# Simultaneous Equations

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## Simultaneous Equations

Simultaneous equations are equations that share the same variables. There will be a solution, or solutions, that work for all equations. In A level maths, you will only see simultaneous equations in two variables, e.g. $x$ and $y$.

There are 2 main methods used to solve simultaneous equations.

A Level   ## Method 1: Elimination

The method of elimination is used when there are two linear simultaneous equations. We eliminate one variable by subtracting one equation from the other.

Example: Solve the following equations:

\begin{aligned} 3x + 7y -44 &= 0 \\ 2x + 3y &= 21 \end{aligned}

Step 1: Write both equations in the form $ax+by=c$, if necessary:

\begin{aligned} 3x + 7y &= 44 \\ 2x + 3y &= 21 \end{aligned}

Step 2: Manipulate the equations so that the coefficients match – multiply the equations to make either the $x$‘s or $y$‘s equal in size (ignoring the signs). Always multiply to get the LCM of the coefficients.

$\,(\times2) \,\,\,\,\,\,\,\,\,3x + 7y = 44\,\,\, \Rightarrow \,\,\, 6x + 14y = 88$

$(\times3)\,\,\,\,\,\,\,\,\, 2x+3y = 21 \,\,\, \Rightarrow \,\,\, 6x+9y=63$

Step 3: Add or subtract the equations to eliminate variable that has terms with equal coefficients, so that you can find the other variable.

In this case, both equations have $+6x$, so we need to subtract:

\begin{aligned}6x + 14y &= 88 \\ (-)\,\,\,\,\,\,\,\,\, 6x+9y&=63 \\ \hline 5y&=25\end{aligned}

Step 4: Solve the resulting equation.

In this case, we need to solve the equation to find $y$

\begin{aligned}(\div 5)\,\,\,\,\,\,\,\,\,5y&=25 \\ \textcolor{red}{y} &\textcolor{red}{= 5} \end{aligned}

Step 5: Find the variable that you eliminated.

In this case, replace (substitute) $y = 5$ into one of the equations and solve to find $x$:

\begin{aligned} 2x + 3y &= 21 \\ 2x + 3(5) &= 21 \\ 2x + 15 &= 21 \\ 2x &= 6 \\ \textcolor{blue}{x} &\textcolor{blue}{= 3} \end{aligned}

Hence,

$\textcolor{blue}{x = 3}$  and  $\textcolor{red}{y = 5}$

A Level   ## Method 2: Substitution

We use the method of substitution when one of the simultaneous equations is quadratic (non-linear), since we can’t use the method of elimination.

Example: Solve

\begin{aligned} x^2 + 2y &= 31 \\ x+y &= 14 \end{aligned}

Step 1: Rearrange the linear equation so that one of the variables is on its own (in this case it will be easier to get $y$ on its own so that we don’t have to do any squaring of brackets)

\begin{aligned} (-x) \,\,\,\,\,\, x+y &= 14 \\ y &= -x + 14 \end{aligned}

Step 2: Substitute this variable into the quadratic equation, so that there is an equation in only one variable.

$y = -x + 14$ so replace $y$ with $-x + 14$:

$x^2 + 2(-x+14) = 31$

Step 3: Expand and solve to find the values for one variable.

\begin{aligned} x^2 + 2(-x+14) &= 31 \\ x^2 - 2x + 28 &= 31 \\ x^2 - 2x - 3 &= 0 \\ (x-3)(x+1) &= 0 \end{aligned}

Hence,

$\textcolor{red}{x = 3}$  and  $\textcolor{red}{x = -1}$

Step 4: Substitute these values into the linear equation (since this will be easier) and solve to find the corresponding values of the other variable.

When $\textcolor{red}{x = 3}$,   $\textcolor{blue}{y} = -3+14 = \textcolor{blue}{11}$

When $\textcolor{red}{x = -1}$,   $\textcolor{blue}{y} = -(-1) + 14 = \textcolor{blue}{15}$

So, there are two pairs of solutions

$\textcolor{red}{x=3}$, $\textcolor{blue}{y = 11}$  and  $\textcolor{red}{x = -1}$, $\textcolor{blue}{y = 15}$

Note: You may need to use the quadratic formula if you get a quadratic equation that is too difficult to solve by factorising.

A Level   ## Interpreting Simultaneous Equations Geometrically

To interpret simultaneous equations geometrically, we need to draw a sketch of the two functions and describe what we see.

The number of solutions is equal to the number of intersections between the graphs:

Two solutions – the graphs intersect twice

One solution – the graphs meet at a single point – the graph is a tangent to the curve at this point

No solutions – the graphs do not intersect A Level   A Level   ## Example: Interpreting Simultaneous Equations Geometrically

Interpret the following geometrically:

$y = x^2 - 3x$  and  $y = x - 4$

[4 marks] Substitute $y=x-4$ into $y = x^2 - 3x$, and then solve for $x$:

\begin{aligned} x - 4 &= x^2 - 3x \\ x^2 - 4x + 4 &= 0 \\ (x-2)^2 &= 0 \\ \textcolor{red}{x} &\textcolor{red}{= 2} \end{aligned}

Then, substitute $\textcolor{red}{x = 2}$ into $y = x - 4$ and solve for $y$:

\begin{aligned} y &= x - 4 \\ y &= 2 - 4 \\ \textcolor{blue}{y} &\textcolor{blue}{= -2} \end{aligned}

Hence, there is only one solution $\textcolor{red}{x = 2}$, $\textcolor{blue}{y = -2}$

Therefore, the graphs will meet at a single point: $(\textcolor{red}{2}, \textcolor{blue}{-2})$

So, the straight line is a tangent to the curve at the point $(\textcolor{red}{2},\textcolor{blue}{-2})$

A Level   ## Example Questions

Multiply the first equation by $2$ so that they coefficients of $y$ match:

\begin{aligned} 8x + 2y &= 36 \\ 3x + 2y &= 21 \end{aligned}

Subtract the second equation from the first, to eliminate the $y$ variable:

$5x = 15$

And then solve:

$x = 3$

Then, substitute $x =3$ into either equation and solve:

\begin{aligned} 4(3) + y &= 18 \\ 12 + y &= 18 \\ y &= 6 \end{aligned}

Hence,

$x = 3$  and  $y = 6$

Rearrange $x-y=2$ so that $x$ is on its own:

$x = y+2$

Then, substitute this into $x^2 + y^2 = 100$ and expand and solve for $y$:

\begin{aligned} (y+2)^2 + y^2 &= 100 \\ y^2 + 4y + 4 + y^2 &= 100 \\ 2y^2 + 4y - 96 &= 0 \\ (2y-12)(y+8) &= 0 \end{aligned}

Hence, $2y - 12 = 0$  and  $y + 8 = 0$ $\Rightarrow y = 6$  and  $y = -8$

Substitute these values into the non-linear equation, to find the values of $x$:

$y = 6 \Rightarrow x = 6+2 = 8$

$y = -8 \Rightarrow x = -8 + 2 = -6$

Hence, the points of intersection are $(8, 6)$ and $(-6, -8)$

Substitute $y = x$ into $y = x^2 + 5x + 5$, and solve:

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

You cannot get a real number from square rooting a negative number. Therefore there are no real solutions.

You could have found the discriminant of $x^2 +4x + 5$:

$b^2 - 4ac = 4^2 - 4 \times 1 \times 5 = 16 - 20 = -4$

The discriminant is $<0$, so there are no real roots, and therefore no real solutions.

A Level

A Level

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