2}{4} \\ y &= 0 \end{align*} This gives the point \((-1;0)\).

If \(x = \frac{1}{4}\): \begin{align*} y &= \dfrac{2(\frac{1}{4}) + 2}{4} \\ y &= \dfrac{\frac{5}{2}}{4} \\ &= \frac{5}{8} \end{align*} This gives the point \((\frac{1}{4};\frac{5}{8})\).

The solution is \(x = -1 \text{ and } y = 0\) or \(x = \frac{1}{4} \text{ and } y = \frac{5}{8}\). These are the coordinate pairs for the points of intersection.

\(a - 2b - 3 = 0\); \(a-3{b}^{2}+4=0\)

We make \(a\) the subject of each equation:

\[a = 2b + 3\] \begin{align*} a - 3{b}^{2} + 4 &= 0 \\ a &= 3b^{2} - 4 \end{align*}

Next equate the two equations and solve for \(b\):

\begin{align*} 2b + 3 &= 3b^{2} - 4 \\ 0 &= 3b^{2} - 2b - 7 \\ 0 &= \dfrac{-(-2) \pm \sqrt{(-2)^{2} - 4(3)(-7)}}{2(3)} \\ b &= \dfrac{2 \pm \sqrt{4 + 84}}{6} \\ b &= \dfrac{2 \pm \sqrt{88}}{6} \end{align*}

Now we substitute the values for \(b\) back into the first equation to calculate the corresponding \(a\)-values

If \(b = \dfrac{2 + \sqrt{88}}{6}\): \begin{align*} a &= 2\left(\dfrac{2 + \sqrt{88}}{6}\right) + 3 \\ &= \dfrac{2 + \sqrt{88}}{3} + 3 \\ &= \dfrac{11 + \sqrt{88}}{3} \end{align*}

If \(b = \dfrac{2 - \sqrt{88}}{6}\): \begin{align*} a &= 2\left(\dfrac{2 - \sqrt{88}}{6}\right) + 3 \\ &= \dfrac{2 - \sqrt{88}}{3} + 3 \\ &= \dfrac{11 - \sqrt{88}}{3} \end{align*}

The solution is \(b = \dfrac{2 \pm \sqrt{88}}{6} \text{ and } a = \dfrac{11 \pm \sqrt{88}}{6}\)

\(x^2 - y + 2 = 3x\)

\(4x = 8 + y\)

We make \(y\) the subject of each equation:

\begin{align*} x^{2} + y + 2 &= 3x \\ y &= -x^{2} + 3x - 2 \end{align*} \[y = 4x - 8\]

Next equate the two equations and solve for \(x\):

\begin{align*} 4x - 8 &= -x^{2} + 3x - 2 \\ x^{2} + x - 6 &= 0 \\ (x + 3)(x - 2) & = 0 \\ \text{Therefore } x = -3 &\text{ or } x = 2 \end{align*}

Now we substitute the values for \(x\) back into the second equation to calculate the corresponding \(y\)-values

If \(x = -3\): \begin{align*} y &= 4(-3) - 8 \\ y &= -20 \end{align*} This gives the point \((-3;-20)\).

If \(x = 2\): \begin{align*} y &= 4(2) - 8 \\ &= 0 \end{align*} This gives the point \((2;0)\).

The solution is \(x = -3 \text{ and } y = -20\) or \(x = 2 \text{ and } y = 0\). These are the coordinate pairs for the points of intersection.

\(2y + x^2 + 25 = 7x\)

\(3x = 6y + 96\)

We make \(y\) the subject of each equation:

\begin{align*} 2y + x^2 + 25 &= 7x \\ 2y + x^2 + 25 &= 7x - x^{2} + 25 \\ y &= \frac{-x^{2} + 7x + 25}{2} \end{align*} \begin{align*} 2y & = x - 32 \\ y & = \frac{x - 32}{2} \end{align*}

Next equate the two equations and solve for \(x\):

\begin{align*} \frac{x - 32}{2} &= \frac{-x^{2} + 7x + 25}{2} \\ x - 32 &= -x^{2} + 7x + 25 \\ x^{2} - 6x - 57 & = 0 \\ x & = \dfrac{-(-6) \pm \sqrt{(-6)^{2} - 4(1)(-57)}}{2(1)} \\ & = \dfrac{6 \pm \sqrt{36 + 228}}{2} \\ & = \dfrac{6 \pm \sqrt{264}}{2} \end{align*}

Now we substitute the values for \(x\) back into the second equation to calculate the corresponding \(y\)-values

If \(x = \dfrac{6 + \sqrt{264}}{2}\): \begin{align*} 2y &= \dfrac{6 + \sqrt{264}}{2} - 32 \\ y &= \dfrac{70 + \sqrt{264}}{4} \\ &= \dfrac{70 + \sqrt{264}}{4} \end{align*}

If \(x = \dfrac{6 - \sqrt{264}}{2}\): \begin{align*} 2y &= \dfrac{6 - \sqrt{264}}{2} - 32 \\ y &= \dfrac{70 - \sqrt{264}}{4} \\ &= \dfrac{70 - \sqrt{264}}{4} \end{align*}

The solution is \(x = \dfrac{6 \pm \sqrt{264}}{2} \text{ and } y = \dfrac{70 \pm \sqrt{264}}{2}\)

Solve the following systems of equations graphically. Check your solutions by also solving algebraically.

\({x}^{2}-1-y=0\)

\(y+x-5=0\)

Make \(y\) the subject of both equations

For the first equation we have:

\begin{align*} x^{2} - 1 - y &= 0 \\ y &= x^2 - 1 \end{align*}

and for the second equation:

\begin{align*} y + x - 5 &= 0 \\ y &= -x + 5 \end{align*}

Draw the straight line graph and parabola on the same system of axes:

74cfd605ddbad5ba8c750fd3b9152cb2.png

From the diagram we see that the graphs intersect at \((-3;8)\) and \((2;3)\).

We can solve algebraically to check. Doing this we find the same solution.

The solutions to the system of simultaneous equations are \((-3;8)\) and \((2;3)\).

\(x+y-10=0\)

\({x}^{2}-2-y=0\)

Make \(y\) the subject of both equations

For the first equation we have:

\begin{align*} x + y - 10 &= 0 \\ y &= -x + 10 \end{align*}

and for the second equation:

\begin{align*} x^{2} - 2 - y &= 0 \\ y &= x^{2} - 2 \end{align*}

Draw the straight line graph and parabola on the same system of axes:

From the diagram we see that the graphs intersect at \((-4;14)\) and \((3;7)\).

We can solve algebraically to check. Doing this we find the same solution.

The solutions to the system of simultaneous equations are \((-4;14)\) and \((3;7)\).

\(xy = 12\)

\(7 = x+y\)

Make \(y\) the subject of both equations

For the first equation we have:

\begin{align*} xy &= 12 \\ y &= \frac{12}{x} \end{align*}

and for the second equation:

\begin{align*} 7 &= x + y \\ y &= 7 - x \end{align*}

Draw the two graphs on the same system of axes:

We also note that this sytem of equations has the followig restrictions: \(x \ne 0\) and \(y \ne 0\)

From the diagram we see that the graphs intersect at \((3;4)\) and \((4;3)\).

We can solve algebraically to check. Doing this we find the same solution.

The solutions to the system of simultaneous equations are \((3;4)\) and \((4;3)\).

\(6-4x-y=0\)

\(12-2{x}^{2}-y=0\)

Make \(y\) the subject of both equations

For the first equation we have:

\begin{align*} 6 - 4x - y &= 0 \\ y &= -4x + 6 \end{align*}

and for the second equation:

\begin{align*} 12 - 2x^{2} - y &= 0 \\ y &= -2x^{2} + 12 \end{align*}

Draw the two graphs on the same system of axes:

From the diagram we see that the graphs intersect at approximately \((4;-10)\) and \((-2;14)\).

We can solve algebraically to check (and to get a more accurate answer). Doing this gives:

\begin{align*} -4x + 6 &= -2x^{2} + 12 \\ 2x - 3 &= x^{2} - 6 \\ x^{2} - 2x - 9 & = 0 \\ x & = \dfrac{-(-2) \pm \sqrt{(-2)^{2} - 4(1)(-9)}}{2(1)} \\ x & = \dfrac{2 \pm \sqrt{(4 + 36}}{2} \\ x & = \dfrac{2 \pm \sqrt{40}}{2} \end{align*}

Solving for \(y\) gives:

\begin{align*} y & = -4\left(\dfrac{2 \pm \sqrt{40}}{2} \right) + 6 \\ &= -2(2 \pm \sqrt{40}) + 6 \\ & = -4 \pm 2\sqrt{40} + 6 \\ & = 2 \pm 2\sqrt{40} \end{align*}