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:

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

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