Simultaneous equations

Simultaneous equations describe quantities that must satisfy two or more conditions at the same time. A solution is therefore a set of values that makes every equation true.

For two variables, each equation represents a graph. The simultaneous solutions are precisely the points where the graphs intersect. This idea explains why a pair may have one solution, several solutions or no real solution.

Before you start

You should be able to:

• solve linear equations;
• expand and factorise quadratic expressions;
• use the quadratic formula;
• substitute values into expressions;
• recognise equations of straight lines and quadratics.

If any of these feel uncertain, review quadratics and functions first.

Linear simultaneous equations

Consider

$$\begin{aligned} 2x+y&=8,\\ x-y&=1. \end{aligned}$$

The solution must lie on both lines. Since two distinct non-parallel straight lines meet once, this system has one solution.

Elimination

Elimination combines the equations so that one variable disappears. Add the equations above:

$$\begin{aligned} 2x+y&=8,\\ x-y&=1,\\ \hline 3x&=9. \end{aligned}$$

Hence $x=3$. Substitute this into either original equation:

$3-y=1 \quad\Longrightarrow\quad y=2.$

The solution is $(x,y)=(3,2)$.

Always check it in both original equations:

$2(3)+2=8, \qquad 3-2=1.$

Both statements are true, so the solution is valid.

Making coefficients match

The variables will not always cancel immediately. Solve

$$\begin{aligned} 3x+2y&=7,\\ 5x-3y&=4. \end{aligned}$$

The lowest common multiple of $2$ and $3$ is $6$. Multiply the first equation by $3$ and the second by $2$:

$$\begin{aligned} 9x+6y&=21,\\ 10x-6y&=8. \end{aligned}$$

Adding gives

$19x=29 \quad\Longrightarrow\quad x=\frac{29}{19}.$

Use $3x+2y=7$ to find $y$:

$$3\left(\frac{29}{19}\right)+2y=7 \quad\Longrightarrow\quad 2y=\frac{46}{19} \quad\Longrightarrow\quad y=\frac{23}{19}.$$

Therefore

$\boxed{(x,y)=\left(\frac{29}{19},\frac{23}{19}\right)}.$

Fractions are perfectly legitimate solutions. Do not assume that an awkward value signals an error.

Substitution

Substitution is usually efficient when one equation already isolates a variable. Solve

$$\begin{aligned} y&=3x-5,\\ 2x+y&=11. \end{aligned}$$

Replace $y$ in the second equation with $3x-5$:

$$2x+(3x-5)=11.$$

Then

$5x=16 \quad\Longrightarrow\quad x=\frac{16}{5}.$

Now find $y$:

$$y=3\left(\frac{16}{5}\right)-5 =\frac{48}{5}-\frac{25}{5} =\frac{23}{5}.$$

So $(x,y)=\left(\frac{16}{5},\frac{23}{5}\right)$.

Which method should you choose?

• Use elimination when coefficients already match, or can be made to match easily.
• Use substitution when one variable is isolated, or has coefficient $1$ or $-1$.
• Choose the method that reduces algebraic complexity. Both methods are mathematically equivalent.

What the number of solutions means

For two linear equations:

• different gradients give one intersection and one solution;
• equal gradients with different intercepts give parallel lines and no solution;
• the same line written in different forms gives infinitely many solutions.

For example,

$$\begin{aligned} 2x+4y&=6,\\ x+2y&=5 \end{aligned}$$

has no solution. Doubling the second equation gives $2x+4y=10$, which contradicts $2x+4y=6$.

By contrast,

$$\begin{aligned} 2x+4y&=6,\\ x+2y&=3 \end{aligned}$$

has infinitely many solutions because the first equation is exactly twice the second. Elimination produces $0=0$, confirming that the equations describe the same line.

One linear and one quadratic equation

A line and a quadratic curve can meet twice, once, or not at all. Substitution turns the pair into a quadratic equation.

Solve

$$\begin{aligned} y&=x+1,\\ y&=x^2-3x+1. \end{aligned}$$

At an intersection, both expressions for $y$ are equal:

$x+1=x^2-3x+1.$

Rearrange and solve:

$$x^2-4x=0 \quad\Longrightarrow\quad x(x-4)=0.$$

Therefore $x=0$ or $x=4$. Substitute each value into $y=x+1$:

$$x=0 \Longrightarrow y=1, \qquad x=4 \Longrightarrow y=5.$$

There are two solutions:

$\boxed{(0,1)\text{ and }(4,5)}.$

Notice that each $x$ value must be paired with its corresponding $y$ value. Writing separate lists of $x$ and $y$ values can accidentally create false pairings.

A quadratic that does not factorise

Solve

$$\begin{aligned} y&=2x-1,\\ x^2+y^2&=10. \end{aligned}$$

Substitute $y=2x-1$ into the second equation:

$$x^2+(2x-1)^2=10.$$

Expand carefully:

$$x^2+4x^2-4x+1=10 \quad\Longrightarrow\quad 5x^2-4x-9=0.$$

Using the quadratic formula,

$$x=\frac{4\pm\sqrt{(-4)^2-4(5)(-9)}}{10} =\frac{4\pm14}{10}.$$

Thus $x=\frac95$ or $x=-1$. Since $y=2x-1$:

$$x=\frac95 \Longrightarrow y=\frac{13}{5}, \qquad x=-1 \Longrightarrow y=-3.$$

The solutions are

$\boxed{\left(\frac95,\frac{13}{5}\right)\text{ and }(-1,-3)}.$

Tangency and the discriminant

Suppose the line $y=mx+c$ is substituted into a curve and produces

$ax^2+bx+d=0.$

The discriminant $\Delta=b^2-4ad$ reveals the number of real intersections:

$$\begin{array}{c|c} \Delta>0 & \text{two distinct intersections}\\ \Delta=0 & \text{one repeated intersection, so the line is tangent}\\ \Delta<0 & \text{no real intersections} \end{array}$$

Finding a parameter for tangency

Find the values of $k$ for which the line $y=kx+1$ is tangent to the parabola $y=x^2-3x+4$.

At an intersection,

$kx+1=x^2-3x+4,$

so

$x^2-(k+3)x+3=0.$

Tangency requires a repeated root, so set the discriminant equal to zero:

$$(k+3)^2-4(1)(3)=0.$$

Hence

$$(k+3)^2=12 \quad\Longrightarrow\quad k+3=\pm2\sqrt3.$$

Therefore

$\boxed{k=-3\pm2\sqrt3}.$

There are two possible tangents because two lines in this family touch the parabola.

Modelling with simultaneous equations

Translate each condition into an equation before solving. State what your variables represent, preserve units, and reject solutions that are impossible in context.

Example: dimensions

A rectangle has perimeter $34$ cm and area $60$ cm$^2$. Find its dimensions.

Let the length be $x$ cm and the width be $y$ cm. Then

$$2x+2y=34, \qquad xy=60.$$

The first equation gives $y=17-x$. Substitute into the area equation:

$$x(17-x)=60.$$

Therefore

$$x^2-17x+60=0 \quad\Longrightarrow\quad (x-5)(x-12)=0.$$

So $x=5$ or $x=12$. The corresponding pairs are $(5,12)$ and $(12,5)$, which describe the same rectangle. Its dimensions are

$\boxed{5\text{ cm by }12\text{ cm}}.$

Example: rejecting a mathematical solution

The height of an object after $t$ seconds is $h=20t-5t^2$ metres. Find when it is at height $15$ m.

Solve simultaneously with $h=15$:

$$20t-5t^2=15 \quad\Longrightarrow\quad t^2-4t+3=0.$$

Thus $(t-1)(t-3)=0$, so $t=1$ or $t=3$. Both values are non-negative and lie during the motion, so both are valid. The object passes 15 m once while rising and once while falling.

If the same algebra produced $t=-1$, that root would solve the equations but would normally be rejected because the model begins at $t=0$.

Common mistakes

Changing only part of an equation

When multiplying an equation, multiply every term on both sides. From $2x+3y=7$, multiplying by $4$ gives $8x+12y=28$.

Losing a minus sign during elimination

If subtracting one equation from another, use brackets or write the subtraction term by term. Adding equations with opposite coefficients is often safer.

Forgetting the second coordinate

Solving the resulting equation gives one variable only. Substitute back to find the other variable and report ordered pairs.

Dividing by an expression that might be zero

Dividing by $x$ can lose the solution $x=0$. Factorise instead. For example, $x^2=4x$ should become $x(x-4)=0$, giving $x=0$ or $x=4$.

Accepting every algebraic root in a model

Check restrictions such as positive length, non-negative time, integer counts and the stated domain. Rejection needs a contextual reason, not merely that a value looks inconvenient.

Rounding too early

Keep exact fractions and surds throughout the calculation. Round only the final answer, and only when the question requests it.

A reliable solving routine

1. Identify the type of equations and choose elimination or substitution.
2. Reduce the system to one equation in one unknown.
3. Solve completely, remembering that a quadratic may give two roots.
4. Substitute each root back to find its matching second coordinate.
5. Check every ordered pair in both original equations.
6. Apply any domain or modelling restrictions.
7. Present exact values unless a stated degree of accuracy is required.

Check your understanding

1. Solve $3x+y=13$ and $2x-y=2$.
2. Solve $2x+3y=1$ and $5x-2y=16$.
3. Solve $y=x+2$ and $y=x^2-4$.
4. Determine the number of real solutions of $y=2x+5$ and $y=x^2+2x+8$ without finding the intersection points.
5. Find $k$ if $y=4x+k$ is tangent to $y=x^2+2$.

Answers

1. $(x,y)=(3,4)$.
2. $(x,y)=(2,-1)$.
3. Equating gives $x^2-x-6=0$, so the solutions are $(-2,0)$ and $(3,5)$.
4. Substitution gives $x^2+3=0$, so there are no real solutions.
5. Substitution gives $x^2-4x+(2-k)=0$. Tangency requires $16-4(2-k)=0$, so $k=-2$.

What to learn next

Continue with inequalities to study regions and solution sets, or functions in modelling to develop the modelling ideas used here.