Simultaneous equations: elimination, substitution and graphs

Simultaneous equations are equations that must be true at the same time. A solution gives one value for each unknown and makes every equation in the system true.

For example,

$$\begin{aligned} x+y&=7,\\ x-y&=1 \end{aligned}$$

has the solution $x=4$, $y=3$, because

$4+3=7\qquad\text{and}\qquad4-3=1.$

The pair $(4,3)$ satisfies both conditions. A value that satisfies only one equation is not a simultaneous solution.

Prerequisites

You should already be able to:

• solve linear equations;
• expand brackets and collect like terms using algebraic fluency;
• substitute numbers and expressions;
• factorise and solve quadratic equations;
• recognise straight line graphs and basic quadratic graphs.

Solutions are intersections

An equation in $x$ and $y$ describes a set of points. For example, $y=2x+1$ describes a straight line. If two equations are drawn on the same axes, a simultaneous solution must lie on both graphs, so it is an intersection.

This gives three possibilities for two linear equations:

• non-parallel lines meet once, so there is one solution;
• distinct parallel lines never meet, so there is no solution;
• two equations may describe the same line, giving infinitely many solutions.

This geometric interpretation is also a useful check on algebra. Two ordinary non-parallel straight lines cannot produce two different solutions.

Elimination

Elimination combines equations so that one unknown disappears. You may add or subtract whole equations because equal quantities remain equal when they are added or subtracted.

Worked example 1: coefficients already cancel

Solve

\begin{aligned} 2x+y&=11,\tag{1}\\ 3x-y&=9.\tag{2} \end{aligned}

The coefficients of $y$ are $+1$ and $-1$, so add the equations:

$$\begin{aligned} 2x+y&=11\\ 3x-y&=9\\ \hline 5x&=20. \end{aligned}$$

Hence $x=4$. Substitute into either original equation, for example equation $(1)$:

$2(4)+y=11\quad\Longrightarrow\quad y=3.$

Therefore

$\boxed{(x,y)=(4,3)}.$

Check in both originals:

$2(4)+3=11,\qquad 3(4)-3=9.$

Add or subtract?

• Opposite signs, such as $+3y$ and $-3y$: add.
• Same signs, such as $+3y$ and $+3y$: subtract.

This is a guide, not a separate rule. The aim is simply to make one coefficient zero.

Worked example 2: subtract equal coefficients

Solve

\begin{aligned} 5x+2y&=19,\tag{1}\\ 3x+2y&=13.\tag{2} \end{aligned}

Subtract equation $(2)$ from equation $(1)$:

$$(5x+2y)-(3x+2y)=19-13.$$

Therefore $2x=6$, so $x=3$. Then

$3(3)+2y=13\quad\Longrightarrow\quad2y=4\quad\Longrightarrow\quad y=2.$

Thus $(x,y)=(3,2)$.

Notice the brackets in the subtraction. Every term in the second equation is subtracted.

Worked example 3: make coefficients equal

Solve

\begin{aligned} 3x+2y&=7,\tag{1}\\ 5x-3y&=4.\tag{2} \end{aligned}

Neither variable cancels immediately. The lowest common multiple of $2$ and $3$ is $6$, so multiply the whole of equation $(1)$ by $3$ and equation $(2)$ by $2$:

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

Add:

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

Substitute into equation $(1)$:

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

Therefore

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

Fractions are valid solutions. Keep them exact rather than replacing them with rounded decimals.

Misconception: multiply only the chosen term

If $3x+2y=7$ is multiplied by $3$, the result is

$9x+6y=21,$

not $3x+6y=7$. Multiplying an equation means multiplying every term on both sides.

Self-check 1

Solve by elimination.

1. $2x+y=13$ and $x-y=2$.
2. $4x+3y=18$ and $4x-y=10$.
3. $2x+3y=12$ and $5x-2y=11$.

Answers

1. Add the equations: $3x=15$, so $x=5$ and $y=3$.
2. Subtract the second equation from the first: $4y=8$, so $y=2$ and $x=3$.
3. Multiply the first equation by $2$ and the second by $3$. Adding $4x+6y=24$ and $15x-6y=33$ gives $19x=57$. Thus $x=3$ and $y=2$.

Substitution

Substitution replaces an unknown with an equal expression. It is usually quickest when one equation already has the form $x=\cdots$ or $y=\cdots$.

Worked example 4: an unknown already isolated

Solve

\begin{aligned} y&=3x-4,\tag{1}\\ 2x+y&=11.\tag{2} \end{aligned}

Equation $(1)$ says that $y$ and $3x-4$ have the same value. Replace $y$ in equation $(2)$:

$$\begin{aligned} 2x+(3x-4)&=11,\\ 5x-4&=11,\\ 5x&=15,\\ x&=3. \end{aligned}$$

Now use equation $(1)$:

$y=3(3)-4=5.$

So $(x,y)=(3,5)$.

Worked example 5: rearrange before substituting

Solve

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

Equation $(2)$ is easy to rearrange because $y$ has coefficient $-1$:

$3x-y=3\quad\Longrightarrow\quad y=3x-3.$

Substitute this into equation $(1)$:

$$\begin{aligned} x+2(3x-3)&=8,\\ 7x-6&=8,\\ 7x&=14,\\ x&=2. \end{aligned}$$

Then $y=3(2)-3=3$, giving $(2,3)$.

The brackets in $2(3x-3)$ are essential because the whole expression equals $y$.

Which method should I choose?

Use elimination when coefficients already match or have a small common multiple. Use substitution when an unknown is isolated or has coefficient $1$ or $-1$. Either valid method gives the same answer, so choose the route with less arithmetic.

Self-check 2

1. Solve $y=2x+1$ and $x+y=10$.
2. Solve $x=7-2y$ and $3x+y=11$.
3. Solve $2x-y=8$ and $x+3y=9$ by first making $y$ the subject of the first equation.

Answers

1. $x+(2x+1)=10$, so $x=3$ and $y=7$.
2. $3(7-2y)+y=11$, so $-5y=-10$. Hence $y=2$ and $x=3$.
3. $y=2x-8$. Then $x+3(2x-8)=9$, so $7x=33$. Hence $x=\dfrac{33}{7}$ and $y=\dfrac{10}{7}$.

Special linear systems

Elimination also reveals when a system does not have exactly one solution.

No solution

Consider

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

Doubling the second equation gives $2x+4y=10$, contradicting the first equation. Algebra eventually gives a false statement such as $0=4$. The graphs are distinct parallel lines, so there is no solution.

Infinitely many solutions

Now consider

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

The first equation is exactly twice the second. Elimination gives $0=0$, a statement that is always true. Both equations describe the same line, so there are infinitely many solutions.

Do not turn $0=0$ into $x=0$ or $y=0$. It means that elimination has removed all information because the equations were equivalent.

One linear and one quadratic equation

A straight line and a quadratic curve may meet twice, once or not at all. Substitution reduces their simultaneous equations to one quadratic equation.

Worked example 6: two intersections

Solve

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

At an intersection, the two expressions for $y$ are equal:

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

Rearrange and factorise:

$$\begin{aligned} x^2-3x&=0,\\ x(x-3)&=0. \end{aligned}$$

Therefore $x=0$ or $x=3$. Find the corresponding $y$ value for each branch using $y=x+2$:

$$x=0\Longrightarrow y=2, \qquad x=3\Longrightarrow y=5.$$

The solutions are

$\boxed{(0,2)\text{ and }(3,5)}.$

Worked example 7: irrational intersections

Solve

$y=2x+3\qquad\text{and}\qquad y=x^2+2.$

Equate the expressions for $y$:

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

Then

$x^2-2x-1=0,$

so the quadratic formula gives

$x=1\pm\sqrt2.$

Using $y=2x+3$ gives

$y=5\pm2\sqrt2,$

with matching signs. Hence

$\boxed{(1+\sqrt2,\,5+2\sqrt2)\quad\text{and}\quad(1-\sqrt2,\,5-2\sqrt2)}.$

Keep each $x$ value paired with the $y$ value calculated from it. Four combinations are not possible.

Self-check 3

Solve each system.

1. $y=x+1$ and $y=x^2-3x+1$.
2. $y=5-x$ and $y=x^2-x-1$.

Answers

1. $x+1=x^2-3x+1$ gives $x(x-4)=0$. The solutions are $(0,1)$ and $(4,5)$.
2. $5-x=x^2-x-1$ gives $x^2=6$, so $x=\pm\sqrt6$. The solutions are $(\sqrt6,5-\sqrt6)$ and $(-\sqrt6,5+\sqrt6)$.

Forming equations from a context

Worked example 8: ticket prices

Three adult tickets and two child tickets cost £46. Two adult tickets and five child tickets cost £52. Find each price.

Let $a$ be the adult price and $c$ the child price, in pounds. Translate each statement:

$$\begin{aligned} 3a+2c&=46,\\ 2a+5c&=52. \end{aligned}$$

Multiply the first equation by $2$ and the second by $3$:

$$\begin{aligned} 6a+4c&=92,\\ 6a+15c&=156. \end{aligned}$$

Subtract to obtain $11c=64$, so $c=\dfrac{64}{11}$. Then

\quad\Longrightarrow\quad a=\frac{126}{11}.$$ Thus an adult ticket costs $\dfrac{126}{11}$ pounds and a child ticket costs $\dfrac{64}{11}$ pounds. The unusual prices are mathematically valid. In a real modelling problem, they may suggest that the data were artificial or that further practical constraints are needed. ## A reliable solving routine 1. Define unknowns if they arise from a context. 2. Write both equations clearly and label them if useful. 3. Choose elimination or substitution to create one equation in one unknown. 4. Solve that equation accurately. 5. Substitute back to find the other unknown. 6. State an ordered pair or a contextual answer. 7. Check every solution in both original equations. ## Common mistakes - Solving the equations separately. The same pair of values must satisfy both. - Changing only one term when multiplying an equation. Multiply the whole equation. - Losing a minus sign when subtracting equations. Use brackets around the equation being subtracted. - Stopping after finding one unknown. A solution normally requires both values. - Rejecting fractions. Exact fractional solutions are entirely legitimate. - Finding two $x$ values in a linear with quadratic system but only one $y$ value. Substitute each branch separately. - Trusting a graph as exact. A graph can locate or estimate intersections, but algebra is needed for exact values unless the coordinates are unmistakable. ## Final check Without doing full calculations, decide what each result means after elimination. 1. $7x=21$. 2. $0=5$. 3. $0=0$. <details> <summary>Answers</summary> 1. $x=3$. Substitute back to obtain one corresponding $y$, so the system has one solution. 2. This contradiction means there is no solution. The lines are parallel and distinct. 3. The equations are equivalent, so there are infinitely many solutions on the same line. </details> ## Next steps Strengthen the graphical meaning through [graphs of functions](/learn/foundations/graphs-of-functions/) and [straight line graphs](/learn/foundations/straight-line-graphs/). Then study the full A-level treatment of [simultaneous equations](/learn/algebra/simultaneous-equations/), including more demanding line and curve systems. These methods also support [coordinate geometry](/learn/foundations/coordinate-geometry/) and later algebraic modelling.