P1.2 Simultaneous equations and inequalities
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This page covers solving one linear and one quadratic equation together, linear and quadratic inequalities, set notation and regions on a graph (Pure Mathematics 1, AS). Factorising, completing the square and the discriminant on their own are in P1.1; sketching cubics and counting intersections from a sketch are in P1.3.
What this page covers
8 points
- Solving one linear and one quadratic equation simultaneously by substitution
- Reading the solutions as the intersection points of a line and a curve
- Using the discriminant to decide whether a line crosses, touches or misses a curve
- Linear inequalities, including ones with brackets and fractions
- Quadratic inequalities, including ones that need expanding first
- Inequalities in a parameter such as \(k\) that come from a discriminant condition
- Combining inequalities and writing solution sets in set notation
- Regions on a graph: solid and dashed boundaries, and describing a region with inequalities
In one minute
6 points
- To solve a linear and a quadratic equation together, make one variable the subject of the linear equation, substitute into the quadratic, solve the resulting quadratic, and then use the linear equation to find the partner of each value.
- Each solution pair is a point where the line meets the curve; after substituting, the discriminant \(b^2-4ac\) of the quadratic tells you how many: positive gives two points, zero means the line is a tangent, negative means no points.
- Linear inequalities are solved like equations, except that multiplying or dividing by a negative number reverses the inequality sign.
- Quadratic inequalities: rearrange to \(f(x)>0\) or \(f(x)<0\), find the critical values (the roots), sketch, and read off where the curve is above or below the \(x\)-axis.
- Set notation: \(\{x : x<-4\}\cup\{x : x>2\}\) for "or"; \(\{x : -\tfrac32\le x<1\}\) for a single interval.
- Regions: a solid boundary is included (\(\le\), \(\ge\)); a dashed boundary is not (\(<\), \(>\)).
Contents
13 sections
- How do you solve a linear and a quadratic equation simultaneously?
- What do the solutions tell you about a line and a curve?
- How do you solve a linear inequality?
- How do you solve a quadratic inequality?
- What if the inequality is in a constant such as k?
- How do you combine inequalities and use set notation?
- How do you show inequalities as regions on a graph?
- Exam tips for P1.2
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
How do you solve a linear and a quadratic equation simultaneously?
With two linear equations you can often add or subtract to eliminate a variable. When one equation contains \(x^2\), \(y^2\) or \(xy\), elimination usually fails: the squared and product terms do not cancel. The method that always works is substitution. The linear equation gives one variable in terms of the other with no squares at all, so you can replace that variable everywhere in the quadratic and be left with a single equation in one unknown.
The substitution method
- Rearrange the linear equation to make \(x\) or \(y\) the subject. Choose the one that avoids fractions: if a variable has coefficient 1, use it.
- Substitute into the quadratic equation and expand carefully. Put brackets round the whole substituted expression before squaring or multiplying.
- Solve the resulting quadratic in one variable: factorise, complete the square or use the formula.
- Find the partner of each value by substituting into the linear equation. The linear equation gives exactly one partner each time; the quadratic might give two, and one of them would be false.
- Pair the answers: write them as coordinates or as matched statements \(x=\ldots,\ y=\ldots\).
If both equations already have \(y\) as the subject, step 1 is free: set the two right-hand sides equal to each other.
Solve the simultaneous equations \(y=x^2-4x+5\) and \(y=x+1\).
Both expressions equal \(y\), so set them equal: \(x^2-4x+5=x+1\), which gives \(x^2-5x+4=0\).
Factorise: \((x-1)(x-4)=0\), so \(x=1\) or \(x=4\).
Use the linear equation \(y=x+1\) for the partners: \(x=1\) gives \(y=2\), and \(x=4\) gives \(y=5\).
Check in the quadratic: \(1-4+5=2\) ✓ and \(16-16+5=5\) ✓. \(x=1, y=2\) and \(x=4, y=5\)
Solve the simultaneous equations \(x+3y=4\) and \(x^2+xy+y^2=3\).
Choose the subject. \(x\) has coefficient 1 in the linear equation, so \(x=4-3y\) (no fractions that way).
Substitute, with brackets, and expand. \((4-3y)^2+(4-3y)y+y^2=3\), so \(16-24y+9y^2+4y-3y^2+y^2=3\), which simplifies to \(7y^2-20y+13=0\).
Solve. \((y-1)(7y-13)=0\), so \(y=1\) or \(y=\tfrac{13}{7}\).
Partners from \(x=4-3y\). \(y=1\) gives \(x=1\); \(y=\tfrac{13}{7}\) gives \(x=4-\tfrac{39}{7}=-\tfrac{11}{7}\).
\(x=1, y=1\) and \(x=-\tfrac{11}{7}, y=\tfrac{13}{7}\)
Writing \((4-3y)^2\) as \(16+9y^2\), or dropping the brackets so that \(-3y\times y\) loses its sign. Expand \((a-b)^2=a^2-2ab+b^2\) in full on its own line before collecting terms.
What do the solutions tell you about a line and a curve?
Every real solution pair is a point that lies on both graphs, so it is a point where the line meets the curve. After you substitute, everything about the intersection is stored in one quadratic equation. Its number of real roots is the number of intersection points, and the discriminant \(b^2-4ac\) of that quadratic counts the roots without solving it.
| After substituting | Roots of the quadratic | The line… |
|---|---|---|
| \(b^2-4ac>0\) | two distinct real roots | crosses the curve at two points |
| \(b^2-4ac=0\) | one repeated root | touches the curve at one point: it is a tangent |
| \(b^2-4ac<0\) | no real roots | does not meet the curve |
This is the key idea of the whole page. A geometric question ("does the line touch the curve?") becomes an algebraic one ("is this discriminant zero?"). The idea comes back with circles in P2 and with tangents in calculus.
The curve \(C\) has equation \(y=x^2+x-1\).
(a) Show that the line \(y=3x-2\) is a tangent to \(C\), and find the point of contact.
(b) Find the set of values of \(c\) for which the line \(y=3x+c\) meets \(C\) at two distinct points.
(a) Substitute: \(x^2+x-1=3x-2\), so \(x^2-2x+1=0\). For a tangent we need \(b^2-4ac=0\): here \((-2)^2-4(1)(1)=0\), so there is a repeated root and the line is a tangent.
Point of contact: \((x-1)^2=0\) gives \(x=1\), and then \(y=3(1)-2=1\). Point of contact \((1, 1)\)
(b) Substitute the general line: \(x^2+x-1=3x+c\), so \(x^2-2x-(1+c)=0\). Two distinct points means discriminant \(>0\):
\((-2)^2-4(1)\bigl(-(1+c)\bigr)=4+4(1+c)=8+4c>0\), so \(c>-2\). \(\{c : c>-2\}\)
This is consistent with (a): \(c=-2\) gives the tangent itself, the boundary case.
When a question says "show that the line is a tangent", write the condition in words before the numbers ("for a tangent, \(b^2-4ac=0\)"), then substitute and conclude in a sentence. When the letter \(c\) is already used in the question, rename the quadratic's coefficients (or write the discriminant out in full) so the two meanings do not get mixed up.
How do you solve a linear inequality?
An inequality such as \(3x-1<8\) describes a whole range of values, not a single number. You solve it with the same steps as a linear equation: expand brackets, collect terms, divide. Almost every step keeps the inequality sign exactly as it was:
- Adding or subtracting the same quantity on both sides never changes the sign.
- Multiplying or dividing by a positive number keeps the sign.
- Multiplying or dividing by a negative number reverses it: \(-2x<6\) gives \(x>-3\).
Why the reversal? Multiplying by \(-1\) reflects the number line about 0, so the order of every pair of numbers flips: \(2<5\) but \(-2>-5\). With fractions, multiply every term by the lowest common denominator first; it is a positive number, so the sign is safe. You can avoid dividing by a negative altogether by collecting the \(x\) terms on the side where their coefficient ends up positive.
Find the set of values of \(x\) for which (a) \(5-3(x-1)>2x-7\) (b) \(\dfrac{2x-1}{3}-\dfrac{x+4}{4}\le x-2\).
(a) Expand: \(5-3x+3>2x-7\), so \(8-3x>2x-7\). Collect the \(x\) terms on the right, where they are positive: \(15>5x\). Divide by 5: \(x<3\)
(b) Multiply every term by 12: \(4(2x-1)-3(x+4)\le 12x-24\), so \(8x-4-3x-12\le12x-24\), giving \(5x-16\le12x-24\).
Collect on the right and divide by the positive number 7: \(8\le7x\). \(\{x : x\ge\tfrac87\}\)
Multiplying only the fractions by 12 and forgetting the other terms, or writing \(-3(x+4)\) as \(-3x+12\). Every term on both sides is multiplied, and a minus sign in front of a bracket changes the sign of everything inside it.
How do you solve a quadratic inequality?
You cannot solve \(x^2+2x-8>0\) by "moving things across" in the same way, because the sign of a quadratic changes at its roots. The reliable method uses a sketch.
Four steps
- Rearrange so that one side is zero: \(f(x)>0\), \(<0\), \(\ge0\) or \(\le0\). Expand any brackets first.
- Find the critical values: solve \(f(x)=0\).
- Sketch \(y=f(x)\), marking the roots on the \(x\)-axis.
- Read off where the curve is above the axis (\(f(x)>0\)) or below it (\(f(x)<0\)).
For a positive \(x^2\) coefficient and roots \(\alpha<\beta\), the curve is a ∪-shape, so:
- \(f(x)<0\) for \(\alpha<x<\beta\): one interval, between the roots;
- \(f(x)>0\) for \(x<\alpha\) or \(x>\beta\): two separate intervals, outside the roots.
For "\(\le\)" or "\(\ge\)" the critical values themselves are included, because there \(f(x)=0\).
Solve the inequality \((x-1)(x+5)>2x+3\).
Expand and rearrange to compare with zero: \(x^2+4x-5>2x+3\), so \(x^2+2x-8>0\).
Critical values: \((x+4)(x-2)=0\), so \(x=-4\) or \(x=2\).
The graph is ∪-shaped, and we want the parts above the axis, which are outside the roots. \(x<-4\) or \(x>2\)
If the \(x^2\) coefficient is negative, either sketch the ∩-shape directly or multiply through by \(-1\) first and reverse the sign. For example, \(6+x-x^2>0\) becomes \(x^2-x-6<0\), that is \((x-3)(x+2)<0\), so \(-2<x<3\). Both routes give the same answer, as the sketch shows.
What if the inequality is in a constant such as k?
Discriminant questions often end in a quadratic inequality, not in \(x\) but in a constant such as \(k\). The condition "the line misses the curve" means \(b^2-4ac<0\), and \(b\) or \(c\) contains \(k\), so the discriminant is itself a quadratic in \(k\). Solve it in exactly the same way: critical values, then a sketch, now with \(k\) on the horizontal axis.
The line \(y=kx+1\), where \(k\) is a constant, does not meet the curve \(y=x^2+2x+5\). Find the set of possible values of \(k\).
Substitute and collect into a quadratic in \(x\): \(x^2+2x+5=kx+1\), so \(x^2+(2-k)x+4=0\).
No intersection means no real roots, so \(b^2-4ac<0\): \((2-k)^2-4(1)(4)<0\), which expands to \(k^2-4k-12<0\).
Critical values: \((k-6)(k+2)=0\), so \(k=-2\) or \(k=6\).
"\(<0\)" on a ∪-shaped graph is between the roots. \(\{k : -2<k<6\}\)
At the critical values the line is a tangent: \(k=-2\) touches the curve at \((-2, 5)\) and \(k=6\) touches it at \((2, 13)\).
Writing "\(k<-2\), \(k<6\)" or "\(k<6\) and \(k<-2\)" because the inequality sign was "\(<\)". The sign tells you which part of the graph you want, not which way to point each answer. Sketch it every time.
How do you combine inequalities and use set notation?
When \(x\) must satisfy two inequalities at once, the answer is the overlap of the two solution sets. The safest way to find it is to draw both on number lines stacked above each other and read off where both are shaded. A filled circle marks an end value that is included (\(\le\), \(\ge\)); an open circle one that is not (\(<\), \(>\)).
An answer that is one interval, such as \(-\tfrac32\le x<1\), can be written as a single double inequality. An answer in two separate pieces needs the word "or", or a union symbol in set notation. The braces \(\{x : \ldots\}\) are read "the set of all \(x\) such that…".
| Meaning | Inequality form | Set notation |
|---|---|---|
| one interval | \(-\tfrac32\le x<1\) | \(\{x : -\tfrac32\le x<1\}\) |
| two pieces ("or") | \(x<-4\) or \(x>2\) | \(\{x : x<-4\}\cup\{x : x>2\}\) |
| both at once ("and") | \(x>-3\) and \(x\le4\) | \(\{x : x>-3\}\cap\{x : x\le4\}=\{x : -3<x\le4\}\) |
Find the set of values of \(x\) for which both \(4(x+2)<3x+9\) and \(2x^2-3x-9\le0\). Give your answer in set notation.
Linear inequality. \(4x+8<3x+9\), so \(x<1\).
Quadratic inequality. Critical values: \((2x+3)(x-3)=0\), so \(x=-\tfrac32\) or \(x=3\). "\(\le0\)" on a ∪-shape is between the roots, with the roots included: \(-\tfrac32\le x\le3\).
Overlap. The lower end comes from the quadratic and is included; the upper end comes from the linear inequality and is excluded. \(\{x : -\tfrac32\le x<1\}\)
When the question asks for set notation, give your final answer in set notation, not just as inequalities. Use \(\cup\) between two pieces; writing \(-4>x>2\) says something no number can satisfy.
How do you show inequalities as regions on a graph?
An inequality in \(x\) and \(y\), such as \(y>x^2-4\), is satisfied by a whole region of the plane. To draw it:
- Draw the boundary by replacing the inequality sign with "\(=\)".
- Use a solid boundary for \(\le\) or \(\ge\) (points on it are included) and a dashed boundary for \(<\) or \(>\) (not included).
- \(y>f(x)\) is the region above the curve \(y=f(x)\); \(y<f(x)\) is below it.
- If you are unsure which side, test a point that is not on the boundary; the origin is often easiest.
- Shade and label the region you are asked for, unless the question gives a different instruction.
The same picture answers one-variable questions too: \(f(x)<g(x)\) holds for the \(x\)-values where the graph of \(f\) is below the graph of \(g\), which is either between or outside their intersection points.
The curve \(C\) has equation \(y=x^2-2x-3\) and the line \(L\) has equation \(y=x+1\).
(a) Find the coordinates of the points where \(L\) meets \(C\).
(b) The region \(R\) lies above \(C\) and below \(L\). Points on \(C\) are included in \(R\); points on \(L\) are not. Write down inequalities that define \(R\).
(c) The line \(y=x+k\) is a tangent to \(C\). Find the value of \(k\) and the coordinates of the point of contact.
(d) Hence, or otherwise, find the set of values of \(x\) for which \(x^2-2x-3<x+1\).
(a) \(x^2-2x-3=x+1\), so \(x^2-3x-4=0\), so \((x+1)(x-4)=0\). Then \(y=x+1\) gives \((-1, 0)\) and \((4, 5)\)
(b) Above \(C\) with the curve included (\(\le\)); below \(L\) with the line excluded (\(<\)): \(x^2-2x-3\le y<x+1\)
(c) Substitute \(y=x+k\): \(x^2-3x-(3+k)=0\). Tangent means discriminant \(=0\): \(9+4(3+k)=0\), so \(21+4k=0\) and \(k=-\tfrac{21}{4}\).
The repeated root is \(x=-\dfrac{b}{2a}=\tfrac32\), and \(y=\tfrac32-\tfrac{21}{4}=-\tfrac{15}{4}\). Check on the curve: \(\tfrac94-3-3=-\tfrac{15}{4}\) ✓. \(k=-\tfrac{21}{4}\), point of contact \(\left(\tfrac32, -\tfrac{15}{4}\right)\)
(d) The curve is below the line between the intersection points from (a). \(\{x : -1<x<4\}\)
Describing a region with only one inequality, or with the right inequalities but the wrong strict or non-strict signs. Every boundary needs its own inequality, and the sign must match how that boundary is drawn.
Exam tips for P1.2
Always rearrange the linear equation and substitute into the quadratic, never the other way round. Pick the variable whose coefficient is 1 if there is one, so that no fractions appear.
A simultaneous-equation answer is a set of pairs. Find each partner from the linear equation (substituting into the quadratic can produce extra, false partners) and present the answers as coordinates or as matched \(x=\ldots,\ y=\ldots\).
For any quadratic inequality, write the critical values first, then draw a quick sketch with them marked. A two-second sketch decides between "between the roots" and "outside the roots" far more reliably than a rule remembered under pressure.
In parameter questions, say which condition you are using (\(b^2-4ac>0\), \(=0\) or \(<0\)) and why, before substituting. In a "show that" part, every line of that working should be visible.
"Find the set of values" wants inequalities; "in set notation" wants braces; "state the integer values" wants a list. Check whether the end values are included and use the correct strict or non-strict sign.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| From \(x^2>4x\), divide by \(x\) to get \(x>4\). | \(x\) could be negative or zero; dividing by it may reverse the sign or lose solutions. | \(x^2-4x>0\Rightarrow x(x-4)>0\Rightarrow x<0\) or \(x>4\). |
| \(x^2>9\) means \(x>3\). | Negative values such as \(-5\) also satisfy it. | \(x<-3\) or \(x>3\). |
| Two separate intervals can be written as \(-4>x>2\). | No number is less than \(-4\) and greater than \(2\) at once. | \(x<-4\) or \(x>2\), or a union in set notation. |
| A zero discriminant means the line misses the curve. | Zero means one repeated root, so there is one point of contact. | Zero discriminant ⇔ tangent; negative ⇔ no intersection. |
| Any \(y\)-value that satisfies the quadratic can be paired with any \(x\). | Each solution pair must satisfy both equations. | Find each partner from the linear equation and keep the pairs together. |
| A dashed boundary means the region is on the other side. | Dashed or solid says nothing about which side is the region. | Dashed: boundary not included (\(<\), \(>\)). Solid: boundary included (\(\le\), \(\ge\)). |
How this topic connects
P1.1 Indices, surds and quadratics
Factorising, completing the square and \(b^2-4ac\) are the tools used at every step here; this page applies them to two equations at once.
P1.3 Curve sketching and P1.4 Straight-line graphs
Sketches show how many times graphs meet, and perpendicular lines and intersection points use the same substitution method.
Circles (P2) and linear programming (D1)
A line meets a circle at two, one or no points, decided by exactly the same discriminant test; a tangent to a circle is the case \(b^2-4ac=0\). In Decision Mathematics, a set of linear inequalities defines a feasible region on a graph, with the same boundary and shading conventions.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) | solving the quadratic after substituting, when it does not factorise | No – learn it |
| \(b^2-4ac>0,\ =0,\ <0\) | line crosses (two points), touches (tangent), misses the curve | No – learn it |
| \(x=-\dfrac{b}{2a}\) | the repeated root, giving the point of contact of a tangent | No – derive it from the formula with \(b^2-4ac=0\) |
| \((x-\alpha)(x-\beta)<0\Leftrightarrow\alpha<x<\beta\); \(>0\Leftrightarrow x<\alpha\) or \(x>\beta\) \((\alpha<\beta)\) | reading a quadratic inequality | No – derive it from a sketch |
| \(a<b\) and \(m<0\Rightarrow ma>mb\) | multiplying or dividing an inequality by a negative number | No – learn it |
Check your understanding
1. Solve the simultaneous equations \(y=2x\) and \(x^2+y^2=20\).
Substitute \(y=2x\): \(x^2+4x^2=20\), so \(x^2=4\) and \(x=\pm2\). Using \(y=2x\) for the partners: \((2, 4)\) and \((-2, -4)\).
2. Solve \(3-2x\ge11\).
\(-2x\ge8\). Dividing by \(-2\) reverses the sign: \(x\le-4\).
3. Solve \(x^2<9x\).
\(x^2-9x<0\), so \(x(x-9)<0\). Critical values 0 and 9; the ∪-shaped curve is below the axis between them: \(0<x<9\). (Dividing by \(x\) would have lost the lower limit.)
4. Find the value of \(k\) for which \(y=x+k\) is a tangent to \(y=x^2-3x+1\), and the point of contact.
\(x^2-3x+1=x+k\), so \(x^2-4x+1-k=0\). Tangent: \(16-4(1-k)=0\), so \(12+4k=0\) and \(k=-3\). Then \(x^2-4x+4=0\), \((x-2)^2=0\), \(x=2\), \(y=2-3=-1\). \(k=-3\), point of contact \((2, -1)\).
5. Which of the points \((1, -2)\) and \((3, -1)\) lie in the region defined by \(y\ge x^2-4\) and \(y<0\)?
\((1, -2)\): \(-2\ge1-4=-3\) ✓ and \(-2<0\) ✓, so it is inside. \((3, -1)\): \(-1\ge9-4=5\) is false, so it is outside. Only \((1, -2)\).
6. Find the set of values of \(p\) for which the equation \(x^2+px+p+3=0\) has two distinct real roots.
Two distinct roots: \(p^2-4(p+3)>0\), so \(p^2-4p-12>0\), \((p-6)(p+2)>0\). Critical values \(-2\) and \(6\); "\(>0\)" on a ∪-shape is outside the roots: \(\{p : p<-2\}\cup\{p : p>6\}\).
Specification checklist
Tick every box before the exam. The wording is ours; the numbers refer to the P1 specification sections.
- (P1 1.6) I can solve a pair of simultaneous equations, one linear and one quadratic, by substitution and give every solution pair.
- (P1 1.6) I can explain the solutions as the points where a line meets a curve, and use the discriminant to decide whether a line crosses, touches or misses a curve.
- (P1 1.7) I can solve linear inequalities, including ones with brackets and fractions, reversing the sign when I multiply or divide by a negative number.
- (P1 1.8) I can solve quadratic inequalities by finding the critical values and using a sketch, including inequalities in a parameter that come from a discriminant.
- (P1 1.8) I can combine two inequalities and write the solution set in set notation, using \(\cup\) and \(\cap\) correctly.
- (P1 1.9) I can show inequalities in \(x\) and \(y\) as regions on a graph, with solid boundaries for \(\le\) and \(\ge\) and dashed boundaries for \(<\) and \(>\), and describe a given region with inequalities.