Simultaneous Equations

Simultaneous equations are two (or more) equations that must be true at the same time. When there are two unknowns, you need two equations to pin them down, and the solution is the pair of values that satisfies both. A single equation in two unknowns, such as $x + y = 5$x+y=5, has infinitely many solutions; it is only when a second, independent equation is added that the answer is fixed to one pair. This lesson covers the two standard algebraic methods — elimination and substitution — for a pair of linear equations, how to form simultaneous equations from a worded context, and the Higher-tier case where one equation is linear and the other quadratic. Executing each method is AO1; setting up equations from a real situation and interpreting the answer is AO3, with clear layout earning AO2.

Key Vocabulary

Term	Meaning
Simultaneous equations	Two or more equations that are all true together
Elimination	Adding or subtracting equations to remove one unknown
Substitution	Rearranging one equation and substituting into the other
Coefficient	The number multiplying an unknown, e.g. the $3$3 in $3x$3x
Solution	The pair of values $(x, y)$(x,y) that satisfies both equations
Point of intersection	Where the two lines (or curves) meet on a graph
Consistent	Having at least one solution

Geometrically, a pair of linear simultaneous equations represents two straight lines, and the solution is the single point where they cross. This is why a consistent pair has exactly one solution: two non-parallel lines meet at one point. Keeping this picture in mind also explains the Higher-tier linear–quadratic case, where a line can cross a curve at two points, giving two solution pairs. Whichever algebraic method you use, you are really just calculating those points of intersection without drawing the graph.

Solving by Elimination

In elimination you make the coefficients of one unknown match (in size), then add or subtract the equations to remove that unknown. A handy rule: Same Signs Subtract; different signs add. The aim of every elimination is to reduce two equations in two unknowns down to a single equation in one unknown, which you already know how to solve. Once you have one value, substitute it back to find the other.

Worked Example 1

Solve $3x + y = 11$3x+y=11 and $x + y = 5$x+y=5.

The $y$y-coefficients are both $+1$+1. Same signs, so subtract: $(3x + y) - (x + y) = 11 - 5$(3x+y)−(x+y)=11−5, giving $2x = 6$2x=6, so $x = 3$x=3. Substitute into $x + y = 5$x+y=5: $3 + y = 5$3+y=5, so $y = 2$y=2.

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

Worked Example 2

Solve $2x + 3y = 16$2x+3y=16 and $2x - y = 4$2x−y=4.

The $x$x-coefficients match at $+2$+2. Same signs, so subtract: $(2x + 3y) - (2x - y) = 16 - 4$(2x+3y)−(2x−y)=16−4, giving $4y = 12$4y=12, so $y = 3$y=3. Substitute into $2x - y = 4$2x−y=4: $2x - 3 = 4$2x−3=4, so $2x = 7$2x=7 and $x = 3.5$x=3.5.

Answer: $x = 3.5$x=3.5, $y = 3$y=3

Worked Example 3 — multiplying first

Solve $3x + 2y = 19$3x+2y=19 and $x + 4y = 13$x+4y=13.

Multiply the first equation by $2$2 so the $y$y-coefficients match: $6x + 4y = 38$6x+4y=38. Now subtract the second: $(6x + 4y) - (x + 4y) = 38 - 13$(6x+4y)−(x+4y)=38−13, giving $5x = 25$5x=25, so $x = 5$x=5. Substitute: $5 + 4y = 13$5+4y=13, so $4y = 8$4y=8 and $y = 2$y=2.

Answer: $x = 5$x=5, $y = 2$y=2

Common error: subtracting when the matched terms have opposite signs (or adding when they are the same). Decide add-or-subtract by looking at the signs of the matched coefficients.

Solving by Substitution

Rearrange one equation to make a single unknown the subject, then substitute that expression into the other equation. Substitution is usually the better choice when one equation already has a single $x$x or $y$y by itself (for example $y = 2x + 1$y=2x+1), because the rearranging is done for you. It is also the method you must use for the Higher-tier linear–quadratic case, where one equation contains a squared term and elimination will not work.

Worked Example 4

Solve $y = 2x + 1$y=2x+1 and $3x + y = 16$3x+y=16.

The first equation already gives $y$y. Substitute into the second: $3x + (2x + 1) = 16$3x+(2x+1)=16, so $5x + 1 = 16$5x+1=16, giving $5x = 15$5x=15 and $x = 3$x=3. Then $y = 2(3) + 1 = 7$y=2(3)+1=7.

Answer: $x = 3$x=3, $y = 7$y=7

Worked Example 5

Solve $x - 2y = 1$x−2y=1 and $4x + 3y = 26$4x+3y=26.

Rearrange the first: $x = 1 + 2y$x=1+2y. Substitute: $4(1 + 2y) + 3y = 26$4(1+2y)+3y=26, so $4 + 8y + 3y = 26$4+8y+3y=26, giving $11y = 22$11y=22 and $y = 2$y=2. Then $x = 1 + 2(2) = 5$x=1+2(2)=5.

Answer: $x = 5$x=5, $y = 2$y=2

Common error: forgetting to multiply the whole bracket when substituting — here $4(1 + 2y)$4(1+2y) must give $4 + 8y$4+8y, not $4 + 2y$4+2y.

Forming Simultaneous Equations from Context

Many worded problems hide two relationships, each connecting the same two unknown quantities. The strategy is to define two variables clearly, write one equation per relationship, and then solve the pair by elimination or substitution. Problems about the cost of two different items, or two unknown numbers linked by two facts, are the classic setting. As always, finish by answering the actual question with the correct units (and pounds written outside the maths).

Worked Example 6

Three apples and two bananas cost £$1.30$1.30; one apple and four bananas cost £$1.20$1.20. Work out the cost of one apple and one banana.

Let an apple cost $a$a pence and a banana cost $b$b pence. Then $3a + 2b = 130$3a+2b=130 and $a + 4b = 120$a+4b=120. Multiply the second by $3$3: $3a + 12b = 360$3a+12b=360. Subtract the first: $10b = 230$10b=230, so $b = 23$b=23. Then $a + 4(23) = 120$a+4(23)=120, giving $a = 28$a=28.

Answer: an apple costs £$0.28$0.28 and a banana costs £$0.23$0.23.

Linear–Quadratic Simultaneous Equations [Higher]

This whole section is Higher tier. When one equation is linear and one quadratic, substitute the linear equation into the quadratic, which produces a single quadratic to solve. There are usually two solution pairs.

Worked Example 7

Solve $y = x + 1$y=x+1 and $y = x^{2} - 5$y=x2−5.

Set them equal: $x^{2} - 5 = x + 1$x2−5=x+1, so $x^{2} - x - 6 = 0$x2−x−6=0, giving $(x - 3)(x + 2) = 0$(x−3)(x+2)=0, so $x = 3$x=3 or $x = -2$x=−2. Find $y$y from $y = x + 1$y=x+1: when $x = 3$x=3, $y = 4$y=4; when $x = -2$x=−2, $y = -1$y=−1.

Answer: $(3, 4)$(3,4) and $(-2, -1)$(−2,−1)

Worked Example 8 [Higher]

Still Higher tier. Solve $y = 2x - 3$y=2x−3 and $x^{2} + y^{2} = 5$x2+y2=5.

Substitute $y = 2x - 3$y=2x−3 into the circle: $x^{2} + (2x - 3)^{2} = 5$x2+(2x−3)2=5. Expand: $x^{2} + 4x^{2} - 12x + 9 = 5$x2+4x2−12x+9=5, so $5x^{2} - 12x + 4 = 0$5x2−12x+4=0. Factorise: $(5x - 2)(x - 2) = 0$(5x−2)(x−2)=0, so $x = \dfrac{2}{5}$x=52​ or $x = 2$x=2. Then $y = 2x - 3$y=2x−3 gives $y = -\dfrac{11}{5}$y=−511​ or $y = 1$y=1.

Answer: $\left(\dfrac{2}{5}, -\dfrac{11}{5}\right)$(52​,−511​) and $(2, 1)$(2,1)

Extended Worked Examples

Worked Example 9 — Both equations need scaling

Solve $4x + 3y = 18$4x+3y=18 and $3x + 5y = 19$3x+5y=19.

Multiply the first by $5$5 and the second by $3$3 to match the $y$y-terms: $20x + 15y = 90$20x+15y=90 and $9x + 15y = 57$9x+15y=57. Subtract: $11x = 33$11x=33, so $x = 3$x=3. Then $4(3) + 3y = 18$4(3)+3y=18, giving $3y = 6$3y=6 and $y = 2$y=2.

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

Worked Example 10 — A money-and-quantity problem

A theatre sells $5$5 adult tickets and $3$3 child tickets for £$129$129, and $2$2 adult tickets and $4$4 child tickets for £$74$74. Work out the price of each ticket.