Add Maths · 5 Simultaneous equations

5 Simultaneous equations

Before you start

Solving a pair of linear simultaneous equations.
Squaring a bracket in full: ${(5 - 2 x)}^{2} = 25 - 20 x + 4 x^{2}$ .
Solving a quadratic by factorising or by the quadratic formula (which is on the List of formulas), and the discriminant $b^{2} - 4 a c$ .
Rearranging a linear equation such as $y - x + 3 = 0$ into $y = x - 3$ .

Syllabus objectives covered (1)

Solve simultaneous equations in two unknowns by elimination or substitution.

Part 1 of 6

What a solution is

Two equations in $x$ and $y$ are simultaneous when both must be true at once. A solution is a pair of values, one $x$ and one $y$ , that satisfies both equations.
Each equation is a graph, so a solution is a point where the two graphs meet. Two straight lines usually meet once. A line and a curve can meet twice, once or not at all, so when one equation is not linear, expect up to two pairs.
That brings a trap the linear case does not have: you can find the right numbers and still pair them wrongly, or lose one pair altogether. Keep it in mind through every method below.

Part 2 of 6

Two linear equations: elimination or substitution

Elimination. Multiply one or both equations so that one unknown has the same size of coefficient in each. Then subtract if the signs are the same, or add if they are opposite, to remove it.
For $3 x + 2 y = 12$ and $5 x - 3 y = 1$ , make the $y$ coefficients $6$ and $- 6$ by multiplying the first equation by $3$ and the second by $2$ .
$9 x + 6 y = 36, 10 x - 6 y = 2$
The signs are opposite, so add: $19 x = 38$ and $x = 2$ . Put that into the first original equation: $6 + 2 y = 12$ , so $y = 3$ .
Substitution. Make one unknown the subject of one equation, then replace it everywhere in the other. This method still works when an equation is not linear, so it is the one to be fluent in.
Choose the rearrangement that avoids fractions. From $2 x + y = 5$ , $y = 5 - 2 x$ is clean; $x = \frac{5 - y}{2}$ is not.

Part 3 of 6

A line and a curve: substitute from the linear equation

When one equation is linear and the other is not, always rearrange the linear one and substitute it into the non-linear one. Going the other way (making $x$ the subject of a curve) usually needs a square root and splits into cases.
Take $y = x + 1$ and $x^{2} + y^{2} = 25$ . Replace $y$ by $(x + 1)$ , with the bracket, because the whole expression gets squared.
$x^{2} + {(x + 1)}^{2} = 25$
Expand and make one side zero: $x^{2} + x^{2} + 2 x + 1 = 25$ , so $2 x^{2} + 2 x - 24 = 0$ . Divide by the common factor $2$ to keep the numbers small.
$x^{2} + x - 12 = 0$
Factorise: $(x + 4) (x - 3) = 0$ , so $x = - 4$ or $x = 3$ . Two values of $x$ mean two solutions to find, not one.
Find each partner by substituting into the linear equation $y = x + 1$ . $x = 3$ gives $y = 4$ , and $x = - 4$ gives $y = - 3$ . Those are the two points where the line cuts the circle.
Two intersections, so two solution pairs: $(3, 4)$ and $(- 4, - 3)$ . Each $x$ has its own $y$ .
Why the linear equation and not the circle? Putting $x = 3$ into $x^{2} + y^{2} = 25$ gives $y^{2} = 16$ , so $y = 4$ or $y = - 4$ . But $(3, - 4)$ is not on the line. The curve cannot rule out that false pair; the linear equation gives exactly one $y$ for each $x$ .
Write the answers as pairs so they cannot be mismatched: $x = 3, y = 4$ and $x = - 4, y = - 3$ .

Part 4 of 6

How many solutions, and the pair you must not drop

The quadratic you get after substituting tells you how many pairs there are. Two different roots mean the line crosses the curve twice: two pairs, and both are required. Finding one pair and stopping is the most common way to lose marks in this topic.
A repeated root means one pair: the line touches the curve without crossing it, so it is a tangent. For $y = 4 x - 4$ and $y = x^{2}$ , substituting gives $x^{2} - 4 x + 4 = 0$ , which is ${(x - 2)}^{2} = 0$ . The only point is $(2, 4)$ .
No real roots ( $b^{2} - 4 a c < 0$ ) means the line misses the curve, and the equations have no real solution. That is a correct answer, not a mistake.
Watch for a square root inside the working: $x^{2} = 16$ means $x = 4$ or $x = - 4$ . Write both down before you go on.

Part 5 of 6

Both equations non-linear: eliminate a whole term

Substitution is not the only route. Sometimes adding, subtracting or dividing the equations removes a whole term, just as elimination removes a letter from linear equations.
For $x^{2} + y^{2} = 20$ and $x^{2} - y^{2} = 12$ , add them: $2 x^{2} = 32$ , so $x^{2} = 16$ . Subtract them: $2 y^{2} = 8$ , so $y^{2} = 4$ .
Now count carefully. $x = \pm 4$ and $y = \pm 2$ , and every combination works, because both equations contain only $x^{2}$ and $y^{2}$ . That gives four pairs: $(4, 2)$ , $(4, - 2)$ , $(- 4, 2)$ and $(- 4, - 2)$ .
When both equations are products of powers, divide one by the other so that common factors cancel. For $x y^{2} = 18$ and $x^{2} y = 12$ , dividing cancels one $x$ and one $y$ .
$\frac{x y^{2}}{x^{2} y} = \frac{18}{12} \Rightarrow \frac{y}{x} = \frac{3}{2}$
Dividing is safe here because neither right-hand side is zero, so neither $x$ nor $y$ can be $0$ . If a right-hand side were $0$ , dividing could lose solutions, so check before you divide.
Dividing leaves a linear link, $y = \frac{3}{2} x$ , and you are back to substitution. That example is finished in the worked examples.

Part 6 of 6

Equations with x and y in denominators

An equation such as $\frac{x}{y} + \frac{y}{x} = \frac{5}{2}$ is not ready for substitution. First clear every fraction by multiplying every term on both sides by the lowest common denominator, here $2 x y$ .
$2 x^{2} + 2 y^{2} = 5 x y$
Now it has no fractions, and the line-and-curve method applies: substitute from the linear equation, get a quadratic in one letter, solve it, and pair.
At the end, reject any pair that makes an original denominator zero. Multiplying by $x y$ can let in $x = 0$ or $y = 0$ , which the original equation never allowed.
Check every pair in both original equations. With fractions and squared brackets, that is the quickest way to catch a slip before it costs accuracy marks.

Worked examples

Example 1. Solve the simultaneous equations $2 x + y = 5$ and $x^{2} + x y + y^{2} = 7$ .

Show the working

The first equation is linear. Make $y$ its subject, because that avoids fractions: $y = 5 - 2 x$ .
Substitute into the second equation, with a bracket round every $y$ .
$x^{2} + x (5 - 2 x) + {(5 - 2 x)}^{2} = 7$
Expand each term separately: $x (5 - 2 x) = 5 x - 2 x^{2}$ and ${(5 - 2 x)}^{2} = 25 - 20 x + 4 x^{2}$ .
$x^{2} + 5 x - 2 x^{2} + 25 - 20 x + 4 x^{2} = 7$
Collect like terms and make one side zero: $3 x^{2} - 15 x + 18 = 0$ . Divide by $3$ .
$x^{2} - 5 x + 6 = 0$
Factorise: $(x - 2) (x - 3) = 0$ , so $x = 2$ or $x = 3$ . Two values of $x$ , so two pairs to find.
Use the linear equation $y = 5 - 2 x$ for each one: $x = 2$ gives $y = 1$ , and $x = 3$ gives $y = - 1$ .
Check both pairs in the curve: $4 + 2 + 1 = 7$ and $9 - 3 + 1 = 7$ . Both work.

Answer: $x = 2, y = 1$ and $x = 3, y = - 1$ .

Example 2. Solve the simultaneous equations $y = x + 1$ and $\frac{x}{y} + \frac{y}{x} = \frac{5}{2}$ .

Show the working

Clear the fractions first. Multiply every term of the second equation by $2 x y$ .
$2 x^{2} + 2 y^{2} = 5 x y$
Substitute $y = x + 1$ , keeping the brackets.
$2 x^{2} + 2 {(x + 1)}^{2} = 5 x (x + 1)$
Expand: $2 x^{2} + 2 x^{2} + 4 x + 2 = 5 x^{2} + 5 x$ . Move everything to the right-hand side so the $x^{2}$ term stays positive.
$0 = x^{2} + x - 2$
Factorise: $(x + 2) (x - 1) = 0$ , so $x = - 2$ or $x = 1$ .
Pair them using $y = x + 1$ : $x = 1$ gives $y = 2$ , and $x = - 2$ gives $y = - 1$ .
Neither pair makes $x$ or $y$ zero, so both are allowed. Check $(1, 2)$ : $\frac{1}{2} + 2 = \frac{5}{2}$ . Check $(- 2, - 1)$ : $2 + \frac{1}{2} = \frac{5}{2}$ .

Answer: $x = 1, y = 2$ and $x = - 2, y = - 1$ .

Example 3. Solve the simultaneous equations $x y^{2} = 18$ and $x^{2} y = 12$ .

Show the working

Both equations are products and neither right-hand side is zero, so $x \neq 0$ and $y \neq 0$ . That makes dividing one equation by the other safe.
Divide the first equation by the second, and rearrange.
$\frac{x y^{2}}{x^{2} y} = \frac{18}{12} \Rightarrow \frac{y}{x} = \frac{3}{2} \Rightarrow y = \frac{3 x}{2}$
Substitute into $x^{2} y = 12$ . Either equation would work.
$x^{2} \cdot \frac{3 x}{2} = 12 \Rightarrow x^{3} = 8$
A number has only one real cube root, so $x = 2$ , not $\pm 2$ . Then $y = \frac{3 \times 2}{2} = 3$ .
Check: $x y^{2} = 2 \times 9 = 18$ and $x^{2} y = 4 \times 3 = 12$ . Both hold. There is exactly one pair here, and the reason is the cube, not a lost solution.

Answer: $x = 2, y = 3$ .

Where people go wrong

Mistake: Finding both $x$ values but giving only one pair, or giving the $x$ values with no $y$ values.

Instead: Each root of the quadratic gives its own pair. Find the $y$ for each and write them together: $x = \dots, y = \dots$ .

A solution is a point. The second pair carries its own marks, so one pair can only earn part of the credit.

Mistake: Substituting the $x$ values into the curve, rather than the line, to find $y$ .

Instead: Use the linear equation to find each $y$ .

A curve with $y^{2}$ in it gives two $y$ values for one $x$ , which creates pairs that are not on the line, such as $(3, - 4)$ for $y = x + 1$ and $x^{2} + y^{2} = 25$ .

Mistake: Squaring a substituted bracket term by term: ${(5 - 2 x)}^{2} = 25 + 4 x^{2}$ .

Instead: ${(5 - 2 x)}^{2} = (5 - 2 x) (5 - 2 x) = 25 - 20 x + 4 x^{2}$ .

The middle term $- 20 x$ comes from the two cross products. Losing it changes the quadratic, so both pairs come out wrong.

Mistake: Writing $x^{2} = 16$ as $x = 4$ only.

Instead: $x = 4$ or $x = - 4$ .

Both square to $16$ . When both equations contain only $x^{2}$ and $y^{2}$ , missing the negatives loses three of the four pairs.

Mistake: Multiplying only some terms by $2 x y$ when clearing fractions, e.g. leaving the right-hand side as $\frac{5}{2}$ .

Instead: Multiply every term on both sides by the same expression.

The equation stays true only when the whole of each side is multiplied.

Check yourself

Solve $2 x - 3 y = 7$ and $3 x + 4 y = 2$ by elimination.
Answer
Multiply by $4$ and $3$ : $8 x - 12 y = 28$ and $9 x + 12 y = 6$ . Add: $17 x = 34$ , so $x = 2$ . Then $4 - 3 y = 7$ , so $y = - 1$ .
Solve $y = x + 1$ and $x^{2} + y^{2} = 13$ .
Answer
$x^{2} + {(x + 1)}^{2} = 13$ gives $2 x^{2} + 2 x - 12 = 0$ , so $x^{2} + x - 6 = 0$ and $(x + 3) (x - 2) = 0$ . The pairs are $x = 2, y = 3$ and $x = - 3, y = - 2$ .
Solve $x y = 12$ and $x + y = 7$ .
Answer
$y = 7 - x$ , so $x (7 - x) = 12$ , which gives $x^{2} - 7 x + 12 = 0$ and $x = 3$ or $4$ . The pairs are $x = 3, y = 4$ and $x = 4, y = 3$ .
You have found $x = 3$ while solving $y = x + 1$ and $x^{2} + y^{2} = 25$ . Explain why you should not use the circle's equation to find $y$ .
Answer
The circle gives $y^{2} = 16$ , so $y = \pm 4$ , but $(3, - 4)$ does not satisfy $y = x + 1$ . The linear equation gives the one correct partner, $y = 4$ .
Solve $x^{2} + y^{2} = 34$ and $x^{2} - y^{2} = 16$ .
Answer
Adding gives $2 x^{2} = 50$ , so $x^{2} = 25$ . Subtracting gives $2 y^{2} = 18$ , so $y^{2} = 9$ . Four pairs: $(5, 3)$ , $(5, - 3)$ , $(- 5, 3)$ , $(- 5, - 3)$ .
Determine how many times the line $y = 2 x - 1$ meets the curve $y = x^{2}$ .
Answer
$x^{2} = 2 x - 1$ gives $x^{2} - 2 x + 1 = 0$ , which is ${(x - 1)}^{2} = 0$ . There is one repeated root, so they meet once, at $(1, 1)$ , and the line is a tangent.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. The line $y = 2 x - 1$ meets the curve $x^{2} + x y + y^{2} = 3$ at the points $A$ and $B$ .

(a) Find the coordinates of $A$ and $B$ . 5 marks No calculator

Mark scheme

M1 substitutes $y = 2 x - 1$ into the curve to obtain an equation in $x$ only
A1 $7 x^{2} - 5 x - 2 = 0$ oe
M1 solves their three-term quadratic by factorising or the formula
A1 $x = 1$ and $x = - \frac{2}{7}$
A1 $(1, 1)$ and $(- \frac{2}{7}, - \frac{11}{7})$ cao

Question 2. Two equations are $x + 2 y = 7$ and $x y = 6$ .

(a) Show that eliminating $x$ gives $2 y^{2} - 7 y + 6 = 0$ . 2 marks No calculator

Mark scheme

M1 $x = 7 - 2 y$ substituted into $x y = 6$
A1 $7 y - 2 y^{2} = 6$ rearranged to the given equation with no errors; nfww

(b) Hence solve the simultaneous equations. 3 marks No calculator

Mark scheme

M1 solves the given quadratic, e.g. $(2 y - 3) (y - 2) = 0$
A1 $y = 2$ and $y = \frac{3}{2}$
A1 $x = 3, y = 2$ and $x = 4, y = \frac{3}{2}$ , correctly paired

Question 3.

Solve the simultaneous equations $y = x + 2$ and $\frac{1}{x} + \frac{1}{y} = \frac{4}{3}$ . 6 marks No calculator

Mark scheme

M1 multiplies every term by $3 x y$ oe to clear the fractions: $3 y + 3 x = 4 x y$
M1 substitutes $y = x + 2$ into their equation
A1 $2 x^{2} + x - 3 = 0$ oe
M1 solves their three-term quadratic
A1 $x = 1, y = 3$
A1 $x = - \frac{3}{2}, y = \frac{1}{2}$

How this is examined

Finding where a line meets a curve: substitute, solve the quadratic, pair the answers: a five- or six-mark Solve, or Find the coordinates of the points where a line meets a curve The linear equation rearranged and substituted with brackets (M), a correct three-term quadratic equal to zero (A), a valid solving method (M), then both $x$ values and both matching $y$ values (A). The last accuracy mark needs the second pair.
Clearing fractions or dividing the equations before substituting: a Solve where one equation has $x$ or $y$ in a denominator, or where both are products such as $x y^{2}$ and $x^{2} y$ Multiplying every term by the common denominator (or dividing the equations to cancel) shown as its own step, then the substitution method, with any pair that makes a denominator zero rejected.
Showing that eliminating one unknown gives a printed quadratic: a two-mark Show that before a Solve The substitution and every line of expansion written down, arriving at the printed equation exactly. Because the result is given, the accuracy mark needs working that reaches it with no errors along the way.
Presenting solution pairs so they cannot be mismatched: the final accuracy marks of any simultaneous-equations question Answers written as $x = 3, y = 4$ and $x = - 4, y = - 3$ (or as coordinates), each $y$ found from the linear equation and checked in the other equation.

Words to know

Simultaneous equations: Two or more equations that must all be true for the same values of the unknowns.
Solution pair: One value of $x$ with its matching value of $y$ ; on a graph, a single point where the graphs meet.
Substitution: Replacing one unknown by an expression for it taken from the other equation.
Elimination: Adding, subtracting or dividing equations so that an unknown, or a whole term, disappears.
Tangent: A line that touches a curve at a point without crossing it; substituting gives a repeated root.
Discriminant: $b^{2} - 4 a c$ for $a x^{2} + b x + c = 0$ . If it is negative there are no real roots, so the line and curve do not meet.

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