Maths › Algebra and functions › Simultaneous equations and inequalities
Simultaneous equations and inequalities
Two equations sharing their unknowns are a picture of a line crossing a curve, and the algebra lands on the crossings. Those same pictures then answer a further question. An inequality identifies the values for which one expression is greater or less than the other, and the answer becomes a region of the plane.
Builds on Quadratic functions.
IN THIS TOPIC
- Solve one linear and one quadratic equation together by substitution.
- Read the discriminant of the substituted equation as the number of intersections.
- Solve linear, quadratic and fractional inequalities, and write the answers in set notation.
- Shade a region of the plane with the right boundary convention.
COMMON MISCONCEPTION
You can always multiply both sides of an inequality by x.
Where a line meets a curve
Two equations solved simultaneously are asking for the values that satisfy both at once, and geometrically that means the intersection points of their graphs. With one linear and one quadratic equation, the method is substitution. Make the linear equation say y = … or x = …, feed it into the quadratic, and solve the single quadratic that results.
WORKED EXAMPLE
The standard pairing
Solve simultaneously: y = 2x + 3 and y = x2 − 4x + 8.
Equate. x2 − 4x + 8 = 2x + 3, so x2 − 6x + 5 = 0, which factorises as (x − 1)(x − 5) = 0.
x = 1 gives y = 5, and x = 5 gives y = 13. Solutions: (1, 5) and (5, 13).
Answers come in matched pairs and pairing them wrongly costs the marks. Substitute each x into the linear equation, the easier of the two, and say in writing that this is what you are doing.
The discriminant of the substituted quadratic now reads as geometry. Two roots and the line crosses the curve twice. A repeated root and it is a tangent. Nothing real and they never meet. Substitution also handles rougher pairings than y = mx + c, as here.
GUIDED PRACTICE
A tougher pair
Solve simultaneously: 2x − 3y = 6 and x2 − y2 + 3x = 50, before opening the working.
Show the working
From the linear equation, x = 3 + 3y/2. Substituting and clearing fractions gives 5y2 + 54y − 128 = 0.
The discriminant is 542 + 4 × 5 × 128 = 5476 = 742, so y = (−54 ± 74)/10, giving y = 2 or y = −64/5.
y = 2 gives x = 6, and y = −64/5 gives x = −81/5. Solutions: (6, 2) and (−81/5, −64/5), and both check in the original quadratic equation.
Ugly fractions are not a wrong turn. Exact answers often look like this, and substituting back into the original equations is how you know they are right.
Inequalities in one variable
A linear inequality behaves like a linear equation with one rule bolted on. Multiplying or dividing by a negative number reverses it, so 5 − 2x > 1 becomes −2x > −4 and then x < 2. Most candidates do better to move the x term to the side where its coefficient is already positive and dodge the reversal altogether.
A quadratic inequality asks where a parabola sits above or below something. Find the roots, sketch the shape, read the sign. For x2 − 6x + 5 < 0 the roots are 1 and 5 and the parabola opens upward, so it is negative between its roots and the answer is 1 < x < 5. In set language that is {x : x > 1} ∩ {x : x < 5}, an “and”. The outside region x < 1 or x > 5 is {x : x < 1} ∪ {x : x > 5}, an “or”.
A question phrased as x2 − 4x + 8 < 2x + 3 is the same job with one step in front of it. Gather everything on one side and “curve below line” turns into exactly the inequality just solved. Which is neat, because the line and the curve are this lesson's opening pair.
Fractions, and the sign trap
Multiplying an inequality by x reverses it whenever x is negative, and you almost never know the sign of x in advance. Multiply by x2 instead. It is positive for every x ≠ 0, so a/x < b turns safely into ax < bx2.
WORKED EXAMPLE
A fraction handled safely
Solve 3/x < 2.
Multiply both sides by x2, which is positive for every x ≠ 0. That gives 3x < 2x2, so 2x2 − 3x > 0.
Factorise. x(2x − 3) > 0, an upward parabola with roots 0 and 3/2, positive outside them: x < 0 or x > 3/2.
Sanity-check both branches. x = −1 gives −3 < 2, true; x = 1 gives 3 < 2, false; x = 2 gives 3/2 < 2, true. The negative branch is valid because dividing three by a negative number gives a result below any positive bound.
Regions in the plane
Replace the equals sign in y = f(x) with an inequality and the answer stops being a curve and becomes a region. y > f(x) is everything above the graph and y < f(x) everything below. Several inequalities at once intersect their regions, and the drawing conventions carry meaning. Solid boundary for ≤ or ≥, because the edge belongs. Dotted for < or >, because it does not. Shade the region you want.
The region between this lesson's curve and line draws the lesson's ideas together. Its corners are the simultaneous solutions (1, 5) and (5, 13). Its interior is where the line lies above the parabola. Its x-extent, 1 < x < 5, is the quadratic inequality solved two sections ago.
INDEPENDENT PRACTICE
Reading a region back into algebra
Write down the pair of inequalities whose region is the set of points below the solid line y = x + 1 and strictly above the dotted parabola y = x2.
Show the working
Below a solid line gives y ≤ x + 1. Strictly above a dotted parabola gives y > x2.
So the region is {(x, y) : y ≤ x + 1} ∩ {(x, y) : y > x2}, or y ≤ x + 1 and y > x2. The solid line and the dotted parabola are what distinguish the two cases; they are the ≤ and the >.
ASSESSMENT FOCUS
- Substitute using the linear equation, and find the second coordinates from it too. Present answers as coordinate pairs, correctly matched.
- A tangency question is a discriminant question. Substitute, then set b2 − 4ac = 0 for touching, > 0 for two crossings, < 0 for missing.
- Solve every quadratic inequality with a sketch. Roots first, shape second, sign region third. Writing 1 < x < 5 for an outside region is the classic reversed answer.
- Never multiply an inequality by a quantity whose sign you do not know. Multiply by its square, or sketch instead. Mark schemes watch this line specifically.
- Dividing a linear inequality by a negative reverses it, and that small slip loses the whole answer. Rearrange so the x coefficient is positive and the question never comes up.
- In set notation, “between” is an intersection and “outside” is a union. On region diagrams, solid boundaries mean included and dotted mean excluded.
CHECK YOURSELF
Find the set of values of x for which x2 + 3x − 10 ≥ 0, giving the answer in set notation.
Show a hint
Factorise, sketch the upward parabola, and decide between or outside.
Show the answer
Factorise. (x + 5)(x − 2) ≥ 0, an upward parabola with roots −5 and 2.
At or above the axis means outside the roots, boundaries included, so x ≤ −5 or x ≥ 2.
In set notation, {x : x ≤ −5} ∪ {x : x ≥ 2}. An “or”, so a union, with both boundary points kept by the ≥.
Substitute the linear equation into the quadratic. The roots are the crossings and the discriminant counts them.
Solve a quadratic inequality from a sketch. Roots, shape, sign.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the simultaneous equations and inequalities questions page.
CHECK YOUR PROGRESS
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- Solve one linear and one quadratic equation together by substitution.
- Read the discriminant of the substituted equation as the number of intersections.
- Solve linear, quadratic and fractional inequalities, and write the answers in set notation.
- Shade a region of the plane with the right boundary convention.
Open the full revision checklist to see every objective in the course in one place.