Simultaneous equations and inequalities flashcards

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Solution of a pair of simultaneous equations
A pair of values (x, y) that makes both equations true at the same time.
Substitution method (one linear, one quadratic)
Make x or y the subject of the linear equation, substitute into the quadratic, solve, then find each partner from the linear equation.
Tangent (line to a curve), in algebraic terms
A line that meets the curve where the quadratic from substitution has a repeated root: b² − 4ac = 0.
Critical values of a quadratic inequality
The roots of the related equation f(x) = 0; they split the number line into the regions to test.
Set notation for an interval
{x : a < x ≤ b} — the set of all x such that a < x ≤ b.
Union ∪ and intersection ∩ of solution sets
∪ means 'or' (in either set); ∩ means 'and' (in both sets).
Solid and dashed boundaries of a region
Solid: points on the boundary are included (≤, ≥). Dashed: not included (<, >).
Discriminant test after substituting a line into a curve
b² − 4ac > 0: two intersection points; = 0: tangent (one point); < 0: no intersection. (No — learn it)
Quadratic formula
x = (−b ± √(b² − 4ac)) / 2a. (No — learn it)
Repeated root when b² − 4ac = 0
x = −b/2a (gives the point of contact of a tangent). (No — derive it)
(x − α)(x − β) < 0 with α < β
α < x < β (between the roots). (No — derive it)
(x − α)(x − β) > 0 with α < β
x < α or x > β (outside the roots). (No — derive it)
Multiplying an inequality by a negative number m
Reverse the sign: if a < b and m < 0 then ma > mb. (No — learn it)
Reading f(x) > 0 from the graph of y = f(x)
The x-values where the curve is above the x-axis.
Reading f(x) < g(x) from two graphs
The x-values where the graph of f is below the graph of g — between or outside their intersection points.
Number line: open circle vs filled circle
Open circle: end value not included (<, >). Filled circle: included (≤, ≥).
Region y > f(x)
The region above the curve y = f(x), drawn with a dashed boundary.
How to check which side of a boundary a region lies
Test a point not on the boundary (often the origin) in the inequality.
Why find y from the linear equation, not the quadratic?
The linear equation gives exactly one partner for each x; the quadratic can give extra values that do not satisfy both equations.
What does each solution pair of a line–curve system represent?
A point where the line meets the curve.
Why not divide both sides of x² > 4x by x?
x may be negative or zero, so the sign might need reversing or solutions are lost. Rearrange: x(x − 4) > 0, so x < 0 or x > 4.
Why does x² > 9 not mean just x > 3?
Negative numbers below −3 also have squares above 9: x < −3 or x > 3.
Why is '−4 > x > 2' wrong for two separate pieces?
No number is below −4 and above 2 at once. Write x < −4 or x > 2 (a union).
Solve y = x² − 4x + 5 and y = x + 1.
x² − 5x + 4 = 0 ⇒ x = 1 or 4; points (1, 2) and (4, 5).
Show y = 3x − 2 is a tangent to y = x² + x − 1.
x² − 2x + 1 = 0, discriminant 4 − 4 = 0 ⇒ tangent; touches at (1, 1).
Solve (x − 1)(x + 5) > 2x + 3.
x² + 2x − 8 > 0 ⇒ (x + 4)(x − 2) > 0 ⇒ x < −4 or x > 2.
Solve 5 − 3(x − 1) > 2x − 7.
8 − 3x > 2x − 7 ⇒ 15 > 5x ⇒ x < 3.
For which k does y = kx + 1 miss y = x² + 2x + 5?
x² + (2 − k)x + 4 = 0; (2 − k)² − 16 < 0 ⇒ (k − 6)(k + 2) < 0 ⇒ −2 < k < 6.
Values of x with 4(x + 2) < 3x + 9 and 2x² − 3x − 9 ≤ 0.
x < 1 and −3/2 ≤ x ≤ 3 ⇒ {x : −3/2 ≤ x < 1}.
Solve x² < 9x.
x(x − 9) < 0 ⇒ 0 < x < 9.
Find k so that y = x + k is a tangent to y = x² − 3x + 1.
x² − 4x + 1 − k = 0; 16 − 4(1 − k) = 0 ⇒ k = −3; contact at (2, −1).
Dividing an inequality by a negative number
Reverse the inequality sign — or move the x terms to the side where they are positive.
Quadratic inequality: fastest reliable method
Rearrange to f(x) > 0, < 0, ≥ 0 or ≤ 0, find the critical values, sketch, read off the region.
'Show that the line is a tangent'
State 'tangent ⇔ b² − 4ac = 0', substitute, show it equals 0, conclude.
Answer asked 'in set notation'
Use braces: {x : x < −4} ∪ {x : x > 2}; inequality form alone may lose the final mark.
Fractions in a linear inequality
Multiply every term (on both sides) by the LCM of the denominators before collecting terms.
Quadratic inequality in context (e.g. a length)
Reject the part of the solution that is impossible in context, such as a negative length.

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