# Simultaneous Equations

## Simultaneous Equations Revision

**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.

**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}

**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.

**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**

**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})

## Simultaneous Equations Example Questions

**Question 1:** Solve 4x + y = 18 and 3x + 2y = 21

**[4 marks]**

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

**Question 2:** Find the coordinates of the point of intersection of x^2 + y^2 = 100 and x-y=2.

**[6 marks]**

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)

**Question 3:** Show that the pair of equations y = x and y = x^2 + 5x + 5 have no real solutions.

**[3 marks]**

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.

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