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Simultaneous Equations

Elimination Method

The elimination method for solving simultaneous equations is most often used in the scenario where you have a series of linear equation of the form $$ \begin{align} a_1x + a_2y + a_3z &= \alpha \\ b_1x + b_2y + b_3z &= \beta \\ c_1x + c_2y + c_3z &= \gamma \\ \end{align} $$

We proceed by subtracting multiples of one row from another until variables have coefficients of zero and hence disappear from the equation. For example we could eliminate $x$ from the third row using the operation $r_3 \rightarrow r_3 - \frac{a_1}{c_1}r_1$. This would give a new third equation of $$0x + \left(c_2 - \frac{a_1}{c_1}a_2\right)y + \left(c_3 - \frac{a_1}{c_1}a_3 \right)z = \gamma - \frac{a_1}{c_1}\alpha$$

With sufficient patience, this approach generalises to sets of equations of any size and leads nicely into the study of matrices.

Method of substitution

Given a pair of equations, $f(x,y) =p,\; g(x,y) =q$, our aim is to rearrange one of them into the form $x = u(y)$ (or $y = v(x)$). We then substitute u(y) into the second equation to get an equation in a single variable: $g(u(y),y) = q$. This equation is then solved and the value of the other variable calculated.

Elimination 1

Solve the following set of equations using the method of elimination. $$ \begin{align} 2x + 12y &= 18 \\ 5x + 4y &= 19 \end{align} $$
solution - press button to display

We shall eliminate $x$ from the second equation by constructing a new second equation $r_2 \rightarrow r_2 - \frac{5}{2}r_1$.

$$ \begin{align}2x + 12y &= 18 \\ 0x + (4- \frac{5}{2}(12))y &= 19 - \frac{5}{2}(18) \end{align}$$

Simplifying this yields $$ \begin{align} 2x + 12y &= 18\\ -26y &= -26 \end{align}$$

The new second equation gives us $y = 1$, substituting this value of $y$ into the first equation gives $2x + 12 = 18$ and hence $x = 3$

substitution 1

Solve the following pair of simultaneous equations, using the substitution method. $$ \begin{align} x +y &= 1 \\ x^2 + y^2 &= 4 \end{align} $$
solution - press button to display

In this scenario, we first rearrange on equation into the form $x=\dots$ or $y =\dots$:

From equation 1, we get $x = 1- y$. In the second equation, we now replace $x$ with $1-y$ to get

$$ \begin{align} (1-y)^2 + y^2 &= 4 \\ 1 - 2y + 2y^2 &= 4 \\ 2y^2 - 2y - 3 &= 0 \end{align}$$ We now have a quadratic equation to solve.

$$ \begin{align} y &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ &= \frac{2\pm \sqrt{(2)^2 - 4(2)(-3)}}{4} \\ &= \frac{2\pm \sqrt{28}}{4}\\ &= \frac{1\pm \sqrt{7}}{2} \end{align} $$ The corresponding $x$ values are given by $$ x = 1 - \frac{1\pm \sqrt{7}}{2} $$

Our two pairs of solutions are therefore $$ \left(\frac{1}{2}(1 -\sqrt{7}),\frac{1}{2}(1 +\sqrt{7})\right),\;\left(\frac{1}{2}(1 +\sqrt{7}),\frac{1}{2}(1 -\sqrt{7})\right) $$