Equations and inequalities · Year 11
In Equations and inequalities (Year 11) you practise 5 skills: Linear equations · Quadratic equations · A line and a parabola · How many solutions · Inequalities. A solution of an equation is a value that makes it true, and a solution of an inequality is a set of values. Solve a quadratic equation with the quadratic formula; the expression under the root, the discriminant, says how many solutions there are: two when it is positive, one when it is zero, and none when it is negative. Solve a quadratic inequality with the parabola: find where it meets the x axis and see where the graph is above or below the axis.
The solution of a system of equations is where the graphs meet. When you compare two functions, asking where one is larger than the other is asking where their difference is positive. In an equation with algebraic fractions, multiply by the common denominator, and at the end check that no solution makes a denominator zero. In an equation with a root, isolate the root, square both sides, and check every solution in the original equation, because squaring can add a solution that does not work. In an equation that is quadratic in a square, put a new variable in place of the square, solve the quadratic, and go back: a negative value of the new variable gives no solution.
Tip: At the end of every equation, put the solution back in. It takes a moment and catches most mistakes.
What you learn
Linear equations
For example: Solve the equation: 6(x + 5) = −3(x + 3) + 39
Answer and solution
The answer: 0
Expand the brackets: 6x + 30 = −3x − 9 + 39 Collect like terms: 6x + 30 = −3x + 30 Move the terms with x to one side and the numbers to the other: 9x = 0 Divide both sides by 9: x = 0
Quadratic equations
For example: Solve: x² − 2x − 63 = 0
Answer and solution
The answer: x = 9 or x = −7
The expression under the square root in the quadratic formula: (−2)² − 4 × 1 × (−63) = 256 √256 = 16 x = (2 + 16) ÷ 2 = 9 x = (2 − 16) ÷ 2 = −7
A line and a parabola
For example: Here are the simultaneous equations: y = x² + 2x − 3 y = −5x − 15 Find the values of x where the parabola and the line cross.
Answer and solution
The answer: x = −4 or x = −3
Set the two expressions for y equal: x² + 2x − 3 = −5x − 15 Gather everything on one side: x² + 7x + 12 = 0 The expression under the square root in the quadratic formula: 7² − 4 × 1 × 12 = 1 √1 = 1 x = (−7 + 1) ÷ 2 = −3 x = (−7 − 1) ÷ 2 = −4
How many solutions
For example: Without solving it: how many solutions does the equation have? 3x² − 7x + 10 = 0
Answer and solution
The answer: 0
The expression under the square root in the quadratic formula: (−7)² − 4 × 3 × 10 = −71 It is negative, so the equation has no solution. The number of solutions: 0
Inequalities
For example: Solve the inequality: (x + 7) ÷ 4 ≥ −4
Answer and solution
The answer: x ≥ −23
Multiply both sides by 4. The number is positive, so the sign stays the same: x + 7 ≥ −16 Subtract 7 from both sides: x ≥ −23 The solution: x ≥ −23
The exercises are made by rules and the answers are worked out in code, so there are new exercises every time. The topic follows the national curriculum in England: mathematics (key stages 3 and 4).