Maths › Algebra and functions › Simultaneous equations and inequalities

Simultaneous equations and inequalities

Two equations sharing their unknowns are a picture: a line crossing a curve, and the algebra lands exactly on the crossings. Then the same pictures answer a subtler question, not where expressions are equal, but where one beats the other, and whole regions of the plane become the answer.

Year 12-13EDEXCEL 9MA0 2.4, 2.5

Builds on Quadratic functions.

IN THIS TOPIC

  • Solve simultaneous equations by substitution and elimination, including linear with quadratic.
  • Solve linear and quadratic inequalities, including with brackets and fractions, and write the answers in set notation.
  • Represent inequalities graphically: regions, shading, dotted and solid boundaries.

WHAT YOU PROBABLY THINK

You can always multiply both sides of an inequality by x.

Where a line meets a curve

Two equations solved simultaneously are being asked 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 = …), substitute 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; 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 so.

Solving y equals 2x plus 3 with y equals x squared minus 4x plus 8 simultaneously finds where the line meets the parabola: the points 1, 5 and 5, 13(1, 5)(5, 13)y = x² − 4x + 8y = 2x + 3the algebra's two solutions are the picture's two meetings
FIG. 1The algebra, drawn: the line y = 2x + 3 crosses the parabola y = x² − 4x + 8 at exactly (1, 5) and (5, 13), the two solutions of the substituted quadratic.

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, no real roots and they never meet. And substitution handles rougher pairings than y = mx + c, including this rougher pair:

YOUR TURN

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: y = 2 or y = −64/5.

y = 2 gives x = 6; 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 on a line

A quadratic inequality asks where a parabola sits above or below the axis, so the method is: 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: 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”.

Fractions bring the lie above to the surface. Multiplying an inequality by x flips it when x is negative, and you rarely know the sign, so multiply by x2, which is never negative, so a/x < b becomes ax < bx2.

WORKED EXAMPLE

A fraction handled safely

Solve 3/x < 2.

Multiply both sides by x2 (positive for every x ≠ 0): 3x < 2x2, so 2x2 − 3x > 0.

Factorise: x(2x − 3) > 0. An upward parabola with roots 0 and 3/2 is 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 survives precisely because dividing three by a negative goes 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, y < f(x) everything below. Several inequalities at once intersect their regions, and the drawing conventions carry meaning: solid boundary for ≤ or ≥ (the edge belongs), dotted for < or > (it does not), with the wanted region shaded.

The region where y is greater than the parabola and less than the line: shaded between the two curves, with both boundaries dotted because neither is includedy = x² − 4x + 8y = 2x + 3the regiondotted boundaries: strict inequalities leave the edges out
FIG. 2The region satisfying y > x² − 4x + 8 and y < 2x + 3 at once: shaded between the parabola and the line, closing at their two crossings, with both boundaries dotted because both inequalities are strict.

The region between this lesson's curve and line ties the whole story together: its corners are the simultaneous solutions (1, 5) and (5, 13), its interior is where the line beats the parabola, and its x-extent, 1 < x < 5, is the same quadratic inequality x2 − 6x + 5 < 0 solved above.

TRY IT UNSEEN

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: y ≤ x + 1. Strictly above a dotted parabola: y > x2.

So the region is {(x, y) : y ≤ x + 1} ∩ {(x, y) : y > x2}: y ≤ x + 1 and y > x2. The line's solidity and the parabola's dots are not decoration; they are the ≤ and the >.

THE EXAM BIT

  • 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 of unknown sign; multiply by its square, or sketch instead. The mark scheme watches this line specifically.
  • In set notation, “between” is an intersection and “outside” is a union; and on region diagrams, solid boundaries mean included, 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: x ≤ −5 or x ≥ 2.

Set notation: {x : x ≤ −5} ∪ {x : x ≥ 2}: an “or”, so a union, with both boundary points kept by the ≥.

Substitute the linear into the quadratic: the roots are the crossings, the discriminant the count.

Inequalities are sketches: roots, shape, sign, and regions shade above or below with dotted-or-solid edges.

CHECK YOUR PROGRESS

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  • Solve simultaneous equations by substitution and elimination, including linear with quadratic.
  • Solve linear and quadratic inequalities, including with brackets and fractions, and write the answers in set notation.
  • Represent inequalities graphically: regions, shading, dotted and solid boundaries.

No animated video for this topic yet; these notes stand alone.