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Simultaneous Equations

Q: Which equation do I substitute back into?

The linear one. It gives exactly one y for each x, so your coordinate pairs are matched correctly. The quadratic can give two y-values and lead to wrongly paired points.

Solving equations simultaneously means finding the values that satisfy all of them at once — geometrically, the points where their graphs cross. At A-Level the key new skill is handling one linear and one quadratic equation by substitution.

28 min Video by TLMaths Algebra and Functions

The big picture

A single equation usually pins down a whole curve or line of possibilities; a second equation narrows those down to the specific points that work for both. That is why simultaneous equations are really a question about $intersections$ — where does this line meet that curve? Seeing it geometrically explains everything that follows: two lines meet at one point (usually), but a line and a parabola can meet twice, once (a tangent), or not at all. The algebra of substitution and the geometry of intersection are two views of the same thing, and connecting them is what turns this from a procedure into understanding.

A-Level Maths B4-03: One Linear and One Quadratic EquationWatch the walkthrough, then work through the notes below.

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What you'll be able to do

Solve two linear simultaneous equations by elimination or substitution
Solve one linear and one quadratic equation by substitution
Give solutions as coordinate pairs, not just $x$ -values
Interpret the number of solutions as the number of intersection points

The geometric meaning

Every equation in $x$ and $y$ draws a graph. A solution to two equations $simultaneously$ is a point $(x, y)$ lying on $both$ graphs — an intersection. Two straight lines usually cross once; a line and a parabola can cross twice, touch once, or miss entirely.

This is why solutions come in pairs: each intersection has both an $x$ - and a $y$ -coordinate. Reporting only the $x$ -values answers half the question.

Why this works

Because solving simultaneously is finding intersections, the number of solutions and the geometry always agree. If your algebra gives two solutions, the line really does cut the curve twice — a built-in sanity check.

Two linear equations

With two straight lines, either $eliminate$ a variable (scale the equations so one variable matches, then add or subtract) or $substitute$ (rearrange one equation and put it into the other). Both reach the single crossing point.

1Rearrange the second:

x = y + 1

2Substitute into the first:

3 (y + 1) + 2 y = 12 \Rightarrow 5 y + 3 = 12

5 y = 9 \Rightarrow y = 1.8

4Back-substitute:

x = 1.8 + 1 = 2.8

Answer

x = 2.8, y = 1.8

— the point

(2.8, 1.8)

Tip — Substitution is usually cleaner when one equation already has a variable with coefficient 1; elimination is quicker when coefficients line up neatly.

One linear and one quadratic

This is the A-Level case, and $substitution is essential$ — you cannot eliminate a squared term by adding. Rearrange the $linear$ equation for one variable, substitute it into the quadratic, and you get a single quadratic in one variable. Solve it, then back-substitute into the $linear$ equation to find the partner values.

1Substitute the linear equation into the quadratic:

x^{2} + (x + 1)^{2} = 25

2Expand:

x^{2} + x^{2} + 2 x + 1 = 25 \Rightarrow 2 x^{2} + 2 x - 24 = 0

3Divide by 2 and factorise:

x^{2} + x - 12 = 0 \Rightarrow (x + 4) (x - 3) = 0

4So

x = - 4

x = 3

. Back-substitute into

y = x + 1

y = - 3

y = 4

Answer

(- 4, - 3)

and

(3, 4)

— the line cuts the circle at two points.

Why this works

Always back-substitute into the $linear$ equation, never the quadratic — the quadratic gives two $y$ -values for some $x$ -values, and you would risk pairing the wrong ones. The linear equation gives one clean partner each time.

How many intersections?

After substituting, you are left with a quadratic — so the discriminant tells you how the line and curve meet. Two solutions means the line cuts the curve twice; a repeated solution ( $Δ = 0$ ) means the line is a $tangent$ , touching once; no real solutions means they never meet.

This ties the chapter together: substitution reduces the geometry to a quadratic, and the discriminant reads off the geometry.

line meets curve once \Leftrightarrow Δ = 0 (tangent)

The discriminant of the substituted quadratic counts the intersections.

Think like an examiner

Always present answers as coordinate $pairs$ . Solving for $x$ alone typically caps the marks even if the $x$ -values are correct.
Substitute the found values back into the $linear$ equation to pair them up correctly.
Keep the two solutions matched: the $x = - 4$ goes with $y = - 3$ , not with the other $y$ . Muddling pairs is a common lost mark.
If asked to “show the line is a tangent”, substitute and show the resulting quadratic has discriminant zero.

Common misconceptions

Giving only the

x

-values as the final answer.

Each solution is a pair

(x, y)

— find the matching

y

for every

x

Back-substituting into the quadratic equation.

Use the linear equation, which gives exactly one

y

per

x

and pairs them correctly.

Trying to eliminate a variable from a linear/quadratic pair by adding.

Use substitution — you cannot cancel a squared term by adding equations.

Expanding

(x + 1)^{2}

x^{2} + 1

(x + 1)^{2} = x^{2} + 2 x + 1

— do not drop the middle term.

Method summary

linear \to rearrange, substitute

Reduce to one quadratic.

back-substitute into the linear eqn

Pair up the coordinates.

Δ = 0 \Rightarrow tangent

Number of intersections.

Stretch yourself

The line $y = m x + 1$ is a tangent to the parabola $y = x^{2} + 3$ . Find the possible values of the gradient $m$ .

Hint — Set the expressions equal, form a quadratic in $x$ , and impose the tangency condition (discriminant $= 0$ ) to get an equation in $m$ .

Questions students ask

What does it mean to solve equations “simultaneously”?

It means finding the values of the unknowns that satisfy all the equations at the same time. Graphically, those values are the coordinates of the point(s) where the graphs intersect.

Why must I use substitution for a linear and quadratic pair?

Elimination works by adding or subtracting to cancel a variable, but you cannot cancel an

x^{2}

term that way. Substituting the linear equation into the quadratic reduces everything to a single quadratic you can solve.

Which equation do I substitute back into?

The linear one. It gives exactly one

y

for each

x

, so your coordinate pairs are matched correctly. The quadratic can give two

y

-values and lead to wrongly paired points.

How do I know how many solutions to expect?

After substituting you get a quadratic; its discriminant tells you. Two real roots means two intersection points, a repeated root means the line is a tangent (one point), and no real roots means the graphs do not meet.

Key takeaways

Solving simultaneously finds the point(s) where graphs intersect.
Two linear equations: eliminate or substitute to one crossing point.
Linear and quadratic: substitute the linear into the quadratic, solve, back-substitute.
Always give coordinate pairs, and back-substitute into the linear equation.
The discriminant of the substituted quadratic counts the intersections (0, 1 tangent, or 2).

How this fits the course

Build on

Straight-Line Graphs (later)soon
Circles (later)soon

Leads to

Inequalities

Test yourself

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